problem solving applications go math

Go Math Review

With a strong, standards-aligned conceptual approach, a wide array of multisensory activities, an understandable teaching style and offering plenty of revision and review, Go Math can be a highly effective K-8 math program to help students develop their math skills in a less stressful and more engaging manner. 

What We Like

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What Is Go Math?

Published by Houghton Mifflin Harcourt, Go Math is a series of math curricula aimed at elementary and middle school students that is available for both homeschools and traditional schools. 

The series teaches common core math with a strong emphasis on understanding, vocabulary and application and does so using a combination of workbook exercises, critical reasoning, hands-on activities, discussions and frequent review. 

What Ages Or Grades is Go Math Intended For?

Go Math is intended for students in K-8 and the standards-aligned program ultimately covers everything from basic numeracy to geometry and statistics. 

Homeschools can, of course, use the program above or below its intended grades, and given its very straightforward instruction, multisensory lessons and strong emphasis on repetition and review, we feel Go Math can actually be a particularly good choice for older students who are a bit behind in their math studies to work with.

The series follows a traditional grade progression and makes obvious references to each grade on the cover, however, which is something that parents of older students using it as a remedial resource might need to think about.

On the other hand, this can make it a lot easier for homeschooling families switching into the curriculum from another to know where to start. 

It should be noted, though, that without a formal placement test those coming from a less traditional or non-common core program may have to spend a little time going through each grade’s scope and sequence and pay attention to each lesson’s pre-learning assessment of knowledge to make sure no gaps exist.

What’s Required To Teach Each Grade? 

Go Math is fairly traditional as a math curriculum in the materials it uses, really only requiring a teacher’s edition and a student edition. 

As a result, it can be considered fairly compact, with not a lot of stuff that parents will have to buy, store and keep track of over the course of a year. 

Teacher’s Edition

Go Math’s Teacher’s Edition is a full color and very visual softcover book that contains pretty much everything a parent would need to teach the program at home. 

The books contain the program’s teaching material, including well-detailed explanations of teaching strategies and tactics, as well as explanations and outlines of lesson activities and games, answers to questions and exercises, troubleshooting help, common mistakes students make with each topic, links to the common core for each concept and even suggested living books for remediation (which can also serve as a good built-in literature-based component for interested homeschools).  

They also contain a fair amount of content for differentiation, providing teaching tips and ideas for activities for ELL, as well as for students who need remediation and advanced students who would benefit from enriched learning.

This can be very helpful for parents learning at home, and broadens the potential appeal of Go Math quite a bit, although homeschools may need to pick up the program’s Reteach/Enrich books to make full use of the differentiation ideas. 

In terms of their overall layout, the teacher’s editions are pretty typical of a standard classroom teacher’s guide, with lesson guidance and suggestions surrounding a smaller copy of relevant student edition pages (with the answers and exercises completed). 

As a result, they keep things pretty well-organized and are easy to follow and work with once parents are familiar with the flow, and there is a substantial amount of information on how to use the guides to teach at the front of each book in case parents are unsure, which is nice. 

On the whole, Go Math’s teacher’s editions are also pretty well scripted. 

Inside, parents teaching at home should find a good amount of detailed guidance on what to do in each lesson and how to do it, usually in plain English, which should help carry parents through lessons pretty easily. 

Further, there are even “Math Talk” exercises, which are little scripted dialogues that are kind of fun and that can help parents and students discuss (and thereby reinforce) what they are learning through a sort of Socratic back and forth, which is kind of cool.

With all that said, parents should note that Go Math teacher’s editions don’t routinely contain an explicit, word-for-word dialogue and assume an ability to teach and move a lesson along, which can make it a little bit more challenging for new homeschooling families to use.  

Homeschooling parents should also note that, as Go Math is largely intended to be used in schools, it does make reference to traditional teaching and a classroom setting, drawing examples therefrom and even making mention or including activities that use multiple students that will need to be adjusted for home use.

Another thing that homeschoolers should note is that the Go Math student editions and teacher’s guides often make reference to digital resources available on the company’s learning platform, such as a personal math trainer, animated models, virtual manipulatives and much more. 

Sadly, these are only accessible to traditional schools with the appropriate licenses. 

Interestingly, however, the homeschool editions do allow parents to scan included QR codes, which then open a link to a suite of helpful videos that briefly explain and demonstrate math concepts.

Called Math on the Spot, these videos are fairly entertaining and are hosted by a cast of muppets and actors. 

Explaining concepts in simple, everyday English, the hosts go through a particular idea and work through it on-screen with an example or two, making them potentially very useful addition to lessons for parents with limited experience teaching math (or who are themselves a bit rusty on the subject). 

An example of Math on the Spot can be seen below.

Student edition

The student editions are where students spend most of their time in Go Math.

They contain the information and diagrams that students and parents go over as part of the instructional component, vocabulary work, as well as a wide variety of exercises, activities and games to use during a lesson. 

The student editions are soft cover, consumable books that are printed in full color and highly illustrated, making them a good deal more interesting to look at and use than many other black and white math workbooks out there. 

One thing to note, however, is that these books are presented differently depending on if a parent is using a classroom or homeschool version of Go Math. 

In editions intended for traditional schools, the student books (like the teacher’s edition) are a single volume intended to be used in sequence over the course of a year. 

Homeschool bundles, on the other hand, break the student edition up into a number of concept/chapter booklets, in a somewhat similar manner as programs like the Math Mammoth Blue series .

This arrangement can make teaching Go Math at home a lot more flexible in terms of scheduling, as parents can arrange and rearrange the topics they wish to introduce or work on as they would like. 

However, it also means that there can be more material for parents to store, organize and keep track of over the course of a year, which can be a little annoying.

Go Math’s Approach To Teaching Math

A spiral curriculum.

Go Math is a spiral math curriculum. 

In other words, it breaks concepts down into smaller topics, introduces them a little at a time and periodically revisits them in greater depth as students progress in the series. 

This is as opposed to a mastery math program, which would tend to have students focus on one particular topic over the course of a number of lessons and would then move on completely once proficiency is determined. 

As we’ve discussed previous articles , this spiral approach can have its benefits. 

For example, students may not feel as bogged down by a concept in a spiral program as they might in a mastery one, which can make learning less frustrating as students move on to new material fairly regularly and don’t spend weeks with a concept.

It can also make math a little less overwhelming for some students, giving them a break of sorts if they are really struggling with a concept, as they can move on and revisit the concept at a later date.

Spiral programs also tend to incorporate more opportunities for review and revision than mastery ones, with topics being reintroduced and practiced fairly regularly throughout a student’s career. 

Go Math in particular also incorporates a good deal of spiral review exercises into its lessons.

In these exercises students are given an opportunity to review and practice previously learned (often related) concepts alongside new material, which can help in keeping things fresher and prevent students from forgetting important ideas.

Conceptual Math Focus

Go Math is also a conceptual math program. 

Like programs such as Singapore Math or RightStart , Go Math places a particular emphasis on helping students develop a stronger understanding of why the math works and why certain strategies can help solve problems, as opposed to simply learning how to solve problems quickly. 

Consequently, rather than focusing on drilling math facts, the program spends a good deal more time in explaining math topics, providing visual demonstrations, analyzing word problems and getting students used to using multiple approaches and strategies to solve exercises.

Rather interestingly, and somewhat uniquely compared to even other conceptual programs, Go Math will, at times, even have students write out explanations of the logic behind certain math concepts in their own words.

Parents should be aware, however, that this does mean that students tend not to get quite as much drill with Go Math as they might with a more traditional math program. 

Although students do get a fair amount of guided and independent practice during lessons, Go Math workbooks don’t have page after page of computational drill problems as might be found in programs such as CLE or Saxon .

While some students may appreciate this lack of (sometimes eye-rolling) drill, others may need the extra practice computational drill can provide in order to develop skill fluency, and may need to seek out extra practice materials as a result. 

Common Core Aligned

Although there are older, un-aligned editions of Go Math for sale out there, most newer editions of the program meet Common Core standards for math education. 

Both the teacher’s editions and student books specifically note which topics align with which standards, and on the whole the series can be a good option for homeschoolers who prefer (or need) to align to the Common Core. 

Hands-on, Multisensory Learning

In order to teach its concepts, Go Math makes use of a fairly wide variety of hands-on learning activities. 

During lessons, students have an opportunity to see and model ideas and concepts using math manipulatives, such as counting cubes or fractions strips (which are sold separately), which can help them better understand some of math’s more abstract concepts.

Interestingly, Go Math also includes a number of activities that students can engage in during lessons to reinforce and explore math concepts.

For example, the program often includes various math-related board games that students can play.

In order to move their pieces, students must complete a short math problem, which gets them practicing and thinking about their learning without having to do endless drill.

In addition to hands-on activities, Go Math also includes a good deal of visualization (through lesson pictorials) and even has parents and students engage in discussions (including back and forth Socratic dialogues), which add audio and visual learning components to the program. 

As a result, we feel that Go Math can offer strong multisensory learning, which in turn can get students more excited about and engaged with their learning.

More than that, it can go a long way in helping students remember the information they are learning, as multisensory approaches have been shown to improve information retention in the long run.

On the downside, the inclusion of activities and discussion alongside instruction and practice can make math lessons take more time, which in turn can make them a little harder to fit into a busy homeschool schedule. 

In addition, the inclusion of manipulatives and other hands-on activities in lessons does mean that parents will have to go out and buy (as well as organize and store) items for lessons, which in turn means that they will have to do a bit more prep work compared to a traditional lecture-and-workbook program.

Cross Curricular and Real World Connections

Periodically during lessons, Go Math will make overt connections in a dedicated section between the math and other subjects students might be learning. 

At times, these can include references to science subjects, social studies, English language arts and fine art. 

While perhaps not an extremely in-depth study, these cross curricular connections can be interesting and can help students better appreciate math and get them thinking three dimensionally about their general studies.

In addition to making cross curricular connections, now and again Go Math will also demonstrate math applications in the real world, for example by showing how geometric design is used in engineering.

While these real world connections are fairly brief, they can be a great way to use things students are familiar with to show math as a practical and usable subject to study, rather than just something to suffer through and memorize. 

An Emphasis on Vocabulary

Finally, and perhaps a bit unusually for a math program, there is a strong focus on vocabulary development with Go Math. 

Each chapter helps students learn proper math terms through dedicated instruction, practice and games.

Math Go even includes math journaling exercises, which allow students to freely discuss, write or even draw their understanding of math terminology in their own words and which can also be a good point of integration for homeschoolers following a Charlotte Mason approach. 

Although not every student is a fan of vocabulary drill, its addition to lessons can ensure that students develop a stronger understanding of math terms and their proper use, which tends to be important in future math classes and in other STEM courses students will take, and is an often overlooked aspect of many homeschool math programs in our opinion.

How It Works

Go Math is a teacher-led program. 

Parents follow lesson guidelines set out in the teacher’s editions while students follow along and do their work in the student edition.

The program’s books are divided into a number of chapters, each centering on a particular topic or concept in math and containing a number (5-10) of lessons that relate to it. 

For example, one chapter may be about fractions and its lessons may involve comparing and ordering them, multiplying them, simplifying them and so on.

Chapters and lessons in Go Math are fairly consistent and follow a pretty similar pattern.

Chapters begin with a “Show What You Know” exercise as part of a lesson introduction, which evaluates the current knowledge and skill level of a student on related or pre-requisite information before beginning the lessons.

In this way, skill and knowledge gaps are identified and parents can be more sure that their student is ready to learn a particular concept, which is helpful. 

Students are then introduced to key vocabulary and do some exercises to help familiarize themselves with the proper terms that the unit will cover. 

Somewhat similar to English language exercises, students may be asked to define words, fill in the blanks, complete charts, cut out and do memory work using flash cards and so on. 

Following the chapter introduction and vocabulary work, parents and students begin approaching the new information.

Lessons in a chapter start off by introducing a concept, often with an overarching/guiding question or by explaining a concept by modeling it or working through an example problem.

At this point the teacher’s guide may point out or highlight important information or provide tips to help students understand things.

There may also be some hands-on learning, such as a math game, or a discussion to help reinforce the learning.

Once a concept is introduced and explained, parents and students work together on some guided practice questions, with the teacher’s guide offering “go deeper” questions to challenge students to further explore concepts. 

Following this, and if everything is progressing smoothly, students can engage in independent practice in their workbook, such as by doing computational exercises, problem solving, written work to test their understanding and so on. 

At the end of each lesson there tends to be a short assessment (called a Lesson Check ) to make sure students really get what they’ve learned, as well as a spiral review, which brings in some previously learned concepts and gets students to work on them a bit to freshen up their memory. 

After a few lessons (about half-way through a chapter) there is a mid-chapter checkpoint, which is a kind of test lasting a couple of pages that assesses a student’s learning to that point using some relatively traditional short answer and multiple choice questions. 

At the end of each chapter, meanwhile, students are usually given a review lesson, where they can practice everything they’ve learned in that chapter, before being tested with a fairly lengthy and comprehensive end-of-chapter cumulative assessment (about 4-6 pages of work).

Our Thoughts On Go Math Lessons 

Overall, we feel that Go Math lessons are pretty straightforward and easy to teach, even for parents with limited homeschool teaching experience. 

The lessons are very well-organized and sequentially laid out, making them very easy to follow and teach with, especially as the information is written in a pretty understandable, common sense manner. 

In addition, the teacher’s guides offer tons of helpful resources and ideas for conveying the information to students, with a good deal of troubleshooting and differentiation ideas that parents can make use of if needed.

For a program designed for (and frequently used by) traditional schools, Go Math includes an impressive amount of activities, games and multisensory exercises in its lessons.

At any given time, students might be asked to use manipulatives, discuss ideas, write things down, play board games and much, much more, all of which can make learning a lot more fun and engaging than most traditional, and even many homeschool, math programs. 

Lessons in Go Math also include a lot of built-in review and practice, which is something that we appreciated. 

Each lesson contains a fair amount of exercises, both guided and independent, as well dedicated spiral review, discussions between parents and students, a lesson check and, of course, an assessment of prior knowledge and a cumulative assessment at the start and end of the chapter, respectively. 

All this can really help reinforce and solidify concepts for students, and the frequent repetition of both current and previous learning can be highly effective for many students, particularly those with a tendency to develop skill gaps over time. 

Finally, we liked the strong conceptual focus of Go Math, particularly its frequent emphasis on having students practice alternative strategies for solving problems and checking their work. 

Often seen in highly-respected conceptual programs such as Art of Problem Solving , this approach can get kids thinking more critically and analytically about the problems they encounter, helping them develop the skills and fluency they might need to tackle unusual or novel problems in the future. 

That said, it is important to note that we don’t believe Go Math to be the most rigorous or advanced math program out there.

Aimed at least partially at the public school system, the main program is designed to really make math more understandable and approachable.

While the program does introduce concepts pretty well and allows students to explore different strategies and hone their analytic skills, there aren’t quite as many tricky or complex, multi-step word problems as in some more challenging programs aimed at advanced math students, such as Beast Academy . 

As a result, students and parents who are looking for greater challenges may need to pick up the program’s enrichment books or other supplements. 

Another thing parents might want to consider is the length of lessons in Go Math. 

Should parents go through the lessons as intended, there can be a considerable time spent explaining and discussing topics, doing hands-on activities, playing games, writing things down and, of course, reviewing concepts. 

Lessons, therefore, aren’t exactly all that short and can take up 4-6 pages, or over an hour depending on a student’s willingness (or enthusiasm) to discuss things and/or go off track. 

As a result, lessons may not be quite as easy to fit into a busy homeschool schedule as some other programs. 

Pros and Cons of Go Math

Makes math learning more fun and approachable .

With a plethora of activities and discussions to try, and an easy to understand method of teaching, Go Math can make learning math a little more interesting and approachable for students.

Materials are colorful and highly visual

Go Math’s teacher and student editions are full color and use a wide variety of illustrations and visuals to help explain concepts, making them a lot more interesting to look at and use than some other programs out there. 

Lessons are easy to use

By and large, Go Math’s teacher’s editions are very well organized and laid out and teaching lessons tends to be simply a matter of following each step in sequence. 

The program also offers a large number of tips for teaching, advice for troubleshooting lessons and ideas for differentiating learning. 

Lessons are also multisensory and fairly activity-rich

Go Math isn’t simply a lecture-based math program. Its lessons are very multisensory, containing discussions, manipulatives, visual diagrams, hands-on activities and even games, which keeps learning engaging and able to help students with different learning styles better connect to the program, 

The program offers strong, conceptual math learning

Go Math is a strongly conceptual math program that teaches for understanding, rather than simple computational skill. 

It carefully explains why math works the way it does, encourages students to dive deeper into math ideas and helps them critically analyze and work on math problems using different strategies and approaches. 

All of this can help students develop a more solid background in math and become more confident when dealing with unfamiliar or unusual math exercises. 

There is a lot of opportunity for review and repetition

Go Math offers students a lot of opportunity to revisit and practice math concepts, both new and previously learned, during lessons with several dedicated review sections before, during and after each lesson, as well as before each comprehensive chapter test.  

Can take some effort and time on the part of parents

Go Math is a fairly parent intensive math program and its lessons can involve parents leading discussions, guiding practice, assessing student knowledge, setting up games, participating in activities and more. 

Written for the classroom

Go Math’s books are largely designed for classroom use, making frequent mention of the traditional school setting, teachers, classmates and more, which can be a bit of a turnoff for dedicated homeschooling families. 

Who Is Go Math Ideal For?

Homeschools looking for a multisensory, activity-rich math curriculum.

Go Math includes a wide variety of multisensory methods of teaching to help students learn math, including visual diagrams, hands-on activities, journaling, active discussions, videos and even games.

As a result, it can be a very engaging and even fun curriculum for students to use and can be more effective for students with different learning preferences and styles.  

Homeschools looking for an approachable, conceptual math program

Like math curricula such as Singapore Math or Math U See, Go Math is a conceptual math program that tries to help students better understand why math works the way it does, how math works and why certain approaches are effective at solving problems.

It can, therefore, help them develop a deeper and more analytical understanding of math. 

At the same time, the program uses common sense explanations, plenty of visuals, hands-on activities and even games to help make learning a little easier and less intimidating for students. 

Homeschools looking for a common-core aligned program

Go Math is a Common Core aligned math curriculum that is used by schools across the US and can therefore be an excellent option for those interested in such an approach to teaching.

Students that need a good amount of repetition and review to master skills

With chapter pre-assessments, dedicated lesson reviews, spiral reviews, parent-student discussions and more, Go Math offers homeschooling students a lot of opportunity to practice what they’re learning and review previously learned concepts. 

This can be of great help to students who need a good deal of consistent practice to prevent skill and knowledge gaps from forming. 

Those interested in a “school at home” approach

Some families prefer to try and replicate the experience of a public school system as much as possible in their homeschool. 

As a curriculum written for and used by schools across the US, Go Math can fit this approach to at home teaching pretty well.  

Who Is It Not Ideal For?

Those looking for a program specifically designed for homeschools.

Go Math’s teacher’s guides are more or less written for a classroom setting and tend to make references to traditional schooling from time to time. 

Although sold to and fairly frequently used by homeschools, it is not specifically and exclusively designed for homeschooling families, which can bother some parents.  

Students who pick things up quickly and get frustrated with lots of review

Go Math tends to review and revisit material a little more frequently than some other programs, with pre-lesson assessments, lesson exercises, lesson checks, spiral reviews, various games and activities, math talk discussions and more.

Those who pick up and master math concepts quickly may find it a little much.

Those looking for a more advanced homeschool math program

Go Math is a math program that tends to focus on making conceptual math more approachable and understandable for students and generally tracks along grade level in terms of rigor and pace.

As such, it may not be the most ideal program for those with highly advanced or talented math students. 

Note: Prices correct as of writing. All prices in USD. 

There are a couple ways in which homeschooling families can pick up Go Math for home use. 

The program is offered as a complete and fairly expansive homeschool package that, as we’ve mentioned, breaks the student edition up into a number of booklets based around individual concepts for that grade. 

These packages also come with the teacher’s edition, a lesson planning guide (to help parents schedule and pace lessons), an assessment guide with copies of tests and rubrics for correcting them, and a reusable and plasticized chart with math facts, questions and rules that students should remember.

Alternatively, parents can pick up the student and teacher’s editions (with the student edition being sold as a single book), as well the assessment guides (should they feel like doing so) as individual items, which can be an option for those on tighter budgets.

Although the exact price for a curriculum varies by grade, complete homeschool packages tend to cost somewhere between $250-300.

Individual student editions tend to cost around $45, teacher’s editions around $62.40, and the assessment guides around $15.

As always, we encourage parents to check the latest prices for the program, as well as for any discounts or sales that might be on offer. 

Is It Worth The Price?

Although perhaps not the cheapest homeschooling math program around, we feel that Go Math can be an excellent math program for the right homeschooling families. 

Go Math offers parents and students a comprehensive and standards-aligned conceptual math program that helps students understand the math they are learning a little more deeply and can get them thinking far more strategically and critically about problem solving, something that will stand them in good stead in their high school math and STEM courses and beyond.

Perhaps more than that, the curricula explain concepts quite clearly and understandably, frequently connecting the learning to subjects and real-life objects/situations that students may more readily understand. 

It also provides students with ample opportunity for review and revision, often making these interesting through the use of games and hands-on exercises. 

Finally, Go Math is also a very multisensory math program. 

Lessons can involve hands-on activities, visual diagrams, videos, discussions and even games, all of which allow the program to suit a wide array of learners and learning styles and be more engaging in the process.

Bottom Line

About the Author

David Belenky is a freelance writer, former science and math tutor and a tech enthusiast. When he’s not writing about educational tech, he likes to chill out with his family and dog at home.  

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Google Docs will now solve simple math problems faster than you can

The new Google Workspace feature is rolling out now.

What you need to know

• The Smart Compose feature in Google Docs is now able to solve simple math problems as you type them. 
• The tool is rolling out now for Google Workspace business and education users on the rapid release schedule. Users with scheduled releases will see the feature starting Dec. 18. 
• Aside from Google Docs, the feature also works in Slides, Sheets, and Drawings. 

Google is using its machine learning tech to make Smart Compose in Google Workspace more powerful, it announced today. Now, when you enter a simple math equation in a Google Doc, Slide, Sheet, or Drawing, the Smart Compose assistant will automatically suggest the answer. It's rolling out now for paid Google Workspace business and education users, but not everyone will see the feature immediately. 

Smart Compose was originally released to help users write in Google Workspace apps, and it's based on Google's machine learning. That's not to be confused with "Help me write," which is Google's writing assistant with artificial intelligence . With both tools, Google warns that it cannot guarantee accuracy. 

However, the Smart Compose feature should be a lot more helpful with math than words. It's essentially looking for an equal ( = ) symbol in text with a preceding math equation. After Smart Compose detects the equation, it will use ML to solve the equation like a calculator would. But this happens almost immediately, as soon as you press the equals key. 

Like other Smart Compose applications, the solutions will be presented as a suggestion. That means it will appear greyed-out, and users need to press the tab key to insert the solution. To ignore it, simply press the space bar instead.

Google says that the Smart Compose feature will work with "simple" math problems, and we don't know exactly what that means. However, Google's demo of Smart Compose solved an equation with multiple steps using the correct order of operations. 

The feature will be turned on by default for paid Google Workspace business and education customers. It's rolling out today for users whose organizations are on rapid release schedules, but it can take 15 days for the tool to reach all users. Smart Compose won't start rolling out for users with scheduled releases until December 18 and won't finish until 15 days after that. 

Our Google Workspace account didn't have access to the new Smart Compose tools just yet. You can check to make sure Smart Compose is turned on by opening a document, clicking the tools tab, selecting preferences , and then looking for the checkbox that says Show Smart Compose suggestions . To stop seeing the suggestions, simply click the checkbox. 

Google says the new Smart Compose functionality "will increase productivity and accuracy when solving math equations across Workspace." This is probably true, but we suspect teachers and educators might not be thrilled with their students using this feature. Surprisingly, Google says there is no admin control over Smart Compose.

This feature won't be available in personal Google Workspace accounts, at least for now. 

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Brady is a tech journalist covering news at Android Central. He has spent the last two years reporting and commenting on all things related to consumer technology for various publications. Brady graduated from St. John's University in 2023 with a bachelor's degree in journalism. When he isn't experimenting with the latest tech, you can find Brady running or watching sports.

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• fuzzylumpkin A computer that can solve simple maths problems faster than a human? Never going to happen! Reply
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Applications of Solving Equations

• Study Guide

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Solving Word Problems

Strategies for solving word problems with variables.

• Read the problem.
• Determine what is known and what needs to be found (what is unknown).
• Try a few numbers to get a general idea of what the solution could be.
• Write an equation.
• Solve the equation by inverse operations or by plugging in values.
• Check your solution--does it satisfy the equation? Does it make sense in the context of the problem? (e.g. A length should not be negative.)
• What is known? Matt has 12(5) = 60 cents in nickels. Matt has 2(95) = 190 cents total. What needs to be found? The number of dimes that Matt has.

5 dimes? 10(5) + 60 = 110. Too low. 10 dimes? 10(10) + 60 = 160. Still too low. 20 dimes? 10(20) + 60 = 260. Too high. So we know the answer is between 10 and 20.

• Write an equation: 10 d + 60 = 190 where d is the number of dimes Matt has.

10 d + 60 - 60 = 190 - 60 10 d = 130 = d = 13

• Check: 10(13) + 60 = 190? Yes. Does 13 dimes make sense in the context of the problem? Yes.

5 shots? = . Not enough misses. 10 shots? = . Too many misses. So we know the answer is between 5 and 10.

= 51( ) = 51( ) 51 + x = 60 51 + x - 51 = 60 - 51 x = 9

• What is known? The area of the square is 2 times its perimeter. The formula for area is A = x 2 and the formula for perimeter is p = 4 x . What needs to be found? The length of a side.

x = 5 ? A = 5 2 = 25 , p = 4(5) = 20 . Area too small. x = 10 ? A = 10 2 = 100 , p = 4(10) = 40 . Area too large. So we know the answer is between 5 and 10.

• Write an equation: x 2 = 2(4 x ) . x 2 = 8 x
• Check: 8 2 = 8(8) ? Yes. Does 8 make sense in the context of the problem? Yes.

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• 4.2 Solve Applications with Systems of Equations
• Introduction
• 1.1 Use the Language of Algebra
• 1.2 Integers
• 1.3 Fractions
• 1.4 Decimals
• 1.5 Properties of Real Numbers
• Key Concepts
• Review Exercises
• Practice Test
• 2.1 Use a General Strategy to Solve Linear Equations
• 2.2 Use a Problem Solving Strategy
• 2.3 Solve a Formula for a Specific Variable
• 2.4 Solve Mixture and Uniform Motion Applications
• 2.5 Solve Linear Inequalities
• 2.6 Solve Compound Inequalities
• 2.7 Solve Absolute Value Inequalities
• 3.1 Graph Linear Equations in Two Variables
• 3.2 Slope of a Line
• 3.3 Find the Equation of a Line
• 3.4 Graph Linear Inequalities in Two Variables
• 3.5 Relations and Functions
• 3.6 Graphs of Functions
• 4.1 Solve Systems of Linear Equations with Two Variables
• 4.3 Solve Mixture Applications with Systems of Equations
• 4.4 Solve Systems of Equations with Three Variables
• 4.5 Solve Systems of Equations Using Matrices
• 4.6 Solve Systems of Equations Using Determinants
• 4.7 Graphing Systems of Linear Inequalities
• 5.1 Add and Subtract Polynomials
• 5.2 Properties of Exponents and Scientific Notation
• 5.3 Multiply Polynomials
• 5.4 Dividing Polynomials
• Introduction to Factoring
• 6.1 Greatest Common Factor and Factor by Grouping
• 6.2 Factor Trinomials
• 6.3 Factor Special Products
• 6.4 General Strategy for Factoring Polynomials
• 6.5 Polynomial Equations
• 7.1 Multiply and Divide Rational Expressions
• 7.2 Add and Subtract Rational Expressions
• 7.3 Simplify Complex Rational Expressions
• 7.4 Solve Rational Equations
• 7.5 Solve Applications with Rational Equations
• 7.6 Solve Rational Inequalities
• 8.1 Simplify Expressions with Roots
• 8.2 Simplify Radical Expressions
• 8.3 Simplify Rational Exponents
• 8.4 Add, Subtract, and Multiply Radical Expressions
• 8.5 Divide Radical Expressions
• 8.6 Solve Radical Equations
• 8.7 Use Radicals in Functions
• 8.8 Use the Complex Number System
• 9.1 Solve Quadratic Equations Using the Square Root Property
• 9.2 Solve Quadratic Equations by Completing the Square
• 9.3 Solve Quadratic Equations Using the Quadratic Formula
• 9.4 Solve Equations in Quadratic Form
• 9.5 Solve Applications of Quadratic Equations
• 9.6 Graph Quadratic Functions Using Properties
• 9.7 Graph Quadratic Functions Using Transformations
• 9.8 Solve Quadratic Inequalities
• 10.1 Finding Composite and Inverse Functions
• 10.2 Evaluate and Graph Exponential Functions
• 10.3 Evaluate and Graph Logarithmic Functions
• 10.4 Use the Properties of Logarithms
• 10.5 Solve Exponential and Logarithmic Equations
• 11.1 Distance and Midpoint Formulas; Circles
• 11.2 Parabolas
• 11.3 Ellipses
• 11.4 Hyperbolas
• 11.5 Solve Systems of Nonlinear Equations
• 12.1 Sequences
• 12.2 Arithmetic Sequences
• 12.3 Geometric Sequences and Series
• 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

• Solve direct translation applications
• Solve geometry applications

Solve uniform motion applications

Be prepared 4.4.

Before you get started, take this readiness quiz.

The sum of twice a number and nine is 31. Find the number. If you missed this problem, review Example 2.15 .

Be Prepared 4.5

Twins Jon and Ron together earned $96,000 last year. Ron earned $8000 more than three times what Jon earned. How much did each of the twins earn? If you missed this problem, review Example 2.19 .

Be Prepared 4.6

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains. If you missed this problem, review Example 2.43 .

Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

Solve applications with systems of equations.

• Step 1. Read the problem. Make sure all the words and ideas are understood.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for. Choose variables to represent those quantities.
• Step 4. Translate into a system of equations.
• Step 5. Solve the system of equations using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

Example 4.14

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Try It 4.27

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Try It 4.28

The sum of two numbers is −6 . −6 . One number is 10 less than the other. Find the numbers.

Example 4.15

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 plus $15 for each training session. Option B would pay her $ 10,000 + $ 40 $ 10,000 + $ 40 for each training session. How many training sessions would make the salary options equal?

Try It 4.29

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

Try It 4.30

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

Example 4.16

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

Try It 4.31

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Try It 4.32

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

Solve Geometry Applications

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

Complementary and Supplementary Angles

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Example 4.17

Translate to a system of equations and then solve.

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

Try It 4.33

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

Try It 4.34

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

In the next example, we remember that the measures of supplementary angles add to 180.

Example 4.18

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Try It 4.35

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

Try It 4.36

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

Recall that the angles of a triangle add up to 180 degrees. A right triangle has one angle that is 90 degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

Example 4.19

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Try It 4.37

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Try It 4.38

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

Example 4.20

Randall has 125 feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Try It 4.39

Mario wants to put a fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

Try It 4.40

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = r t D = r t where D is the distance traveled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Example 4.21

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

A diagram is useful in helping us visualize the situation.

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k , and Joni, j , will each drive.

Since D = r · t D = r · t we can fill in the Distance column.

Translate into a system of equations.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

Also, since Kelly left later, her time will be 1 2 1 2 hour less than Joni’s time. So,

Try It 4.41

Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

Try It 4.42

Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes ( 1 4 hour ) ( 1 4 hour ) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c .

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c . b + c .

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is b − c . b − c .

We’ll put some numbers to this situation in the next example.

Example 4.22

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Try It 4.43

A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

Try It 4.44

Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind . A wind current blowing against the direction of the plane is called a headwind .

Example 4.23

A private jet can fly 1,095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Try It 4.45

A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1,035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Try It 4.46

A commercial jet can fly 1,728 miles in 4 hours with a tailwind but only 1,536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Access this online resource for additional instruction and practice with systems of equations.

• Systems of Equations

Section 4.2 Exercises

Practice makes perfect.

Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

The sum of two number is 15. One number is 3 less than the other. Find the numbers.

The sum of two number is 30. One number is 4 less than the other. Find the numbers.

The sum of two number is −16. One number is 20 less than the other. Find the numbers.

The sum of two number is −26 . −26 . One number is 12 less than the other. Find the numbers.

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

The sum of two numbers is −27 . −27 . Their difference is −59 . −59 . Find the numbers.

The sum of two numbers is −45 . −45 . Their difference is −89 . −89 . Find the numbers.

Maxim has been offered positions by two car companies. The first company pays a salary of $10,000 plus a commission of $1000 for each car sold. The second pays a salary of $20,000 plus a commission of $500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of $14,000 plus a commission of $100 for each cable package sold. The second pays a salary of $20,000 plus a commission of $25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Amara currently sells televisions for company A at a salary of $17,000 plus a $100 commission for each television she sells. Company B offers her a position with a salary of $29,000 plus a $20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of $12,000 plus a $150 commission for each stove he sells. Company B offers him a position with a salary of $24,000 plus a $50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball? How many calories did he burn per hour when canoeing?

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.

Nancy bought seven pounds of oranges and three pounds of bananas for $17. Her husband later bought three pounds of oranges and six pounds of bananas for $12. What was the cost per pound of the oranges and the bananas?

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water as juice. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

The difference of two complementary angles is 55 degrees. Find the measures of the angles.

The difference of two complementary angles is 17 degrees. Find the measures of the angles.

Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

The difference of two supplementary angles is 8 degrees. Find the measures of the angles.

The difference of two supplementary angles is 88 degrees. Find the measures of the angles.

Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

Wayne is hanging a string of lights 45 feet long around the three sides of his patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his front yard. The length is five feet less than three times the width. Find the length and width of the fencing.

A frame around a family portrait has a perimeter of 90 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

The perimeter of a toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.

Solve Uniform Motion Applications

Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?

College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?

At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?

Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

The Jones family took a 12-mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current.

A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current.

A small jet can fly 1072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A small jet can fly 1435 miles in 5 hours with a tailwind but only 1,215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Writing Exercises

Write an application problem similar to Example 4.14 . Then translate to a system of equations and solve it.

Write a uniform motion problem similar to Example 4.15 that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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Math Review of Solving Application Problems

• July 24, 2014
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Solving application problems is a process that includes understanding the problem, translating it into an equation, solving the equation, checking the answer, and answering the question. This process can be used to solve many different types of problems.

Understanding and Setting up The Application Problem

The first part of the process involves understanding what is being asked. This includes noticing any key words that refer to operations and any quantities that are in relationship to one another. It is important to have an idea of what sort of quantity will represent a solution. Suppose a problem asks how many points the first-place winner had. That would require just one answer. If the problem described that the first-place winner had 10 points more than the second-place winner, and 17 points more than the third-place winner, all three point values would be necessary to completely answer the question.

Figure 1: Setting up the application problem involves finding quantities in relationship to one another.

Translating the Problem into an Equation

The second part of the problem involves putting the problem in symbol form. This includes choosing a letter to represent a variable, and writing down exactly what the variable stands for. Suppose that the point values of the three winners totaled 114. Let the point value of the first-place winner be x, the second-place winner x – 10, and the third-place winner x – 17. Then, x + x – 10 + x – 17 = 114.

Figure 2: Translating the problem into an equation involves putting the problem into symbol form.

Solving and Checking the Equation

If x + x – 10 + x – 17 = 114, then 3x = 114 + 27, or 3x = 141. So, x = 47, x – 10 = 37, and x -17 = 30. To check, 47 + 37 + 30 = 114. The answer makes sense, and fits the parameters of the problem. In this case, the original equation was x + x -10 + x – 17 = 114.

Answer the Question Asked

If the work is done to understand the problem before it is set up and solved, it is easier to answer the appropriate question. In this case, all three point values were necessary to completely answer the question. Suppose that more information were added to the problem. During the same contest last year, the point values of the first-place, second-place and third-place winners totaled 100, but the second place winner had 14 points less than the first-place winner, and the third-place winner had 18 points less than the first place winner. Which year did the third-place winner earn more points, and what was the difference? Last year, the equation was x + x – 14 + x – 18 = 100, so 3x = 132. The first-place winner earned 44 points, the second-place winner earned 44-14 or 30 points, and the third-place winner earned 28 points. However, the question asked has 2 parts. Last year, the third-place winner earned 28 points, and this year, the third-place winner earned 30 points. This year, the third-place winner earned more points, and the difference was a gain of 2 points.

Figure 3: Answering the question asked brings all the parts of the problem into balance.

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Math Review of Changing Application Problems into Equations

Math review of factoring a monomial from a polynomial.

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Go Math Grade 6 Answer Key Chapter 8 Solutions of Equations

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Lesson 1: Solutions of Equations

Share and Show – Page No. 423

Problem solving + applications – page no. 424, solutions of equations – page no. 425, lesson check – page no. 426.

Lesson 2: Write Equations

Share and Show – Page No. 429

Problem solving + applications – page no. 430, write equations – page no. 431, lesson check – page no. 432.

Lesson 3: Investigate • Model and Solve Addition Equations

Share and Show – Page No. 435

Problem solving + applications – page no. 436, model and solve addition equations – page no. 437, lesson check – page no. 438.

Lesson 4: Solve Addition and Subtraction Equations

Share and Show – Page No. 441

Unlock the problem – page no. 442, solve addition and subtraction equations – page no. 443, lesson check – page no. 444.

Lesson 5: Investigate • Model and Solve Multiplication Equations

Share and Show – Page No. 447

Page no. 448, model and solve multiplication equations – page no. 449, lesson check – page no. 450.

Lesson 6: Solve Multiplication and Division Equations

Share and Show – Page No. 453

Problem solving + applications – page no. 454, solve multiplication and division equations – page no. 455, lesson check – page no. 456.

Lesson 7: Problem Solving • Equations with Fractions

Share and Show – Page No. 459

On your own – page no. 460, problem solving equations with fractions – page no. 461, lesson check – page no. 462.

Mid-Chapter Checkpoint

Mid-Chapter Checkpoint – Vocabulary – Page No. 463

• Mid-Chapter Checkpoint – Page No. 464

Lesson 8: Solutions of Inequalities

Share and Show – Page No. 467

• Problem Solving + Applications – Page No. 468

Solutions of Inequalities – Page No. 469

Lesson check – page no. 470.

Lesson 9: Write Inequalities

Share and Show – Page No. 473

Make generalizations – page no. 474, write inequalities – page no. 475, lesson check – page no. 476.

Lesson 10: Graph Inequalities

Share and Show – Page No. 479

Problem solving + applications – page no. 480, graph inequalities – page no. 481, lesson check – page no. 482.

Chapter 8 Review/Test

Chapter 8 Review/Test – Page No. 483

• Chapter 8 Review/Test – Page No. 484

Chapter 8 Review/Test Page No. 485

Chapter 8 review/test – page no. 486, chapter 8 review/test page no. 487, chapter 8 review/test page no. 488.

Determine whether the given value of the variable is a solution of the equation.

Question 1. x + 12 = 29; x = 7 The variable is __________

Answer: not a solution

Explanation: Substitute the value in the given equation x + 12 = 29 If x = 7 7 + 12 = 29 19 ≠ 29 Thus the variable is not a solution.

Question 2. n − 13 = 2; n = 15 The variable is __________

Answer: a solution

Explanation: Substitute the value in the given equation n = 15 n − 13 = 2 15 – 13 = 2 The variable is a solution.

Question 3. \(\frac{1}{2}\)c = 14; c = 28 The variable is __________

Explanation: Substitute the value in the given equation c = 28 \(\frac{1}{2}\)c = 14 \(\frac{1}{2}\) × 28 = 14 14 = 14 Thus the variable is a solution.

Question 4. m + 2.5 = 4.6; m = 2.9 The variable is __________

Explanation: Substitute the value in the given equation m + 2.5 = 4.6 m = 2.9 2.9 + 2.5 = 4.6 5.4 ≠ 4.6 Thus the variable is not a solution.

Question 5. d − 8.7 = 6; d = 14.7 The variable is __________

Explanation: Substitute the value in the given equation d = 14.7 d − 8.7 = 6 14.7 – 8.7 = 6 6 = 6 Thus the variable is a solution.

Question 6. k − \(\frac{3}{5}\) = \(\frac{1}{10}\); k = \(\frac{7}{10}\) The variable is __________

Explanation: Substitute the value in the given equation k = \(\frac{7}{10}\) k − \(\frac{3}{5}\) = \(\frac{1}{10}\) \(\frac{7}{10}\) – \(\frac{3}{5}\) = \(\frac{1}{10}\) \(\frac{7}{10}\) – \(\frac{6}{10}\) = \(\frac{1}{10}\) \(\frac{1}{10}\) = \(\frac{1}{10}\) Thus the variable is a solution.

On Your Own

Question 7. 17.9 + v = 35.8; v = 17.9 The variable is __________

Explanation: Substitute the value in the given equation 17.9 + v = 35.8 v = 17.9 17.9 + 17.9 = 35.8 35.8 = 35.8 Thus the variable is a solution.

Question 8. c + 35 = 57; c = 32 The variable is __________

Explanation: Substitute the value in the given equation c + 35 = 57 c = 32 32 + 35 = 57 67 ≠ 57 Thus the variable is not a solution.

Question 9. 18 = \(\frac{2}{3}\)h; h= 12 The variable is __________

Explanation: Substitute the value in the given equation 18 = \(\frac{2}{3}\)h h = 12 \(\frac{2}{3}\) × 12 = 8 18 ≠ 8 Thus the variable is not a solution.

Question 10. In the equation t + 2.5 = 7, determine whether t = 4.5, t = 5, or t = 5.5 is a solution of the equation. The solution is ________.

Answer: t = 4.5

Explanation: Substitute the value in the given equation t = 4.5 t + 2.5 = 7 4.5 + 2.5 = 7 7 = 7 t = 5 t + 2.5 = 7 5 + 2.5 = 7 7.5 ≠ 7 Not a solution t = 5.5 t + 2.5 = 7 5.5 + 2.5 = 7 8 ≠ 7 Not a solution

Question 11. Antonio ran a total of 9 miles in two days. The first day he ran 5 \(\frac{1}{4}\) miles. The equation 9 – d = 5 \(\frac{1}{4}\) can be used to find the distance d in miles Antonio ran the second day. Determine whether d = 4 \(\frac{3}{4}\), d = 4, or d = 3 \(\frac{3}{4}\) is a solution of the equation, and tell what the solution means. The solution is ________ \(\frac{□}{□}\)

Answer: 3 \(\frac{3}{4}\)

Explanation: 9 – d = 5 \(\frac{1}{4}\) Substitute d = 4 \(\frac{3}{4}\) in the above equation 9 – 4 \(\frac{3}{4}\) = 5 \(\frac{1}{4}\) 4 \(\frac{1}{4}\) ≠ 5 \(\frac{1}{4}\) Not a solution Substitute d = 4 9 – 4 = 5 \(\frac{1}{4}\) 5 ≠ 5 \(\frac{1}{4}\) Not a solution Substitute d = 3 \(\frac{3}{4}\) 9 – 3 \(\frac{3}{4}\) = 5 \(\frac{1}{4}\) 5 \(\frac{1}{4}\) = 5 \(\frac{1}{4}\) 9 – d = 5 \(\frac{1}{4}\); d = 3 \(\frac{3}{4}\) is a solution.

Question 12. Connect Symbols and Words The length of a day on Saturn is 14 hours less than a day on Mars. The equation 24.7 − s = 14 can be used to find the length in hours s of a day on Saturn. Determine whether s = 9.3 or s = 10.7 is a solution of the equation, and tell what the solution means. Type below: _____________

Answer: s = 10.7

Explanation: The length of a day on Saturn is 14 hours less than a day on Mars. The equation 24.7 − s = 14 can be used to find the length in hours s of a day on Saturn. 24.7 − s = 14 Substitute s = 9.3 in the equation 24.7 – 9.3 = 14 15.4 ≠ 14 Not a solution Substitute s = 10.7 in the equation 24.7 – 10.7 = 14 14 = 14 Therefore s = 10.7 is a solution to the equation.

Question 13. A storm on one of the planets listed in the table lasted for 60 hours, or 2.5 of the planet’s days. The equation 2.5h = 60 can be used to find the length in hours h of a day on the planet. Is the planet Earth, Mars, or Jupiter? Explain. Type below: _____________

Answer: Earth

Explanation: A storm on one of the planets listed in the table lasted for 60 hours, or 2.5 of the planet’s days. 2.5h = 60 h = 60/2.5 h = 24 hours By seeing the above table we can say that Earth is the answer.

Question 14. A day on Pluto is 143.4 hours longer than a day on one of the planets listed in the table. The equation 153.3 − p = 143.4 can be used to find the length in hours p of a day on the planet. What is the length of a storm that lasts \(\frac{1}{3}\) of a day on this planet? ________ hours

Answer: 3.3 hours

Explanation: A day on Pluto is 143.4 hours longer than a day on one of the planets listed in the table. 153.3 − p = 143.4 153.3 – 143.4 = p p = 153.3 – 143.4 p = 9.9 Now p with \(\frac{1}{3}\) to find the length of a storm that lasts of a day on this planet 9.9 × \(\frac{1}{3}\) = 3.3 hours

Question 15. What’s the Error? Jason said that the solution of the equation 2m = 4 is m = 8. Describe Jason’s error, and give the correct solution. Type below: _____________

Answer: m = 2

Explanation: Jason said that the solution of the equation 2m = 4 is m = 8. 2m = 4 m = 4/2 = 2 The error of Jason is he multiplied 2 and 4 but he should divide 4 by 2.

Question 16. The marking period is 45 school days long. Today is the twenty-first day of the marking period. The equation x + 21 = 45 can be used to find the number of days x left in the marking period. Using substitution, Rachel determines there are _____ days left in the marking period. Rachel determines there are _____________ days left.

Explanation: The marking period is 45 school days long. Today is the twenty-first day of the marking period. The equation x + 21 = 45 x = 45 – 21 = 24 days Using substitution, Rachel determines there are 24 days left in the marking period. Thus Rachel determines there are 24 days left.

Question 1. x − 7 = 15; x = 8 The variable is __________

Explanation: Substitute the value in the given equation. x = 8 8 – 7 = 15 1 ≠ 15 Therefore the variable is not a solution.

Question 2. c + 11 = 20; c = 9 The variable is __________

Explanation: Substitute the value in the given equation. c = 9 9 + 11 = 20 20 = 20 Therefore the variable is a solution.

Question 3. \(\frac{1}{3}\)h = 6; h = 2 The variable is __________

Explanation: Substitute the value in the given equation. \(\frac{1}{3}\)h = 6 h = 2 \(\frac{1}{3}\) × 2 = 6 \(\frac{2}{3}\) ≠ 6 Therefore the variable is not a solution.

Question 4. 16.1 + d = 22; d = 6.1 The variable is __________

Explanation: Substitute the value in the given equation. 16.1 + d = 22 d = 6.1 16.1 + 6.1 = 22 22.2 ≠ 22 Therefore the variable is not a solution.

Question 5. 9 = \(\frac{3}{4}\)e; e = 12 The variable is __________

Explanation: Substitute the value in the given equation. 9 = \(\frac{3}{4}\)e e = 12 9 = \(\frac{3}{4}\)(12) 9 = 3 × 3 9 = 9 Therefore the variable is a solution.

Question 6. 15.5 – y = 7.9; y = 8.4 The variable is __________

Explanation: Substitute the value in the given equation. 15.5 – y = 7.9 y = 8.4 15.5 – 8.4 = 7.9 7.1 ≠ 7.9 Therefore the variable is not a solution.

Problem Solving

Question 7. Terrance needs to score 25 points to win a game. He has already scored 18 points. The equation 18 + p = 25 can be used to find the number of points p that Terrance still needs to score. Determine whether p = 7 or p = 13 is a solution of the equation, and tell what the solution means. Type below: _____________

Answer: p = 7

Explanation: Terrance needs to score 25 points to win a game. He has already scored 18 points. The equation is 18 + p = 25 Substitute p = 7 in the above equation. 18 + 7 = 25 25 = 25 The variable is a solution. Substitute p = 13 18 + p = 25 18 + 13 = 25 31 ≠ 25 The variable is not a solution. Therefore p = 7 is a solution for the equation.

Question 8. Madeline has used 50 sheets of a roll of paper towels, which is \(\frac{5}{8}\) of the entire roll. The equation \(\frac{5}{8}\)s = 50 can be used to find the number of sheets s in a full roll. Determine whether s = 32 or s = 80 is a solution of the equation, and tell what the solution means. Type below: _____________

Answer: Madeline has used 50 sheets of a roll of paper towels, which is \(\frac{5}{8}\) of the entire roll. \(\frac{5}{8}\)s = 50 s = 50 × \(\frac{8}{5}\) s = 80 because 80 × 5 = 400 400 ÷ 8 = 50

Question 9. Use mental math to find the solution to 4x = 36. Then use substitution to check your answer. Type below: _____________

Answer: x = 9

Explanation: 4x = 36 x = 36/4 x = 9

Question 1. Sheena received a gift card for $50. She has already used it to buy a lamp for $39.99. The equation 39.99 + x = 50 can be used to find the amount x that is left on the gift card. What is the solution of the equation? _____

Answer: 10.01

Explanation: Given: Sheena received a gift card for $50. She has already used it to buy a lamp for $39.99. The equation 39.99 + x = 50 39.99 + x = 50 x = 50 – 39.99 x = 50.00 – 39.99 x = 10.01 Thus $10.01 is left on the gift card.

Question 2. When Pete had a fever, his temperature was 101.4°F. After taking some medicine, his temperature was 99.2°F. The equation 101.4 – d = 99.2 can be used to find the number of degrees d that Pete’s temperature decreased. What is the solution of the equation? _____

Answer: 2.2

Explanation: Given, When Pete had a fever, his temperature was 101.4°F. After taking some medicine, his temperature was 99.2°F. The equation 101.4 – d = 99.2 104.4 – 99.2 = d d = 104.4 – 99.2 d = 2.2

Spiral Review

Question 3. Melanie has saved $60 so far to buy a lawn mower. This is 20% of the price of the lawn mower. What is the full price of the lawn mower that she wants to buy? $ _____

Answer: 300

Explanation: Melanie has saved $60 so far to buy a lawn mower. This is 20% of the price of the lawn mower. 60 ÷ 20% 60 ÷ 20/100 60 × 100/20 = 6000/20 = 300 She wants to buy a $300 price of the lawn mower.

Question 4. A team of scientists is digging for fossils. The amount of soil in cubic feet that they remove is equal to 6³. How many cubic feet of soil do the scientists remove? _____ cubic feet

Answer: 216

Explanation: A team of scientists is digging for fossils. The amount of soil in cubic feet that they remove is equal to 6³. 6 × 6 × 6 = 216 Thus the scientists remove 216 cubic feet of soil.

Question 5. Andrew made p picture frames. He sold 2 of them at a craft fair. Write an expression that could be used to find the number of picture frames Andrew has left. Type below: _____________

Answer: p – 2

Explanation: Andrew made p picture frames. He sold 2 of them at a craft fair. The expression is the difference of 9 and 2 The equation is p – 2

Question 6. Write an expression that is equivalent to 4 + 3(5 + x). Type below: _____________

Answer: 4 + 15 + 3x

Explanation: 4 + 3(5 + x) = 4 + 15 + 3x 3x + 19 Thus the expression 4 + 3(5 + x) is equivalent to 4 + 15 + 3x or 3x + 19

Question 1. Write an equation for the word sentence “25 is 13 more than a number.” Type below: _____________

Answer: Let n represents the unknown number. The phrase ‘more than’ indicates addition operation. Thus the equation is 25 = 13 + n.

Write an equation for the word sentence.

Question 2. The difference of a number and 2 is 3 \(\frac{1}{3}\). Type below: _____________

Answer: Let n represents the unknown number. The phrase “difference” indicates the subtraction operation. The equation is n – 2 = 3 \(\frac{1}{3}\)

Question 3. Ten times the number of balloons is 120. Type below: _____________

Answer: Let n represents the unknown number. The phrase “times” indicates multiplication operation. The equation is 10 × n = 120

Write a word sentence for the equation.

Question 4. x − 0.3 = 1.7 Type below: _____________

Answer: The difference of x and 0.3 is 1.7

Question 5. 25 = \(\frac{1}{4}\)n Type below: _____________

Answer: 25 is n times \(\frac{1}{4}\)

Question 6. The quotient of a number and 20.7 is 9. Type below: _____________

Answer: Let n represents the unknown number. The phrase “quotient” indicates the division operation. Thus the equation is n ÷ 20.7 = 9.

Question 7. 24 less than the number of snakes is 35. Type below: _____________

Answer: Let n represents the unknown number. The phrase “less than” indicates subtraction operation. Thus the equation is n – 24 = 35

Question 8. 75 is 18 \(\frac{1}{2}\) more than a number. Type below: _____________

Answer: Let n represents the unknown number. The phrase “more than” indicates addition operation. 75 = 18 \(\frac{1}{2}\) + n

Question 9. d degrees warmer than 50 degrees is 78 degrees. Type below: _____________

Answer: Let n represents the unknown number. The phrase “warmer than” indicates addition operation. The equation is d + 50 = 78 degrees

Question 10. 15g = 135 Type below: _____________

Answer: g times 15 is 135

Question 11. w ÷ 3.3 = 0.6 Type below: _____________

Answer: The quotient of w and 3.3 is 0.6

Question 12. Write an equation that could be used to find how many miles a hybrid SUV can travel in the city on 20 gallons of gas. Type below: _____________

Answer: From table 36 miles per gallon in the city. A hybrid SUV uses 36 miles per gallon in the city. So, no. of miles = y x = no. of gallons So, y = 36 × x x = 20 gallons Thus y = 36 × 20

Question 13. A sedan traveled 504 miles on the highway on a full tank of gas. Write an equation that could be used to find the number of gallons the tank holds. Type below: _____________

Answer: A sedan uses 28 miles per gallon on the highway. The equation that could be used to find the number of gallons the tank holds is 504 = 28g

Question 14. Connect Symbols to Words Sonya was born in 1998. Carmen was born 11 years after Sonya. If you wrote an equation to find the year in which Carmen was born, what operation would you use in your equation? Type below: _____________

Answer: In this equation, I would use addition or subtraction operation.

Question 15. A magazine has 110 pages. There are 23 full-page ads and 14 half-page ads. The rest of the magazine consists of articles. Write an equation that can be used to find the number of pages of articles in the magazine. Type below: _____________

Answer: The equation that can be used to find the number of pages of articles in the magazine is 23 + 14/2 + a = 110 where a represents the number of articles.

Question 16. What’s the Error? Tony is traveling 560 miles to visit his cousins. He travels 313 miles the first day. He says that he can use the equation m − 313 = 560 to find the number of miles m he has left on his trip. Describe and correct Tony’s error. Type below: _____________

Answer: Tony subtracted the number of miles traveled from the number of miles left. Tony should have written m + 313 = 560

Question 17. Jamie is making cookies for a bake sale. She triples the recipe in order to have enough cookies to sell. Jamie uses 12 cups of flour to make the triple batch. Write an equation that can be used to find out how much flour f is needed for one batch of cookies. Type below: _____________

Answer: The equation that can be used to find out how much flour f is needed for one batch of cookies is 3f = 12

Question 1. 18 is 4.5 times a number. Type below: _____________

Answer: Let n represents the unknown number. The phrase “times” indicates the multiplication operation. The equation is 18 = 4.5n

Question 2. Eight more than the number of children is 24. Type below: _____________

Answer: Let c represents the number of children. The phrase “more than” indicates addition operation. Thus the equation is 8 + c = 24.

Question 3. The difference of a number and \(\frac{2}{3}\) is \(\frac{3}{8}\). Type below: _____________

Answer: Let n represents the unknown number. The phrase “difference” indicates a subtraction operation. The equation is n – \(\frac{2}{3}\) = \(\frac{3}{8}\)

Question 4. A number divided by 0.5 is 29. Type below: _____________

Answer: Let n represents the unknown number. The phrase divided by indicates division operation. The equation is n ÷ 0.5 = 29

Question 5. x − 14 = 52 Type below: _____________

Answer: 14 less than x is 52 the difference of x and 14 is 52 14 fewer than a number is 52.

Question 6. 2.3m = 0.46 Type below: _____________

Answer: The product of 2.3 and m is 0.46 2.3 times m is .46 2.3 of m is 0.46

Question 7. 25 = k ÷ 5 Type below: _____________

Answer: 25 is the quotient of k and 5.

Question 8. \(4 \frac{1}{3}+q=5 \frac{1}{6}\) Type below: _____________

Answer: The sum of \(4 \frac{1}{3}\) and q is [/latex]5 \frac{1}{6}[/latex] q is more than \(4 \frac{1}{3}\) and [/latex]5 \frac{1}{6}[/latex] \(4 \frac{1}{3}\) increased by a number is [/latex]5 \frac{1}{6}[/latex]

Question 9. An ostrich egg weighs 2.9 pounds. The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds. Write an equation that could be used to find the weight w, in pounds, of the emu egg. Type below: _____________

Answer: 2.9 – w = 1.6

Explanation: An ostrich egg weighs 2.9 pounds. The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds. The phrase “difference” indicates the subtraction operation. The equation will be 2.9 – w = 1.6

Question 10. In one week, the number of bowls a potter made was 6 times the number of plates. He made 90 bowls during the week. Write an equation that could be used to find the number of plates p that the potter made. Type below: _____________

Answer: 6p = 90

Explanation: Given, In one week, the number of bowls a potter made was 6 times the number of plates. He made 90 bowls during the week. The phrase “times” indicates the multiplication operation. The equation to find the number of plates p that the potter made will be 6p = 90

Question 11. When writing a word sentence as an equation, explain when to use a variable. Type below: _____________

Answer: In a word sentence, a variable represents “a number.” The sum of a number and three = n + 3 The difference of five times a number and four = 5n – 4

Question 1. Three friends are sharing the cost of a bucket of popcorn. The total cost of the popcorn is $5.70. Write an equation that could be used to find the amount a in dollars that each friend should pay. Type below: _____________

Answer: 3a = 5.70

Explanation: Three friends are sharing the cost of a bucket of popcorn. The total cost of the popcorn is $5.70. The expression will be “5.70 is the product of 3 and a. The equation is 3a = 5.70

Question 2. Salimah had 42 photos on her phone. After she deleted some of them, she had 23 photos left. What equation could be used to find the number of photos p that Salimah deleted? Type below: _____________

Answer: p + 23 = 42

Explanation: Salimah had 42 photos on her phone. After she deleted some of them, she had 23 photos left. The expression is the sum of p and 23 is 42. Thus the equation is p + 23 = 42

Question 3. A rope is 72 feet long. What is the length of the rope in yards? ______ yards

Answer: 24 yard

Explanation: A rope is 72 feet long. Convert from feet to yards. 1 yard = 3 feet 1 foot = 1/3 yards 72 feet = 72 × 1/3 = 24 yards Thus the length of the rope is 24 yards.

Question 4. Julia evaluated the expression 3 3 + 20 ÷ 2 2 . What value should she get as her answer? ______

Explanation: The equation is 3 3 + 20 ÷ 2 2 . 3 3 = 3 × 3 × 3 = 27 2 2 = 2 × 2 = 4 27 + (20 ÷ 4) 27 + 5 = 32 The answer for the above equation is 32.

Question 5. The sides of a triangle have lengths s, s + 4, and 3s. Write an expression in the simplest form that represents the perimeter of the triangle. Type below: _____________

Answer: 5s + 4

Explanation: The perimeter of the triangle is a + b + c P = a + b + c P = s + s + 4 + 3s P = 5s + 4 Thus the perimeter of the triangle is 5s + 4

Question 6. Gary knows that p = 2 \(\frac{1}{2}\) is a solution to one of the following equations. Which one has p = 2 \(\frac{1}{2}\) as its solution? \(p+2 \frac{1}{2}=5\)        \(p-2 \frac{1}{2}=5\) \(2+p=2 \frac{1}{2}\)       4 – p = 2 \(\frac{1}{2}\) Type below: _____________

Answer: p + 2 \(\frac{1}{2}\) = 5

Explanation: \(p+2 \frac{1}{2}=5\) p + 2 \(\frac{1}{2}\) = 5 p = 5 – 2 \(\frac{1}{2}\) p = 2 \(\frac{1}{2}\) \(p-2 \frac{1}{2}=5\) p – 2 \(\frac{1}{2}\) = 5 p = 5 + 2 \(\frac{1}{2}\) p = 7 \(\frac{1}{2}\) \(2+p=2 \frac{1}{2}\) 2 + p = 2 \(\frac{1}{2}\) p = 2 \(\frac{1}{2}\) – 2 p = \(\frac{1}{2}\) 4 – p = 2 \(\frac{1}{2}\) p = 4 – 2 \(\frac{1}{2}\) p = 1 \(\frac{1}{2}\)

Model and solve the equation by using algebra tiles or iTools.

Question 1. x + 5 = 7 x = ______

Explanation:

• Draw 2 rectangles on your MathBoard to represent the two sides of the equation.
• Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 7 in the right rectangle.
• To solve the equation, get the x tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
• Remove five 1 tiles on the left side and five 1 tiles on the right side.
• The remaining titles will be two 1 tiles on the right sides.

Question 2. 8 = x + 1 x = ______

• Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 8 in the right rectangle.
• Remove one 1 tiles on the left side and one 1 tiles on the right side.
• The remaining titles will be seven 1 tiles on the right sides.

Question 3. x + 2 = 5 x = ______

• Use algebra tiles to model the equation. Model x + 2 in the left rectangle, and model 5 in the right rectangle.
• Remove two 1 tiles on the left side and five 1 tiles on the right side.
• The remaining titles will be three 1 tiles on the right sides.

Question 4. x + 6 = 8 x = ______

• Use algebra tiles to model the equation. Model x + 6 in the left rectangle, and model 8 in the right rectangle.
• Remove six 1 tiles on the left side and six 1 tiles on the right side.

Question 5. 5 + x = 9 x = ______

• Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 9 in the right rectangle.
• The remaining titles will be four 1 tiles on the right sides.

Question 6. 5 = 4 + x x = ______

• Use algebra tiles to model the equation. Model x + 4 in the left rectangle, and model 5 in the right rectangle.
• Remove four 1 tiles on the left side and four 1 tiles on the right side.
• The remaining titles will be one 1 tiles on the right sides.

Solve the equation by drawing a model.

Question 7. x + 1 = 5 x = ______

• Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 5 in the right rectangle.

Question 8. 3 + x = 4 x = ______

• Use algebra tiles to model the equation. Model x + 3 in the left rectangle, and model 4 in the right rectangle.
• Remove three 1 tiles on the left side and three 1 tiles on the right side.

Question 9. 6 = x + 4 x = ______

• Use algebra tiles to model the equation. Model x + 4 in the left rectangle, and model 6 in the right rectangle.

Question 10. 8 = 2 + x x = ______

• Use algebra tiles to model the equation. Model x + 2 in the left rectangle, and model 8 in the right rectangle.
• Remove two 1 tiles on the left side and two 1 tiles on the right side.
• The remaining titles will be six 1 tiles on the right sides.

Question 11. Describe a Method Describe how you would draw a model to solve the equation x + 5 = 10. Type below: _____________

Answer: x = 5

• Use algebra tiles to model the equation. Model x + 5 in the left rectangle, and model 10 in the right rectangle.
• The remaining titles will be five 1 tiles on the right sides.

Question 12. Interpret a Result The table shows how long several animals have lived at a zoo. The giraffe has lived at the zoo 4 years longer than the mountain lion. The equation 5 = 4 + y can be used to find the number of years y the mountain lion has lived at the zoo. Solve the equation. Then tell what the solution means. Type below: _____________

Answer: The table shows how long several animals have lived in a zoo. The giraffe has lived at the zoo 4 years longer than the mountain lion. 5 = 4 + y y = 5 – 4 y = 1 The solution is y = 1 The solution means that the mountain lion has lived at the zoo for 1 year.

Question 13. Carlos walked 2 miles on Monday and 5 miles on Saturday. The number of miles he walked on those two days is 3 miles more than the number of miles he walked on Friday. Write and solve an addition equation to find the number of miles Carlos walked on Friday Type below: _____________

Answer: Given that, Carlos walked 2 miles on Monday and 5 miles on Saturday. The number of miles he walked on those two days is 3 miles more than the number of miles he walked on Friday. The equation is f + 3 = 2 + 5 f + 3 = 7 f = 7 – 3 f = 4 The solution is f = 4 The solution means that Carlos walked 4 miles on Friday.

Question 14. Sense or Nonsense? Gabriela is solving the equation x + 1 = 6. She says that the solution must be less than 6. Is Gabriela’s statement sense or nonsense? Explain. Type below: _____________

Answer: Gabriela’s statement makes sense. x + 1 = 6 x = 6 – 1 x = 5 Thus the solution is less than 6.

Answer: Remove 5 squares from each side. The rectangle is by itself on the left and 7 squares are on the right side. So, the solution is x = 7

Model and solve the equation by using algebra tiles.

Question 1. x + 6 = 9 x = ________

• Use algebra tiles to model the equation. Model x + 6 in the left rectangle, and model 9 in the right rectangle.

Question 2. 8 + x = 10 x = ________

• Use algebra tiles to model the equation. Model x + 8 in the left rectangle, and model 10 in the right rectangle.
• Remove eight 1 tiles on the left side and eight 1 tiles on the right side.

8 + x = 10 x = 10 – 8 = 2 x = 2

Question 3. 9 = x + 1 x = ________

• Use algebra tiles to model the equation. Model x + 1 in the left rectangle, and model 9 in the right rectangle.
• Remove 1 tile on the left side and 1 tile on the right side.
• The remaining titles will be eight 1 tiles on the right sides.

Question 4. x + 4 = 7 x = ________

Question 5. x + 6 = 10 x = ________

Question 6. The temperature at 10:00 was 10°F. This is 3°F warmer than the temperature at 8:00. Model and solve the equation x + 3 = 10 to find the temperature x in degrees Fahrenheit at 8:00. Type below: _____________

Answer: x = 7

Explanation: The temperature at 10:00 was 10°F. This is 3°F warmer than the temperature at 8:00. The equation is x + 3 = 10 x = 10 – 3 = 7

Question 7. Jaspar has 7 more checkers left than Karen does. Jaspar has 9 checkers left. Write and solve an addition equation to find out how many checkers Karen has left. Type below: _____________

Answer: c = 2

Explanation: Jaspar has 7 more checkers left than Karen does. Jaspar has 9 checkers left. The expression is c + 7 = 9 The equation to find out how many checkers Karen has left is c + 7 = 9.

Question 8. Explain how to use a drawing to solve an addition equation such as x + 8 = 40. Type below: _____________

• Use algebra tiles to model the equation. Model x + 8 in the left rectangle, and model 40 in the right rectangle.
• Remove eight 1 tile on the left side and eight 1 tile on the right side.
• The remaining titles will be 32 1 tiles on the right side.

x + 8 = 40 x = 40 – 8 x = 32

The equation is x + 6 = 7 x = 7 – 6 x = 1

Question 2. Alice has played soccer for 8 more years than Sanjay has. Alice has played for 12 years. The equation y + 8 = 12 can be used to find the number of years y Sanjay has played. How long has Sanjay played soccer? ________ years

Answer: 4 years

Explanation: Alice has played soccer for 8 more years than Sanjay has. Alice has played for 12 years. the equation is y + 8 = 12 y = 12 – 8 y = 4 years Sanjay played soccer games for 4 years.

Question 3. A car’s gas tank has a capacity of 16 gallons. What is the capacity of the tank in pints? ________ pints

Answer: 128 pints

Explanation: A car’s gas tank has a capacity of 16 gallons. Convert from gallons to pints. 1 gallon = 8 pints 16 gallons = 16 × 8 = 128 pints Thus the capacity of the tank is 128 pints.

Question 4. Craig scored p points in a game. Marla scored twice as many points as Craig but 5 fewer than Nelson scored. How many points did Nelson score? Type below: _____________

Answer: 2p + 5

Explanation: Craig scored p points in a game. Marla scored twice as many points as Craig but 5 fewer than Nelson score. The equation will be 2p + 5.

Question 5. Simplify 3x + 2(4y + x). Type below: _____________

Answer: 5x + 8y

Explanation: The expression is 3x + 2(4y + x) 3x + 2 × 4y + 2 × x 3x + 8y + 2x Combine the like terms. 5x + 8y 3x + 2(4y + x) = 5x + 8y

Question 6. The Empire State Building in New York City is 443.2 meters tall. This is 119.2 meters taller than the Eiffel Tower in Paris. Write an equation that can be used to find the height h in meters of the Eiffel Tower. Type below: _____________

Answer: 119.2 + h = 443.2

Explanation: The Empire State Building in New York City is 443.2 meters tall. This is 119.2 meters taller than the Eiffel Tower in Paris. Here we have to use the addition operation. The equation is 119.2 + h = 443.2

Question 1. Solve the equation n + 35 = 80. n = ________

Explanation: The given equation is n + 35 = 80 n = 80 – 35 n = 45

Solve the equation, and check the solution.

Question 2. 16 + x = 42 x = ________

Explanation: Given the equation 16 + x = 42 x + 16 = 42 x = 42 – 16 x = 26

Question 3. y + 6.2 = 9.1 y = ________

Answer: 2.9

Explanation: The given equation is y + 6.2 = 9.1 y = 9.1 – 6.2 y = 2.9

Question 4. m + \(\frac{3}{10}=\frac{7}{10}\) m = \(\frac{□}{□}\)

Answer: \(\frac{4}{10}\)

Explanation: The given equation is m + \(\frac{3}{10}=\frac{7}{10}\) m = \(\frac{7}{10}\) – \(\frac{3}{10}\) The denominators are common so subtract the numerators m = \(\frac{4}{10}\)

Question 5. z – \(\frac{1}{3}=1 \frac{2}{3}\) z = ________

Explanation: The given equation is z – \(\frac{1}{3}=1 \frac{2}{3}\) z = \(\frac{1}{3}\) + 1 \(\frac{2}{3}\) z = 1 + \(\frac{1}{3}\) + \(\frac{2}{3}\) z = 1 + \(\frac{3}{3}\) z = 1 + 1 = 2 Thus the value of z is 2.

Question 6. 12 = x − 24 x = ________

Explanation: The given equation is 12 = x − 24 x – 24 = 12 x = 12 + 24 x = 36 Thus the value of x is 36.

Question 7. 25.3 = w − 14.9 w = ________

Answer: 40.2

Explanation: The given equation is 25.3 = w − 14.9 w – 14.9 = 25.3 w = 25.3 + 14.9 w = 40.2 The value of w is 40.2

Practice: Copy and Solve Solve the equation, and check the solution.

Question 8. y − \(\frac{3}{4}=\frac{1}{2}\) y = _______ \(\frac{□}{□}\)

Answer: 1 \(\frac{1}{4}\)

Explanation: The given equation is y − \(\frac{3}{4}=\frac{1}{2}\) y = \(\frac{1}{2}\) + \(\frac{3}{4}\) y = 1 \(\frac{1}{4}\) Therefore the value of y is 1 \(\frac{1}{4}\).

Question 9. 75 = n + 12 n = ________

Explanation: The given equation is 75 = n + 12 n + 12 = 75 n = 75 – 12 n = 63 The value of n is 63.

Question 10. m + 16.8 = 40 m = ________

Answer: 23.2

Explanation: The given equation is m + 16.8 = 40 m = 40 – 16.8 m = 23.2 The value of m is 23.2

Question 11. w − 36 = 56 w = ________

Explanation: The given equation is w − 36 = 56 w = 56 + 36 w = 92 The value of  is 92.

Question 12. 8 \(\frac{2}{5}\) = d + 2\(\frac{2}{5}\) d = ________

Explanation: The given equation is 8 \(\frac{2}{5}\) = d + 2\(\frac{2}{5}\) d + 2\(\frac{2}{5}\) = 8 \(\frac{2}{5}\) d = 8 \(\frac{2}{5}\) – 2\(\frac{2}{5}\) d = 8 + \(\frac{2}{5}\) – 2 – \(\frac{2}{5}\) d = 8 – 2 = 6 Thus the value of d is 6.

Question 13. 8.7 = r − 1.4 r = ________

Answer: 10.1

Explanation: The given equation is 8.7 = r − 1.4 r − 1.4 = 8.7 r = 8.7 + 1.4 r = 10.1 The value of r is 10.1

Question 14. The temperature dropped 8 degrees between 6:00 p.m. and midnight. The temperature at midnight was 26ºF. Write and solve an equation to find the temperature at 6:00 p.m. ________ ºF

Answer: 34ºF

Explanation: The temperature dropped 8 degrees between 6:00 p.m. and midnight. The temperature at midnight was 26ºF. 26ºF + 8ºF = 34ºF The equation to find the temperature at 6:00 p.m is 34ºF

Question 15. Reason Abstractly Write an addition equation that has the solution x = 9. Type below: _____________

Answer: x + 4 = 13

Explanation: Let the equation be x + 4 = 13 x = 13 – 4 x = 9

Answer: We need to find Kimberly’s balance at the beginning of July.

Question 16. b. What information do you need from the bank statement? Type below: _____________

Answer: We need the information about the deposit on July 12 and July 25 from the bank statement.

Question 16. c. Write an equation you can use to solve the problem. Explain what the variable represents. Type below: _____________

Answer: x = bank account balance y = deposit 1 z = deposit 2 x = y + z

Question 16. d. Solve the equation. Show your work and describe each step. Type below: _____________

Answer: 120.62 = y + z Where y is the deposit 1 and z represents the deposit 2. y = $45.50, z = $43.24 45.50 + 43.24 = 88.74 x + 88.74 = 120.62

Question 16. e. Write Kimberly’s balance at the beginning of July. $ _______

Answer: 31.88

Explanation: x + 88.74 = 120.62 x = 120.62 – 88.74 x = $31.88 Kimberly’s balance at the beginning of July is $31.88

Question 17. If x + 6 = 35, what is the value of x + 4? Explain how to find the value without solving the equation. Type below: _____________

Answer: x + 6 = 35 x + 4 + 2 = 35 x  + 4 = 35 – 2 x + 4 = 33 Thus the value of x + 4 = 33

Question 18. Select the equations that have the solution n = 23. Mark all that apply. Options: a. 16 + n = 39 b. n – 4 = 19 c. 25 = n – 2 d. 12 = n – 11

Answer: A, B, D

Explanation: a. 16 + n = 39 n = 23 16 + 23 = 39 39 = 39 The variable is a solution. b. n – 4 = 19 n = 23 23 – 4 = 19 19 = 19 The variable is a solution. c. 25 = n – 2 25 = 23 – 2 25 ≠ 21 The variable is not a solution. d. 12 = n – 11 n = 23 12 = 23 – 11 12 = 12 The variable is a solution. Thus the correct answers are options A, B, D.

Question 1. y − 14 = 23 y = _______

Explanation: y − 14 = 23 y = 23 + 14 y = 37 Thus the solution is 37.

Question 2. x + 3 = 15 x = _______

Explanation: The equation is x + 3 = 15 x = 15 – 3 x = 12 The solution is 12.

Question 3. n + \(\frac{2}{5}=\frac{4}{5}\) n = _______ \(\frac{□}{□}\)

Answer: \(\frac{2}{5}\)

Explanation: The equation is n + \(\frac{2}{5}=\frac{4}{5}\) n + \(\frac{2}{5}\) = \(\frac{4}{5}\) n = \(\frac{4}{5}\) – \(\frac{2}{5}\) n = (4 – 2)/5 n = \(\frac{2}{5}\) Thus the solution is \(\frac{2}{5}\)

Question 4. 16 = m − 14 m = _______

Explanation: The equation is 16 = m − 14 m – 14 = 16 m = 16 + 14 m = 30 The solution is m = 30

Question 5. w − 13.7 = 22.8 w = _______

Answer: 36.5

Explanation: The equation is w − 13.7 = 22.8 w = 22.8 + 13.7 w = 36.5 The solution is w = 36.5

Question 6. s + 55 = 55 s = _______

Explanation: The equation is s + 55 = 55 s = 55 – 55 s = 0 The solution is s = 0

Question 7. 23 = x − 12 x = _______

Explanation: The given equation is 23 = x – 12 x – 12 = 23 x = 23 + 12 x = 35 The solution is x = 35.

Question 8. p − 14 = 14 p = _______

Explanation: The given equation is p − 14 = 14 p = 14 + 14 p = 28 The solution is p = 28.

Question 9. m − \(2 \frac{3}{4}=6 \frac{1}{2}\) m = _______ \(\frac{□}{□}\)

Answer: 9 \(\frac{1}{4}\)

Explanation: The given equation is m − \(2 \frac{3}{4}=6 \frac{1}{2}\) m – 2 \(\frac{3}{4}\) = 6 \(\frac{1}{2}\) m = 6 \(\frac{1}{2}\) + 2 \(\frac{3}{4}\) m = 6 + 2 + \(\frac{1}{2}\) + \(\frac{3}{4}\) m = 8 + 1 \(\frac{1}{4}\) m = 9 \(\frac{1}{4}\)

Question 10. A recipe calls for 5 \(\frac{1}{2}\) cups of flour. Lorenzo only has 3 \(\frac{3}{4}\) cups of flour. Write and solve an equation to find the additional amount of flour Lorenzo needs to make the recipe. Type below: _____________

Answer: 1 \(\frac{3}{4}\)

Explanation: A recipe calls for 5 \(\frac{1}{2}\) cups of flour. Lorenzo only has 3 \(\frac{3}{4}\) cups of flour. x + 3 \(\frac{3}{4}\) = 5 \(\frac{1}{2}\) x = 5 \(\frac{1}{2}\) – 3 \(\frac{3}{4}\) x =  1 \(\frac{3}{4}\)

Question 11. Jan used 22.5 gallons of water in the shower. This amount is 7.5 gallons less than the amount she used for washing clothes. Write and solve an equation to find the amount of water Jan used to wash clothes. Type below: _____________

Explanation: Jan used 22.5 gallons of water in the shower. This amount is 7.5 gallons less than the amount she used for washing clothes. Let the amount of water Jan used to wash clothes be x x – 7.5 = 22.5 x = 22.5 + 7.5 x = 30 Therefore the amount of water Jan used to wash clothes is 30 gallons.

Question 12. Explain how to check if your solution to an equation is correct. Type below: _____________

Answer: i. Evaluate the left-hand side expression at the given value to get a number. ii. Evaluate the right-hand side expression at the given value to get a number. iii. See if the numbers match.

Question 1. The price tag on a shirt says $21.50. The final cost of the shirt, including sales tax, is $23.22. The equation 21.50 + t = 23.22 can be used to find the amount of sales tax t in dollars. What is the sales tax? $ _______

Answer: 1.72

Explanation: The price tag on a shirt says $21.50. The final cost of the shirt, including sales tax, is $23.22. The equation is 21.50 + t = 23.22 t = 23.22 – 21.50 t = 1.72 Therefore the sales tax is $1.72 dollars.

Question 2. The equation l – 12.5 = 48.6 can be used to find the original length l in centimeters of a wire before it was cut. What was the original length of the wire? _______ centimeters

Answer: 61.1 centimeters

Explanation: The equation l – 12.5 = 48.6 can be used to find the original length l in centimeters of a wire before it was cut. l – 12.5 = 48.6 l = 48.6 + 12.5 l = 61.1 centimeters Thus the original length of the wire is 61.1 centimeters.

Question 3. How would you convert a mass in centigrams to a mass in milligrams? Type below: _____________

Answer: The conversion factor is 10; so 1 centigram = 10 milligrams. In other words, the value in cg multiplies by 10 to get a value in mg.

Question 4. In the expression 4 + 3x + 5y, what is the coefficient of x? The coefficient is _______

Answer: A numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Thus the coefficient of 3x is 3.

Question 5. Write an expression that is equivalent to 10c. Type below: _____________

Answer: -2(-5c) expand the brackets -2 × -5c = 10c

Question 6. Miranda bought a $7-movie ticket and popcorn for a total of $10. The equation 7 + x = 10 can be used to find the cost x in dollars of the popcorn. How much did the popcorn cost? $ _______

Explanation: Miranda bought a $7-movie ticket and popcorn for a total of $10. The equation is 7 + x = 10 x = 10 – 7 x = 3 Therefore the cost of the popcorn is $3.

Question 1. 4x = 16 x = _______

• Draw 2 rectangles on your Mathboard to represent the two sides of the equation.
• Use algebra tiles to model the equation. Model 4x in the left rectangle, and model 16 in the right rectangle.
• There are four x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into 4 equal groups.

Question 2. 3x = 12 x = _______

• Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 12 in the right rectangle.
• There are three x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into 3 equal groups.

Question 3. 4 = 4x x = _______

• Use algebra tiles to model the equation. Model 4x in the left rectangle, and model 4 in the right rectangle.

Question 4. 3x = 9 x = _______

• Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 9 in the right rectangle.

Question 5. 2x = 10 x = _______

• Use algebra tiles to model the equation. Model 2x in the left rectangle, and model 10 in the right rectangle.
• There are two x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into two equal groups.

Question 6. 15 = 5x x = _______

• Use algebra tiles to model the equation. Model 5x in the left rectangle, and model 15 in the right rectangle.
• There are five x tiles on the left side of your model. To solve the equation by using the model, you need to find the value of one x tile.
• To do this, divide each side of your model into five equal groups.

Question 7. 4x = 8 x = _______

Question 8. 3x = 18 x = _______

Problem Solving + Applications

Question 9. Communicate Explain the steps you use to solve a multiplication equation with algebra tiles. Type below: _____________

Answer: To solve an equation, model the terms of the equation on both sides of an equals sign. Isolate the variable on one side by adding opposites and creating zero pairs. To remove a factor from the variable, divide the sides into rows equal to the factor, and distribute the terms equally among all the rows.

Question 10. Naomi is doing a report about the 1900 and 1904 Olympic Games. Each page will contain info7rmation about 4 of the countries that competed each year. Write and solve an equation to find the number of pages Naomi will need. _______ pages

Answer: 9 pages

Explanation: By seeing the above table we can say that the equation is 4x = 36 The number of countries that competed in the 1900 summer Olympic games is 24. The number of countries that competed in the 1904 summer Olympic games is 12. The total number of countries competed in total is 36. Each page of Naomi’s report contains information about 4 of the countries that competed each year. 4x = 36 x = 36/4 x = 9 Thus Naomi would require 9 pages to complete her report.

Question 11. Pose a Problem Use the information in the bar graph to write and solve a problem involving a multiplication equation. Type below: _____________

Answer: By seeing the above table we can say that the equation is 4x = 72 The number of countries that competed in the 1900 summer Olympic games is 24. The number of countries that competed in the 1904 summer Olympic games is 12. The number of countries that competed in the 1896 summer Olympic games is 14. The number of countries that competed in the 1908 summer Olympic games is 22. The total number of countries competed in total is 72. 4x = 72 x = 72/4 x = 18

Question 12. The equation 7s = 21 can be used to find the number of snakes s in each cage at a zoo. Solve the equation. Then tell what the solution means. s = _______

Explanation: The equation 7s = 21 can be used to find the number of snakes s in each cage at a zoo. Solve the equation. 7 × s = 21 s = 21/7 = 3 The solution s is 3.

Question 13. A choir is made up of 6 vocal groups. Each group has an equal number of singers. There are 18 singers in the choir. Solve the equation 6p = 18 to find the number of singers in each group. Use a model. _______ singers

Answer: 3 singers

Explanation: A choir is made up of 6 vocal groups. Each group has an equal number of singers. There are 18 singers in the choir. The equation 6p = 18 p = 18/6 = 3 p = 3 The solution p is 3.

Question 1. 2x = 8 x = _______

• Use algebra tiles to model the equation. Model 2x in the left rectangle, and model 8 in the right rectangle.

Question 2. 5x = 10 x = _______

• Use algebra tiles to model the equation. Model 5x in the left rectangle, and model 10 in the right rectangle.

Question 3. 21 = 3x x = _______

• Use algebra tiles to model the equation. Model 3x in the left rectangle, and model 21 in the right rectangle.
• To do this, divide each side of your model into three equal groups.

Question 4. 6 = 3x

Question 5. 4x = 12 x = _______

Question 6. A chef used 20 eggs to make 5 omelets. Model and solve the equation 5x = 20 to find the number of eggs x in each omelet. _______ eggs

Explanation: A chef used 20 eggs to make 5 omelets. The equation is 5x = 20 x = 50/5 = 4 Thus there are 4 eggs in each omelet.

Question 7. Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games. Write and solve an equation to find the number of video games Scott played. _______ video games

Explanation: Last month, Julio played 3 times as many video games as Scott did. Julio played 18 video games. The equation will be 3x = 18 x = 18/3 = 6 x = 6 The number of video games Scott played is 6.

Question 8. Write a multiplication equation, and explain how you can solve it by using a model. Type below: _____________

Answer: 15 = 5x Explanation:

Explanation: The equation for the above figure is 3x = 3 Substitute x = 1 3(1) = 3 3/3 = 1 Thus the solution is 1.

Question 2. Carlos bought 5 tickets to a play for a total of $20. The equation 5c = 20 can be used to find the cost c in dollars of each ticket. How much does each ticket cost? $ _______

Explanation: Carlos bought 5 tickets to a play for a total of $20. The equation is 5c = 20 c = 20/5 = 4 c = 4 The cost of each ticket is $4.

Question 3. A rectangle is 12 feet wide and 96 inches long. What is the area of the rectangle? _______ square feet

Answer: 1152

Explanation: A rectangle is 12 feet wide and 96 inches long. Area of rectangle is l × w A = 12 × 96 A = 1152 square feet. Thus the area of the rectangle is 1152 square feet.

Question 4. Evaluate the algebraic expression 24 – x ÷ y for x = 8 and y = 2. _______

Explanation: 24 – x ÷ y for x = 8 and y = 2. Substitute the value of x and y in the equation. 24 – (8 ÷ 2) 24 – 4 = 20

Question 5. Ana bought a 15.5-pound turkey at the grocery store this month. The equation p – 15.5 = 2.5 can be used to find the weight p, in pounds, of the turkey she bought last month. What is the solution of the equation? p = _______

Explanation: Ana bought a 15.5-pound turkey at the grocery store this month. The equation is p – 15.5 = 2.5 p = 2.5 + 15.5 p = 18 The solution for the equation is 18.

Question 6. A pet store usually keeps 12 birds per cage, and there are 7 birds in the cage now. The equation 7 + x = 12 can be used to find the remaining number of birds x that can be placed in the cage. What is the solution of the equation? x = _______

Explanation: A pet store usually keeps 12 birds per cage, and there are 7 birds in the cage now. The equation is 7 + x = 12 x = 12 – 7 x = 5 Thus the solution of the equation is 5.

Question 1. Solve the equation 2.5m = 10. m = _______

Explanation: 2.5m = 10 m = 10/2.5 m = 4

Question 2. 3x = 210 x = _______

Explanation: 3x = 210 x = 210/3 x = 70

Question 3. 2.8 = 4t t = _______

Answer: 0.7

Explanation: 2.8 = 4t 4t = 2.8 t = 2.8/4 t = 0.7

Question 4. \(\frac{1}{3}\)n = 15 n = _______

Explanation: \(\frac{1}{3}\)n = 15 n = 15 × 3 n = 45

Question 5. \(\frac{1}{2}\)y = \(\frac{1}{10}\) y = _______

Answer: \(\frac{1}{5}\)

Explanation: \(\frac{1}{2}\)y = \(\frac{1}{10}\) y = \(\frac{1}{10}\) × 2 y = \(\frac{1}{5}\)

Question 6. 25 = \(\frac{a}{5}\) a = _______

Answer: 125

Explanation: 25 = \(\frac{a}{5}\) a = 25 × 5 a = 125

Question 7. 1.3 = \(\frac{c}{4}\) c = _______

Answer: 5.2

Explanation: 1.3 = \(\frac{c}{4}\) c = 1.3 × 4 c = 5.2

Question 8. 150 = 6m m = _______

Explanation: 6m = 150 m = 150/6 m = 25

Question 9. 14.7 = \(\frac{b}{7}\) b = _______

Answer: 102.9

Explanation: 14.7 = \(\frac{b}{7}\) b = 14.7 × 7 b = 102.9

Question 10. \(\frac{1}{4}\) = \(\frac{3}{5}\)s s = \(\frac{□}{□}\)

Answer: \(\frac{5}{12}\)

Explanation: \(\frac{1}{4}\) = \(\frac{3}{5}\)s \(\frac{1}{4}\) × \(\frac{5}{3}\) = s s = \(\frac{5}{12}\)

Question 11. There are 100 calories in 8 fluid ounces of orange juice and 140 calories in 8 fluid ounces of pineapple juice. Tia mixed 4 fluid ounces of each juice. Write and solve an equation to find the number of calories in each fluid ounce of Tia’s juice mixture. _______ calories

Answer: 15 calories

Explanation: Number of calories in 8 ounces of orange juice = 100 Number of calories in 1 ounce of juice = 100/8 Number of calories in 4 ounces of juice 100/8 × 4 = 50 calories Number of calories in 8 ounces of pineapple juice = 140 Number of calories in 1 ounce of juice = 140/8 Number of calories in 4 ounces of pineapple juice = 140/8 × 4 =70 calories Now the mixture has 50 + 70 calories = 120 calories in 8 ounces So, 1 ounce of the mixture has 120/8 = 15 calories.

Question 12. Write a division equation that has the solution x = 16. Type below: _____________

Answer: 2x = 32 x = 32/2 x = 16 Thus the equation is x = 16.

What’s the Error?

Question 13. Melinda has a block of clay that weighs 14.4 ounces. She divides the clay into 6 equal pieces. To find the weight w in ounces of each piece, Melinda solved the equation 6w = 14.4. Look at how Melinda solved the equation. Find her error. 6w = 14.4 \(\frac{6 w}{6}\) = 6 × 14.4 w = 86.4 Correct the error. Solve the equation, and explain your steps. Describe the error that Melinda made Type below: _____________

Answer: Melinda has a block of clay that weighs 14.4 ounces. She divides the clay into 6 equal pieces. The equation is 6w = 14.4 The error of Melinda is she used the multiplication equation to solve the equation. She must have used the division equation to get the solution. 6w = 14.4 w = 14.4/6 w = 2.4

Question 14. For numbers 14a−14d, choose Yes or No to indicate whether the equation has the solution x = 15. 14a. 15x = 30 14b. 4x = 60 14c. \(\frac{x}{5}\) = 3 14d. \(\frac{x}{3}\) = 5 14a. _____________ 14b. _____________ 14c. _____________ 14d. _____________

Answer: Given the value of x is 15 14a. 15x = 30 15 × 15 = 30 225 ≠ 30 The answer is No. 14b. 4x = 60 4 × 15 = 60 60 = 60 The answer is yes. 14c. \(\frac{x}{5}\) = 3 x/5 = 3 15/5 = 3 3 = 3 The answer is yes. 14d. \(\frac{x}{3}\) = 5 x/3 = 5 15/3 = 5 5 = 5 The answer is yes.

Question 1. 8p = 96 p = ________

Explanation: 8p = 96 8 × p = 96 p = 96/8 p = 12 The solution is 12

Question 2. \(\frac{z}{16}\) = 8 z = ________

Answer: 128

Explanation: The given equation is \(\frac{z}{16}\) = 8 z = 8 × 16 z = 128 The solution is 128.

Question 3. 3.5x = 14.7 x = ________

Answer: 4.2

Explanation: The given equation is 3.5x = 14.7 x = 14.7/3.5 x = 4.2 The solution x is 4.2

Question 4. 32 = 3.2c c = ________

Explanation: The given equation is 32 = 3.2c 3.2 × c = 32 c = 32/3.2 c = 1/0.1 = 10 The solution c is 10.

Question 5. \(\frac{2}{5}\)w = 40 w = ________

Answer: 100

Explanation: The given equation is \(\frac{2}{5}\)w = 40 \(\frac{2}{5}\) × w = 40 w = 40 × 5/2 w = 200/2 w = 100

Question 6. \(\frac{a}{14}\) = 6.8 a = ________

Answer: 95.2

Explanation: The given equation is \(\frac{a}{14}\) = 6.8 a = 6.8 × 14 a = 95.2

Question 7. 1.6x = 1.6 x = ________

Explanation: The given equation is 1.6x = 1.6 x = 1.6/1.6 x = 1 The solution x is 1

Question 8. 23.8 = 3.5b b = ________

Answer: 6.8

Explanation: The given equation is 23.8 = 3.5b 3.5b = 23.8 b = 23.8/3.5 b = 6.8 Thus the solution of the variable b is 6.8

Question 9. \(\frac{3}{5}\) = \(\frac{2}{3}\)t t = \(\frac{□}{□}\)

Answer: \(\frac{9}{10}\)

Explanation: The given equation is \(\frac{3}{5}\) = \(\frac{2}{3}\)t t = \(\frac{3}{5}\) × \(\frac{3}{2}\) t = \(\frac{9}{10}\) Thus the solution of the variable t is \(\frac{9}{10}\)

Question 10. Anne runs 6 laps on a track. She runs a total of 1 mile, or 5,280 feet. Write and solve an equation to find the distance, in feet, that she runs in each lap. ________ feet

Answer: 880

Explanation: Anne runs 6 laps on a track. She runs a total of 1 mile, or 5,280 feet. Let the l represents the runs in each lap. 6 × l = 5280 feet l = 5280/6 l = 880 feet Therefore Anne runs 880 feets in each lap.

Question 11. In a serving of 8 fluid ounces of pomegranate juice, there are 32.8 grams of carbohydrates. Write and solve an equation to find the amount of carbohydrates in each fluid ounce of the juice. ________ grams

Answer: 4.1

Explanation: Given, In a serving of 8 fluid ounces of pomegranate juice, there are 32.8 grams of carbohydrates. Let c represents the amount of carbohydrates in each fluid ounce of the juice 8 × c = 32.8 grams c = 32.8/8 c = 4.1 grams

Question 12. Write and solve a word problem that can be solved by solving a multiplication equation. Type below: _____________

Answer: The quotient of 6 and p is 12 6 ÷ p = 12 p = 6/12 p = 1/2

Question 1. Estella buys 1.8 pounds of walnuts for a total of $5.04. She solves the equation 1.8p = 5.04 to find the price p in dollars of one pound of walnuts. What does one pound of walnuts cost? $ ________

Answer: 2.8

Explanation: Given that, Estella buys 1.8 pounds of walnuts for a total of $5.04. p represents the price in dollars of one pound of walnuts. The equation to find one pound of walnuts cost is 1.8p = 5.04 1.8p = 5.04 p = 5.04/1.8 p = 2.8 Therefore the cost of one pound of walnuts is $2.8

Question 2. Gabriel wants to solve the equation \(\frac{5}{8}\)m = 25. What step should he do to get m by itself on one side of the equation? Type below: _____________

Explanation: Gabriel wants to solve the equation \(\frac{5}{8}\)m = 25. \(\frac{5}{8}\)m = 25 5m = 25 × 8 5 × m = 200 m = 200/5 = 40 Thus m = 40

Question 3. At top speed, a coyote can run at a speed of 44 miles per hour. If a coyote could maintain its top speed, how far could it run in 15 minutes? ________ miles

Explanation: At top speed, a coyote can run at a speed of 44 miles per hour. Convert from minutes to hour. 60 minutes = 1 hour 15 minutes = 15 × 1/60 = 0.25 = 1/4 44 × 1/4 = 11 miles A coyote can run at a speed of 11 miles for 15 minutes.

Question 4. An online store sells DVDs for $10 each. The shipping charge for an entire order is $5.50. Frank orders d DVDs. Write an expression that represents the total cost of Frank’s DVDs. Type below: _____________

Answer: 10d + $5.50

Explanation: An online store sells DVDs for $10 each. The shipping charge for an entire order is $5.50. Frank orders d DVDs. The expression will be the product of 10 and d more than 5.50 The expression is 10d + $5.50

Question 5. A ring costs $27 more than a pair of earrings. The ring costs $90. Write an equation that can be used to find the cost c in dollars of the earrings. Type below: _____________

Answer: $90 – $27 = c

Explanation: A ring costs $27 more than a pair of earrings. The ring costs $90. c represents the cost in dollars of the earrings. Thus the equation is c + $27 = $90 c = $90 – $27.

Question 6. The equation 3s = 21 can be used to find the number of students s in each van on a field trip. How many students are in each van? ________ students

Answer: 7 students

Explanation: The equation 3s = 21 can be used to find the number of students s in each van on a field trip. 3s = 21 s = 21/3 = 7 s = 7 Thus there are 7 students in each van.

Question 1. Connor ran 3 kilometers in a relay race. His distance represents \(\frac{3}{10}\) of the total distance of the race. The equation \(\frac{3}{10}\)d = 3 can be used to find the total distance d of the race in kilometers. What was the total distance of the race? ________ kilometers

Explanation: Connor ran 3 kilometers in a relay race. His distance represents \(\frac{3}{10}\) of the total distance of the race. \(\frac{3}{10}\)d = 3 3 × d = 3 × 10 3 × d = 30 d = 30/3 = 10 kilometers Therefore the total distance of the race is 10 kilometers.

Question 2. What if Connor’s distance of 3 kilometers represented only \(\frac{2}{10}\) of the total distance of the race. What would the total distance of the race have been? ________ kilometers

Explanation: Connor’s distance of 3 kilometers represented only \(\frac{2}{10}\) of the total distance of the race. \(\frac{2}{10}\) × d = 3 2 × d = 3 × 10 d = 30/2 d = 15 kilometers Therefore the total distance of the race has been 15 kilometers.

Question 3. The lightest puppy in a litter weighs 9 ounces, which is \(\frac{3}{4}\) of the weight of the heaviest puppy. The equation \(\frac{3}{4}\)w = 9 can be used to find the weight w in ounces of the heaviest puppy. How much does the heaviest puppy weigh? ________ ounces

Explanation: The lightest puppy in a litter weighs 9 ounces, which is \(\frac{3}{4}\) of the weight of the heaviest puppy. \(\frac{3}{4}\)w = 9 3 × w = 9 × 4 3 × w = 36 w = 36/3 w = 12 The heaviest puppy weighs 12 ounces.

Question 4. Sophia took home \(\frac{2}{5}\) of the pizza that was left over from a party. The amount she took represents \(\frac{1}{2}\) of a whole pizza. The equation \(\frac{2}{5}\)p = \(\frac{1}{2}\) can be used to find the number of pizzas p left over from the party. How many pizzas were left over? _______ \(\frac{□}{□}\) pizzas

Answer: 1 \(\frac{1}{4}\) pizzas

Explanation: Sophia took home \(\frac{2}{5}\) of the pizza that was left over from a party. The amount she took represents \(\frac{1}{2}\) of a whole pizza. \(\frac{2}{5}\)p = \(\frac{1}{2}\) p = \(\frac{1}{2}\) × \(\frac{5}{2}\) p = \(\frac{5}{4}\) p = 1 \(\frac{1}{4}\) pizzas 1 \(\frac{1}{4}\) pizzas were leftover.

Question 5. A city received \(\frac{3}{4}\) inch of rain on July 31. This represents \(\frac{3}{10}\) of the total amount of rain the city received in July. The equation \(\frac{3}{10}\)r = \(\frac{3}{4}\) can be used to find the amount of rain r in inches the city received in July. How much rain did the city receive in July? _______ \(\frac{□}{□}\) inches of rain

Answer: 2 \(\frac{1}{2}\) inches of rain

Explanation: A city received \(\frac{3}{4}\) inch of rain on July 31. This represents \(\frac{3}{10}\) of the total amount of rain the city received in July. \(\frac{3}{10}\)r = \(\frac{3}{4}\) r = \(\frac{3}{4}\) × \(\frac{10}{3}\) r = \(\frac{30}{12}\) r = \(\frac{5}{2}\) r = 2 \(\frac{1}{2}\) The city received 2 \(\frac{1}{2}\) inches of rain in July.

Question 6. Carole ordered 4 dresses for $80 each, a $25 sweater, and a coat. The cost of the items without sales tax was $430. What was the cost of the coat? $ _______

Explanation: Carole ordered 4 dresses for $80 each, a $25 sweater, and a coat. The cost of the items without sales tax was $430. Cost of 4 dresses is 4 × 80 = $320 $320 + $25 = $345 c + 345 = 430 c = 430 – 345 c = 85 Therefore the cost of the coat is $85

Question 7. A dog sled race is 25 miles long. The equation \(\frac{5}{8}\)k = 25 can be used to estimate the race’s length k in kilometers. Approximately how many hours will it take a dog sled team to finish the race if it travels at an average speed of 30 kilometers per hour? _______ \(\frac{□}{□}\) hours

Answer: 1 \(\frac{1}{3}\) hours

Explanation: A dog sled race is 25 miles long. The equation \(\frac{5}{8}\)k = 25 k represents race length in kilometers. \(\frac{5}{8}\)k = 25 5 × k = 25 × 8 5k = 200 k = 200/5 = 40 k = 40 Average speed is k/30 40/30 = 4/3 The average speed of 30 kilometers per hour is 1 \(\frac{1}{3}\) hours.

Question 8. Explain a Method Explain how you could use the strategy solve a simpler problem to solve the equation \(\frac{3}{4}\)x = \(\frac{3}{10}\). Type below: _____________

Answer: x = \(\frac{2}{5}\)

Explanation: \(\frac{3}{4}\)x = \(\frac{3}{10}\) x = \(\frac{3}{10}\) × \(\frac{4}{3}\) x = \(\frac{12}{30}\) x = \(\frac{2}{5}\)

Question 9. In a basket of fruit, \(\frac{5}{6}\) of the pieces of fruit are apples. There are 20 apples in the display. The equation \(\frac{5}{6}\)f = 20 can be used to find how many pieces of fruit f are in the basket. Use words and numbers to explain how to solve the equation to find how many pieces of fruit are in the basket. _______ pieces of fruit

Explanation: In a basket of fruit, \(\frac{5}{6}\) of the pieces of fruit are apples. There are 20 apples in the display. \(\frac{5}{6}\)f = 20 5 × f = 20 × 6 5 × f = 120 f = 120/5 f = 24 There are 24 pieces of friut in the basket.

Read each problem and solve.

Question 1. Stu is 4 feet tall. This height represents \(\frac{6}{7}\) of his brother’s height. The equation \(\frac{6}{7}\)h = 4 can be used to find the height h, in feet, of Stu’s brother. How tall is Stu’s brother? ______ \(\frac{□}{□}\) feet

Answer: 4 \(\frac{2}{3}\) feet

Explanation: Stu is 4 feet tall. This height represents \(\frac{6}{7}\) of his brother’s height. The equation \(\frac{6}{7}\)h = 4 6/7 × h = 4 6 × h = 4 × 7 6 × h =28 h = 28/6 h = 14/3 h = 4 \(\frac{2}{3}\) feet Thus the height of Stu’s brother in feet is 4 \(\frac{2}{3}\) feet.

Question 2. Bryce bought a bag of cashews. He served \(\frac{7}{8}\) pound of cashews at a party. This amount represents \(\frac{2}{3}\) of the entire bag. The equation \(\frac{2}{3}\)n = \(\frac{7}{8}\) can be used to find the number of pounds n in a full bag. How many pounds of cashews were in the bag that Bryce bought? ______ \(\frac{□}{□}\) pounds

Answer: 1 \(\frac{5}{16}\)

Explanation: Bryce bought a bag of cashews. He served \(\frac{7}{8}\) pound of cashews at a party. This amount represents \(\frac{2}{3}\) of the entire bag. \(\frac{2}{3}\)n = \(\frac{7}{8}\) n = \(\frac{7}{8}\) × \(\frac{3}{2}\) n = \(\frac{21}{16}\) n = 1 \(\frac{5}{16}\) Bryce bought 1 \(\frac{5}{16}\) pounds of cashews were in the bag.

Question 3. In Jaime’s math class, 9 students chose soccer as their favorite sport. This amount represents \(\frac{3}{8}\) of the entire class. The equation \(\frac{3}{8}\)s = 9 can be used to find the total number of students s in Jaime’s class. How many students are in Jaime’s math class? ______ students

Answer: 24 students

Explanation: In Jaime’s math class, 9 students chose soccer as their favorite sport. This amount represents \(\frac{3}{8}\) of the entire class. \(\frac{3}{8}\)s = 9 3 × s = 9 × 8 3 × s = 72 s = 72/3 s = 24 students 24 students are in Jaime’s math class.

Question 4. Write a math problem for the equation \(\frac{3}{4}\)n = \(\frac{5}{6}\). Then solve a simpler problem to find the solution. Type below: _____________

Answer: 1 \(\frac{1}{9}\)

Explanation: \(\frac{3}{4}\)n = \(\frac{5}{6}\) n = \(\frac{5}{6}\) × \(\frac{4}{3}\) n = \(\frac{20}{18}\) n = \(\frac{10}{9}\) n = 1 \(\frac{1}{9}\)

Question 1. Roger served \(\frac{5}{8}\) pound of crackers, which was \(\frac{2}{3}\) of the entire box. What was the weight of the crackers originally in the box? \(\frac{□}{□}\) pounds

Answer: \(\frac{15}{16}\) pounds

Explanation: Roger served \(\frac{5}{8}\) pound of crackers, which was \(\frac{2}{3}\) \(\frac{2}{3}\) × p = \(\frac{5}{8}\) p = \(\frac{5}{8}\) × \(\frac{3}{2}\) p = \(\frac{15}{16}\) pounds \(\frac{15}{16}\) was the weight of the crackers originally in the box.

Question 2. Bowser ate 4 \(\frac{1}{2}\) pounds of dog food. That amount is \(\frac{3}{4}\) of the entire bag of dog food. How many pounds of dog food were originally in the bag? ______ pounds

Answer 6 pounds

Explanation: Bowser ate 4 \(\frac{1}{2}\) pounds of dog food. That amount is \(\frac{3}{4}\) of the entire bag of dog food. 4 \(\frac{1}{2}\) = \(\frac{9}{2}\) \(\frac{3}{4}\) p = \(\frac{9}{2}\) p = \(\frac{9}{2}\) × \(\frac{4}{3}\) p = 6 pounds 6 pounds of dog food were originally in the bag.

Question 3. What is the quotient 4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\) _______ \(\frac{□}{□}\)

Explanation: 4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\) \(\frac{14}{3}\) ÷ \(\frac{21}{5}\) = \(\frac{70}{63}\) The mixed fraction of \(\frac{70}{63}\) is 1 \(\frac{1}{9}\) 4 \(\frac{2}{3}\) ÷ 4 \(\frac{1}{5}\) = 1 \(\frac{1}{9}\)

Question 4. Miranda had 4 pounds, 6 ounces of clay. She divided it into 10 equal parts. How heavy was each part? _______ ounces

Answer: 7 ounces

Explanation: Miranda had 4 pounds, 6 ounces of clay. She divided it into 10 equal parts. Convert from pounds to ounces We know that 1 pound = 16 ounces 4 pounds = 4 × 16 ounces = 64 ounces 64 ounces + 6 ounces = 70 ounces Now divide 70 ounces into 10 equal parts. 70 ÷ 10 = 7 ounces. Thus each part was 7 ounces.

Question 5. The amount Denise charges to repair computers is $50 an hour plus a $25 service fee. Write an expression to show how much she will charge for h hours of work. Type below: _____________

Answer: 50h + 25

Explanation: The amount Denise charges to repair computers is $50 an hour plus a $25 service fee. The expression will be product of 50 and h more than 25. The expression is 50h + 25.

Question 6. Luis has saved $14 for a skateboard that costs $52. He can use the equation 14 + m = 52 to find how much more money m he needs. How much more does he need? $ _______

Explanation: Luis has saved $14 for a skateboard that costs $52. He can use the equation 14 + m = 52 14 + m = 52 m = 52 – 14 m = 38 He needs $38 more.

Question 1. A(n) _____ is a statement that two mathematical expressions are equal. Type below: _____________

Answer: An equation is a mathematical statement that two expressions are equal.

Question 2. Adding 5 and subtracting 5 are _____. Type below: _____________

Answer: Solution of an equation.

Concepts and Skills

Question 3. The sum of a number and 4.5 is 8.2. Type below: _____________

Answer: The phrase “sum” indicates an addition operation. So, the equation is n + 4.5 = 8.2

Question 4. Three times the cost is $24. Type below: _____________

Answer: The phrase “times” indicates multiplication. Multiply 3 with c. 3c = 24

Question 5. x − 24 = 58; x = 82 The variable is _____________

Explanation: 82 – 24 = 58 58 = 58 Thus the variable is a solution.

Question 6. \(\frac{1}{3}\)c = \(\frac{3}{8}\), c = \(\frac{3}{4}\) The variable is _____________

Explanation: \(\frac{1}{3}\)c = \(\frac{3}{8}\) c = \(\frac{3}{4}\) \(\frac{1}{3}\) × \(\frac{3}{4}\) = \(\frac{3}{8}\) \(\frac{3}{12}\) ≠ \(\frac{3}{8}\)

Question 7. a + 2.4 = 7.8 a = _____

Answer: 5.4

Explanation: Given the equation is a + 2.4 = 7.8 a + 2.4 = 7.8 a = 7.8 – 2.4 a = 5.4

Question 8. \(b-\frac{1}{4}=3 \frac{1}{2}\) b = _______ \(\frac{□}{□}\)

Explanation: Given the equation is \(b-\frac{1}{4}=3 \frac{1}{2}\) b – \(\frac{1}{4}\) = 3 \(\frac{1}{2}\) b = 3 \(\frac{1}{2}\) + \(\frac{1}{4}\) b = 3 + \(\frac{1}{4}\) + \(\frac{1}{2}\) b = 3 \(\frac{3}{4}\)

Question 9. 3x = 27 x = _______

Explanation: Given the equation is 3x = 27 x = 27/3 x = 9

Question 10. \(\frac{1}{3} s=\frac{1}{5}\) s = \(\frac{□}{□}\)

Answer: \(\frac{3}{5}\)

Explanation: Given the equation is \(\frac{1}{3} s=\frac{1}{5}\) \(\frac{1}{3}\)s = \(\frac{1}{5}\) s = \(\frac{3}{5}\)

Question 11. \(\frac{t}{4}\) = 16 t = _______

Explanation: Given the equation is \(\frac{t}{4}\) = 16 t = 16 × 4 t = 64

Question 12. \(\frac{w}{7}\) = 0.3 w = _______

Answer: 2.1

Explanation: \(\frac{w}{7}\) = 0.3 w/7 = 0.3 w = 0.3 × 7 w = 2.1

Page No. 464

Question 13. A stadium has a total of 18,000 seats. Of these, 7,500 are field seats, and the rest are grandstand seats. Write an equation that could be used to find the number of grandstand seats s. Type below: _____________

Answer: s + 7500 = 18000

Explanation: A stadium has a total of 18,000 seats. Of these, 7,500 are field seats, and the rest are grandstand seats. Let s be the number of grandstand seats. s + 7,500 = 18,000

Question 14. Aaron wants to buy a bicycle that costs $128. So far, he has saved $56. The equation a + 56 = 128 can be used to find the amount a in dollars that Aaron still needs to save. What is the solution of the equation? The solution is _______

Explanation: Aaron wants to buy a bicycle that costs $128. So far, he has saved $56. The equation a + 56 = 128 a = 128 – 56 a = 72 The solution of the equation a + 56 = 128 is 72.

Question 15. Ms. McNeil buys 2.4 gallons of gasoline. The total cost is $7.56. Write and solve an equation to find the price p in dollars of one gallon of gasoline. $ _______

Answer: $3.15

Explanation: Ms. McNeil buys 2.4 gallons of gasoline. The total cost is $7.56. 2.4p = 7.56 p = 7.56/2.4 p = $3.15 The price of one gallon of gasoline is $3.15

Question 16. Crystal is picking blueberries. So far, she has filled \(\frac{2}{3}\) of her basket, and the blueberries weigh \(\frac{3}{4}\) pound. The equation \(\frac{2}{3}\)w = \(\frac{3}{4}\) can be used to estimate the weight w in pounds of the blueberries when the basket is full. About how much will the blueberries in Crystal’s basket weigh when it is full? ______ \(\frac{□}{□}\) pounds

Answer: 1 \(\frac{1}{8}\) pounds

Explanation: Crystal is picking blueberries. So far, she has filled \(\frac{2}{3}\) of her basket, and the blueberries weigh \(\frac{3}{4}\) pound. The equation \(\frac{2}{3}\)w = \(\frac{3}{4}\) w = \(\frac{3}{4}\) × \(\frac{3}{2}\) w = \(\frac{9}{8}\) The mixed fraction of \(\frac{9}{8}\) is 1 \(\frac{1}{8}\) pounds

Determine whether the given value of the variable is a solution of the inequality.

Question 1. a ≥ −6, a = −3 The variable is _____________

Explanation: Substitute the solution a in the inequality. a = -3 -3 ≥ -6 -3 is greater than -6 Thus the variable is a solution.

Question 2. y < 7.8, y = 8 The variable is _____________

Explanation: Substitute the solution y in the inequality. y = 8 8 is less than 7.8 8<7.8 The variable is not the solution.

Question 3. c > \(\frac{1}{4}\), c = \(\frac{1}{5}\) The variable is _____________

Explanation: Substitute the solution c in the inequality. c = \(\frac{1}{5}\) \(\frac{1}{5}\) > \(\frac{1}{4}\) \(\frac{1}{5}\) is greater than \(\frac{1}{4}\) \(\frac{1}{5}\) > \(\frac{1}{4}\) Thus the variable is a solution.

Question 4. x ≤ 3, x = 3 The variable is _____________

Explanation: Substitute the solution x in the inequality. x = 3 3 ≤ 3 3 is less than or equal to 3. Thus the variable is a solution.

Question 5. d < – 0.52, d = – 0.51 The variable is _____________

Explanation: Substitute the solution d in the inequality. -0.51 < -0.52 -0.51 is greater than -0.52 The variable is not the solution.

Question 6. t ≥ \(\frac{2}{3}\), t = \(\frac{3}{4}\) The variable is _____________

Explanation: Substitute the solution t in the inequality. t = \(\frac{3}{4}\) \(\frac{3}{4}\) ≥ \(\frac{2}{3}\) \(\frac{3}{4}\) is greater than \(\frac{2}{3}\) Thus the variable is a solution.

Practice: Copy and Solve Determine whether s = \(\frac{3}{5}\), s = 0, or s = 1.75 are solutions of the inequality.

Question 7. s > – 1 Type below: _____________

Answer: s > – 1 s = \(\frac{3}{5}\) \(\frac{3}{5}\) > -1 \(\frac{3}{5}\) is greater than -1. The variable is the solution. s = 0 0 > -1 0 is greater than -1 Thus the variable is a solution. s = 1.75 1.75 > -1 1.75 is greater than -1 s > -1 Thus the variable is a solution.

Question 8. s ≤ 1 \(\frac{2}{3}\) Type below: _____________

Answer: s ≤ 1 \(\frac{2}{3}\) s = \(\frac{3}{5}\) \(\frac{3}{5}\) ≤ 1 \(\frac{2}{3}\) \(\frac{3}{5}\) is less than but not equal to 1 \(\frac{2}{3}\) The variable is not the solution. s ≤ 1 \(\frac{2}{3}\) s = 0 0 ≤ 1 \(\frac{2}{3}\) The variable is not the solution. s = 1.75 1.75 ≤ 1 \(\frac{2}{3}\) The variable is not the solution.

Question 9. s < 0.43 Type below: _____________

Answer: s < 0.43 \(\frac{3}{5}\) < 0.43 \(\frac{3}{5}\) = 0.6 0.6 is not less than 0.43 Thus the variable is not the solution. s = 0 0 < 0.43 0 is less than 0.43 Thus the variable is the solution. s = 1.75 1.75 < 0.43 1.75 is greater than 0.43 Thus the variable is not the solution.

Give two solutions of the inequality.

Question 10. e < 3 Type below: _____________

Answer: The solution to the inequality must be whole numbers less than 3. e = 1 and 2 are the solutions because 1 and 2 are less than 3. Thus the 2 solutions are 1 and 2.

Question 11. p > – 12 Type below: _____________

Answer: The solution to the inequality must be whole numbers greater than -12 p = 0 and -5 are the solutions because 0 and -5 are greater than -12. Thus the 2 solutions are 0 and -5.

Question 12. y ≥ 5.8 Type below: _____________

Answer: The solution to the inequality must be whole numbers greater than or equal to 5.8 y = 5.8 and 5.9 are the solutions because 5.8 and 5.9 greater than or equal to 5.8 Thus the 2 solutions are 5.8 and 5.9

Question 13. Connect Symbols and Words A person must be at least 18 years old to vote. The inequality a ≥ 18 represents the possible ages a in years at which a person can vote. Determine whether a = 18, a = 17\(\frac{1}{2}\), and a = 91.5 are solutions of the inequality, and tell what the solutions mean. Type below: _____________

Answer: a ≥ 18 Substitute the values of a in the inequality a = 18 18 ≥ 18 Thus the variable is the solution. a = 17\(\frac{1}{2}\) 17\(\frac{1}{2}\) ≥ 18 17\(\frac{1}{2}\) is less than 18. The variable is not the solution. a = 91.5 91.5 > 18 The solution is mean.

Problem Solving + Applcations – Page No. 468

Question 14. The inequality p < 4.75 represents the prices p in dollars that Paige is willing to pay for popcorn. The inequality p < 8.00 represents the prices p in dollars that Paige is willing to pay for a movie ticket. At how many theaters would Paige be willing to buy a ticket and popcorn? ______ theater

Explanation: The inequality p < 4.75 represents the prices p in dollars that Paige is willing to pay for popcorn. The inequality p < 8.00 represents the prices p in dollars that Paige is willing to pay for a movie ticket. From the above table, we can see that there is the only theatre with 8.00 and 4.75 So, Paige is willing to buy a ticket and popcorn from 1 theatre.

Question 15. Sense or Nonsense? Edward says that inequality d ≥ 4.00 represents the popcorn prices in the table, where d is the price of popcorn in dollars. Is Edward’s statement sense or nonsense? Explain. Type below: _____________

Answer: Edward’s statement makes sense because all of the popcorn prices in the table are greater than or equal to $4.00.

Question 16. Use Math Vocabulary Explain why the statement t > 13 is an inequality. Type below: _____________

Answer: The statement is equality because it compares two amounts t and 13 using an inequality symbol.

Question 17. The minimum wind speed for a storm to be considered a hurricane is 74 miles per hour. The inequality w ≥ 74 represents the possible wind speeds of a hurricane. Two possible solutions for the inequality w ≥ 74 are _____ and _____. Two possible solutions for the inequality w ≥ 74 are _____ and _____

Answer: 75 and 80

Explanation: Given that w is greater than or equal to 74. The two possible solutions for the inequality w ≥ 74 are 75 and 80.

Question 1. s ≥ – 1, s = 1 The variable is _____________

Explanation: The inequality is s ≥ – 1 s = 1 1 ≥ – 1 1 is a positive number so 1 will be greater than or equal to -1 Thus the variable is a solution.

Question 2. p < 0, p = 4 The variable is _____________

Explanation: The inequality is p < 0 Given p = 4 Substitute p = 4 in the inequality. 4 < 0 4 is not less than 0 Thus the variable is not a solution.

Question 3. y ≤ – 3, y = – 1 The variable is _____________

Explanation: The inequality is y ≤ – 3 y = -1 -1 ≤ – 3 – 1 is greater than -3 Thus the variable is not a solution.

Question 4. u > \(\frac{-1}{2}\), u = 0 The variable is _____________

Explanation: The inequality is u > \(\frac{-1}{2}\) u = 0 0 > \(\frac{-1}{2}\) 0 is greater than \(\frac{-1}{2}\) Thus the variable is a solution.

Question 5. q ≥ 0.6, q = 0.23 The variable is _____________

Explanation: The inequality is q ≥ 0.6 q = 0.23 0.23 is less than 0.6 Thus the variable is a solution.

Question 6. b < 2 \(\frac{3}{4}\), b = \(\frac{2}{3}\) The variable is _____________

Explanation: The inequality is b < 2 \(\frac{3}{4}\) b = \(\frac{2}{3}\) \(\frac{2}{3}\) < 2 \(\frac{3}{4}\) \(\frac{2}{3}\) is less than 2 \(\frac{3}{4}\) Thus the variable is a solution.

Question 7. k < 2 Type below: _____________

Answer: k = 0 and 1 because they are less than 2. Thus the two possible inequalities for k < 2 are 0 and 1.

Question 8. z ≥ – 3 Type below: _____________

Answer: z = -3 and -2 because -3 and -2 are greater than or equal to -3 Thus the two solutions of the inequality are -3 and -2

Question 9. f ≤ – 5 Type below: _____________

Answer: f = -5 and -6 because -5 and -6 are less than or equal to -5 Thus the two solutions of the inequality are -5 and -6.

Question 10. The inequality s ≥ 92 represents the score s that Jared must earn on his next test to get an A on his report card. Give two possible scores that Jared could earn to get the A. Type below: _____________

Answer: Two possible scores that Jared could earn to get the A are 92 and 100.

Question 11. The inequality m ≤ $20 represents the amount of money that Sheila is allowed to spend on a new hat. Give two possible money amounts that Sheila could spend on the hat. Type below: _____________

Answer: Two possible money amounts that Sheilla could spend on the hat are $15 or $10.

Question 12. Describe a situation and write an inequality to represent the situation. Give a number that is a solution and another number that is not a solution of the inequality. Type below: _____________

Answer: In the United States, the minimum age required to run for president is 35. This can be represented by the inequality a ≥ 35. A number that is a solution is 55 and a number that is not a solution is 29.

Question 1. Three of the following are solutions of g < – 1\(\frac{1}{2}\). Which one is not a solution? g = – 4     g = – 7\(\frac{1}{2}\)   g = 0    g = – 2\(\frac{1}{2}\) Type below: _____________

Answer: g = 0

Explanation: g < – 1\(\frac{1}{2}\). g = – 4 -4 < – 1\(\frac{1}{2}\) g = – 7\(\frac{1}{2}\) – 7\(\frac{1}{2}\) < – 1\(\frac{1}{2}\). g = – 2\(\frac{1}{2}\) – 2\(\frac{1}{2}\) < – 1\(\frac{1}{2}\) g = 0 0 < – 1\(\frac{1}{2}\) Thus 0 is not the solution.

Question 2. The inequality w ≥ 3.2 represents the weight of each pumpkin, in pounds, that is allowed to be picked to be sold. The weights of pumpkins are listed. How many pumpkins can be sold? Which pumpkins can be sold? 3.18 lb, 4 lb, 3.2 lb, 3.4 lb, 3.15 lb Type below: _____________

Answer: 3.2 lb, 3.4 lb

Explanation: The inequality w ≥ 3.2 represents the weight of each pumpkin, in pounds, that is allowed to be picked to be sold. Substitute the solutions in the inequality. w = 3.18 3.18 ≥ 3.2 3.18 is less than 3.2 3.18 < 3.2 lb w = 4 lb 4 ≥ 3.2 4 is greater than 3.2 4 > 3.2 w = 3.2 lb 3.2 ≥ 3.2 3.2 lb is greater than 3.2 lb w = 3.4 lb 3.4 ≥ 3.2 3.4 lb is greater than 3.2 lb w = 3.15 lb 3.15 < 3.2 Thus 3.2 lb, 3.4 lb pumpkins can be sold.

Question 3. What is the value of 8 + (27 ÷ 9) 2 ? _______

Explanation: 8 + (27 ÷ 9) 2 ? 8 + (3) 2 8 + 9 = 17

Question 4. Write an expression that is equivalent to 5(3x + 2z). Type below: _____________

Answer: 15x + 10z

Explanation: 5(3x + 2z) 5 × 3x + 5 × 2z 15x + 10z The expression equivalent to 5(3x + 2z) is 15x + 10z

Question 5. Tina bought a t-shirt and sandals. The total cost was $41.50. The t-shirt cost $8.95. The equation 8.95 + c = 41.50 can be used to find the cost c in dollars of the sandals. How much did the sandals cost? $ _______

Answer: $32.55

Explanation: Tina bought a t-shirt and sandals. The total cost was $41.50. The t-shirt cost $8.95. The equation is 8.95 + c = 41.50 c = 41.50 – 8.95 c = $32.55 The cost of the sandal is 32.55

Question 6. Two-thirds of a number is equal to 20. What is the number? _______

Explanation: 2/3 × n = 20 n = 3/2 × 20 n =  3 × 10 n = 30 The number is 30.

Write an inequality for the word sentence. Tell what type of numbers the variable in the inequality can represent.

Question 1. The elevation e is greater than or equal to 15 meters. Type below: _____________

Answer: The phrase greater than or equal to represents “≥” Thus the inequality is e ≥ 15

Question 2. A passenger’s age a must be more than 4 years. Type below: _____________

Answer: The phrase more than represents the greater than symbol “>” Thus the inequality is a > 4

Write a word sentence for the inequality.

Question 3. b < \(\frac{1}{2}\) Type below: _____________

Answer: By seeing the above inequality we can write the word sentence for inequality as, b is less than \(\frac{1}{2}\)

Question 4. m ≥ 55 Type below: _____________

Answer: By seeing the above inequality we can write the word sentence for inequality as, m is greater than or equal to 55.

Question 5. Compare Explain the difference between t ≤ 4 and t < 4. Type below: _____________

Answer: t ≤ 4 is t is less than or equal to 4 which means t is equal to 4 or 3.9. t < 4 is t is less than 4 which means t is equal to 3, 2, or 1 or 0.

Question 6. A children’s roller coaster is limited to riders whose height is at least 30 inches and at most 48 inches. Write two inequalities that represent the height h of riders for the roller coaster. Type below: _____________

Answer: h represents the height of riders for the roller coaster. A children’s roller coaster is limited to riders whose height is at least 30 inches and at most 48 inches. ar least 30 inches means h must be greater than or equal to 30 inches. i.e., h ≥ 30 inches at most 48 inches means h must be less than 48 inches. i.e., h < 48 inches

The reading skill make generalizations can help you write inequalities to represent situations. A generalization is a statement that is true about a group of facts.

Question 8. Write two inequalities that represent generalizations about the sea otter weights. Type below: _____________

Answer: First, list the weights in pounds in order from least to greatest. 50, 51, 54, 58, 61, 61, 62, 62, 66, 68, 69, 71 Next, write an inequality to describe the weights by using the least weight on the list. Let w represent weights of the otters in the pounds. The least weight is 50 pounds, so all of the weights are greater than or equal to 50 pounds. w ≥ 50 Now write an inequality to describe the weights by using the greatest weights in the list. The greatest weight is 71 pounds, so all of the weights are less than or equal to 71 pounds. w ≤ 71

Answer: First, list the number of pups in order from least to greatest. 6, 6, 7, 10, 15, 16, 20, 23 Next, write an inequality to describe the number of pups by using the least number of pups on the list. Let n represent the number of pups. The least weight is 6 pups. So all of the pups will be greater than or equal to 6. n ≥ 6 Now write an inequality to describe the number of pups by using the greatest weights in the list. The greatest weight is 23 pups so all of the weights are less than or equal to 23 pups. n ≤ 23 pups

Question 1. The width w is greater than 4 centimeters. Type below: _____________

Answer: The inequality symbol for “greater than” is >. w > 4, where w is the width in centimeters. w is a positive number.

Question 2. The score s in a basketball game is greater than or equal to 10 points Type below: _____________

Answer: The inequality symbol for “greater than or equal to” is ≥. s ≥ 10, where s is the score in the basketball game. s is a positive number.

Question 3. The mass m is less than 5 kilograms Type below: _____________

Answer: The inequality symbol for “less than” is <. m < 5, where m is the mass in kilograms. m is a positive number.

Question 4. The height h is greater than 2.5 meters Type below: _____________

Answer: The inequality symbol for “greater than” is >. h > 2.5, where h is the height in meters. h is a positive number.

Question 5. The temperature t is less than or equal to −3°. Type below: _____________

Answer: The inequality symbol for “less than or equal to” is ≤. t ≤  −3° where t is the temperature in degrees. t is a negative number.

Question 6.4 k < – 7 Type below: _____________

Answer: The word sentence for the inequality is k is less than -7.

Question 7. z ≥ 2 \(\frac{3}{5}\) Type below: _____________

Answer: The word sentence for the inequality is z is greater than or equal to 2 \(\frac{3}{5}\).

Question 8. Tabby’s mom says that she must read for at least 30 minutes each night. If m represents the number of minutes reading, what inequality can represent this situation? Type below: _____________

Answer: m ≥ 30

Explanation: Tabby’s mom says that she must read for at least 30 minutes each night. m represents the number of minutes of reading. m is greater than or equal to 30. Thus the inequality is m ≥ 30.

Question 9. Phillip has a $25 gift card to his favorite restaurant. He wants to use the gift card to buy lunch. If c represents the cost of his lunch, what inequality can describe all of the possible amounts of money, in dollars, that Phillip can spend on lunch? Type below: _____________

Answer: c ≤ 25

Explanation: Phillip has a $25 gift card to his favorite restaurant. He wants to use the gift card to buy lunch. c represents the cost of his lunch c is less than or equal to 25. Thus the inequality is c ≤ 25.

Question 10. Write a short paragraph explaining to a new student how to write an inequality. Type below: _____________

Answer: Inequality is a statement that two quantities are not equal. To know which direction to shade a graph, I write inequalities with the variable on the left side of the inequality symbol. I know that the symbol has to point to the same number after I rewrite the inequality. For example, I write 4 < y as y > 4 Now the inequality symbol points in the direction that I should draw the shaded arrow on my graph.

Question 1. At the end of the first round in a quiz show, Jeremy has at most −20 points. Write an inequality that means “at most −20”. Type below: _____________

Answer: The phrase at most refers to less than or equal to. Thus the inequality is J ≤ -20

Question 2. Describe the meaning of y ≥ 7.9 in words. Type below: _____________

Answer: y ≥ 7.9 means y is greater than or equal to 7.9

Question 3. Let y represent Jaron’s age in years. If Dawn were 5 years older, she would be Jaron’s age. Which expression represents Dawn’s age? Type below: _____________

Answer: y – 5

Explanation: Let y represent Jaron’s age in years. If Dawn were 5 years older, she would be Jaron’s age. We have to subtract 5 years to know the age of Jaron. Thus the expression is y – 5.

Question 4. Simplify the expression 7 × 3g. Type below: _____________

Answer: 21g

Question 5. What is the solution of the equation 8 = 8f? f = ________

Answer: 8 = 8f f = 8/8 = 1 f = 1 The solution for the equation 8 = 8f is 1.

Question 6. Which of the following are solutions of the inequality k ≤ – 2? k = 0   k = – 2   k = – 4   k = 1   k = – 1 \(\frac{1}{2}\) Type below: _____________

Answer: k = -2 k = -4

Explanation: k = 0 in the inequality k ≤ – 2 0 ≤ – 2 0 is less than but not equal to -2 Thus 0 is not the solution. k = – 2 k ≤ – 2 -2 ≤ – 2 Thus -2 is the solution. k = – 4 k ≤ – 2 -4 ≤ – 2 Thus -4 is the solution. k = 1 1 ≤ – 2 1 ≤ – 2 1 is greater than but not equal to -2 Thus 1 is not the solution. k = – 1 \(\frac{1}{2}\) – 1 \(\frac{1}{2}\) ≤ – 2 – 1 \(\frac{1}{2}\) ≤ – 2 – 1 \(\frac{1}{2}\) is less than but not equal to -2 Thus – 1 \(\frac{1}{2}\) is not the solution.

Graph the inequality.

Question 1. m < 15 Type below: _____________

Question 2. c ≥ – 1.5 Type below: _____________

Question 3. b ≤ \(\frac{5}{8}\) Type below: _____________

Practice: Copy and Solve Graph the inequality.

Question 4. a < \(\frac{2}{3}\) Type below: _____________

Question 5. x > – 4 Type below: _____________

Question 6. k ≥ 0.3 Type below: _____________

Question 7. t ≤ 6 Type below: _____________

Write the inequality represented by the graph.

Answer: m < 6

Answer: n ≥ -7

Question 10. Model Mathematics The inequality w ≥ 60 represents the wind speed w in miles per hour of a tornado. Graph the solutions of the inequality on the number line. Type below: _____________

Question 11. Graph the solutions of the inequality c < 12 ÷ 3 on the number line Type below: _____________

Question 12. Write an inequality representing t, the heights in inches of people who can go on Twirl & Whirl. Type below: _____________

Answer: The minimum height of people who can go on Twirl and Whirl is 48 inches. So, inequality is t ≥ 48.

Question 13. Graph your inequality from Exercise 12. Type below: _____________

Answer: Draw a full circle at 48 to show that 48 is a solution. Shade to the right of 48 to show that values greater than or equal to 48 are solutions.

Question 14. Write an inequality representing r, the heights in inches of people who can go on Race Track. Type below: _____________

Answer: The minimum height of people who can go on Race track is 24 inches. So, the inequality is r ≥ 42.

Question 15. Graph your inequality from Exercise 14. Type below: _____________

Answer: Draw a full circle at 42 to show that 42 is a solution. Shade to the right of 42 to show that values greater than or equal to 48 are solutions.

Question 16. Write an inequality representing b, the heights in inches of people who can go on both River Rapids and Mighty Mountain. Explain how you determined your answer. Type below: _____________

Answer: You need to be at least 38 inches tall to go on River Rapids and at least 44 inches tall to go on Mighty mountain. So, you need to be at least 44 inches tall to go on both rides. The inequality is b ≥ 44.

Answer: Yes I agree with Darius. That dark circle and the arrow to the left indicates that c ≤ 25

Question 1. h ≥ 3 Type below: _____________

Question 2. x < \(\frac{-4}{5}\) Type below: _____________

Question 3. y > – 2 Type below: _____________

Question 4. n ≥ 1 \(\frac{1}{2}\) Type below: _____________

Question 5. c ≤ – 0.4 Type below: _____________

Answer: n > 3

Answer: n > -5

Question 8. The inequality x ≤ 2 represents the elevation x of a certain object found at a dig site. Graph the solutions of the inequality on the number line. Type below: _____________

Question 9. The inequality x ≥ 144 represents the possible scores x needed to pass a certain test. Graph the solutions of the inequality on the number line. Type below: _____________

Question 10. Write an inequality and graph the solutions on a number line. Type below: _____________

Answer: x ≥ -2 The number line at right shows the solutions of the inequality x ≥ -2

Question 2. Describe the graph of g < 0.6. Type below: _____________

Question 3. Write an expression that shows the product of 5 and the difference of 12 and 9. Type below: _____________

Answer: The equation for the product of 5 and the difference of 12 and 9 5 × 12 – 9 The equation is 5(12 – 9).

Question 4. What is the solution of the equation 8.7 + n = 15.1? n = ________

Answer: 6.4

Explanation: The equation is 8.7 + n = 15.1 n + 8.7 = 15.1 n = 15.1 – 8.7 n = 6.4

Question 5. The equation 12x = 96 gives the number of egg cartons x needed to package 96 eggs. Solve the equation to find the number of cartons needed. ________ cartons

Explanation: Given, The equation 12x = 96 gives the number of egg cartons x needed to package 96 eggs. 12x = 96 x = 96/12 = 8 Thus 8 number of cartons are needed.

Question 6. The lowest price on an MP3 song is $0.35. Write an inequality that represents the cost c of an MP3 song. Type below: _____________

Answer: Given that, The lowest price on an MP3 song is $0.35. c ≥ 0.35 That is an inequality to represent the cost of an MP3 song.

Question 1. For numbers 1a–1c, choose Yes or No to indicate whether the given value of the variable is a solution of the equation. 1a. \(\frac{2}{5}\)v=10; v = 25 1b. n + 5 = 15; n = 5 1c. 5z = 25; z = 5 1a. _____________ 1b. _____________ 1c. _____________

Answer: 1a. \(\frac{2}{5}\)v=10; v = 25 \(\frac{2}{5}\) × 25=10 2 × 5 = 10 10 = 10 The variable is a solution. Thus the answer is yes. 1b. n + 5 = 15; n = 5 Substitute n = 5 5 + 5 = 15 10 ≠ 15 The variable is not a solution. The answer is no. 1c. 5z = 25; z = 5 Substitute z = 5 5 × 5 = 25 25 = 25 The variable is a solution. Thus the answer is yes.

Question 2. The distance from third base to home plate is 88.9 feet. Romeo was 22.1 feet away from third base when he was tagged out. The equation 88.9 − t = 22.1 can be used to determine how far he needed to run to get to home plate. Using substitution, the coach determines that Romeo needed to run _____ feet to get to home plate. Using substitution, the coach determines that Romeo needed to run _____________ feet to get to home plate

Answer: 66.8 feet

Explanation: The distance from third base to home plate is 88.9 feet. Romeo was 22.1 feet away from third base when he was tagged out. The equation is 88.9 − t = 22.1 88.9 − t = 22.1 88.9 – 22.1 = t t = 66.8 feet Thus Using substitution, the coach determines that Romeo needed to run 66.8 feet to get to the home plate.

Question 3. There are 84 grapes in a bag. Four friends are sharing the grapes. Write an equation that can be used to find out how many grapes g each friend will get if each friend gets the same number of grapes. Type below: _____________

Answer: 84 = 4g 84 is the total amount of grapes 4 is the number of friends g = how many grapes each friend will get

Chapter 8 Review/Test Page No. 484

Question 5. Frank’s hockey team attempted 15 more goals than Spencer’s team. Frank’s team attempted 23 goals. Write and solve an equation that can be used to find how many goals Spencer’s team attempted. ______ goals

Answer: 8 goals

Explanation: Frank’s hockey team attempted 15 more goals than Spencer’s team. Frank’s team attempted 23 goals. Let x be the Spencer’s team The phrase more than indicates addition operation. x + 15 = 23 x = 23 – 15 x = 8 goals

Answer: y = 7

• Draw 11 rectangles on your MathBoard to represent the two sides of the equation.
• Use algebra tiles to model the equation. Model y + 10 in the left rectangle, and model 17 in the right rectangle.
• To solve the equation, get the y tile by itself on one side. If you remove a tile from one side, you can keep the two sides equal by removing the same type of tile from the other side.
• Remove ten 1 tiles on the left side and ten 1 tiles on the right side.

Thus 10 + y = 17 y = 17 – 10 = 7 y = 7

Question 7. Gabriella and Max worked on their math project for a total of 6 hours. Max worked on the project for 2 hours by himself. Solve the equation x + 2 = 6 to find out how many hours Gabriella worked on the project. ______ hours

Answer: 4 hours

Explanation: Gabriella and Max worked on their math project for a total of 6 hours. Max worked on the project for 2 hours by himself. x + 2 = 6 x = 6 – 2 x = 4 Gabriella worked 4 hours on the project.

Question 8. Select the equations that have the solution m = 17. Mark all that apply. Options: a. 3 + m = 21 b. m − 2 = 15 c. 14 = m − 3 d. 2 = m − 15

Answer: B, C, D

Explanation: a. 3 + m = 21 3 + 17 = 21 20 ≠ 21 b. m − 2 = 15 17 – 2 = 15 15 = 15 c. 14 = m − 3 14 = 17 – 3 14 = 14 d. 2 = m − 15 2 = 17 – 15 2 = 2 Thus the correct answers are B, C and D.

Question 9. Describe how you could use algebra tiles to model the equation 4x = 20. Type below: _____________

Question 10. For numbers 10a–10d, choose Yes or No to indicate whether the equation has the solution x = 12. 10a. \(\frac{3}{4}\)x = 9 10b. 3x = 36 10c. 5x = 70 10d. \(\frac{x}{3}\) = 4 10a. _____________ 10b. _____________ 10c. _____________ 10d. _____________

Answer: 10a. Yes 10b. Yes 10c. No 10d. Yes

Explanation: 10a. \(\frac{3}{4}\)x = 9 \(\frac{3}{4}\) × 12 = 9 3 × 3 = 9 9 = 9 Thus the answer is yes. 10b. 3x = 36 x = 12 3 × 12 = 36 36 = 36 Thus the answer is yes. 10c. 5x = 70 x = 12 5 × 12 = 70 60 ≠ 70 Thus the answer is no. 10d. \(\frac{x}{3}\) = 4 x/3 = 4 x = 4 × 3 x = 12 Thus the answer is yes.

Question 11. Bryan rides the bus to and from work on the days he works at the library. In one month, he rode the bus 24 times. Solve the equation 2x = 24 to find the number of days Bryan worked at the library. Use a model. Type below: _____________

Question 12. Betty needs \(\frac{3}{4}\) of a yard of fabric to make a skirt. She bought 9 yards of fabric. Part A Write and solve an equation to find how many skirts x she can make from 9 yards of fabric. ________ skirts

Answer: 12 skirts

Explanation: Betty needs \(\frac{3}{4}\) of a yard of fabric to make a skirt. She bought 9 yards of fabric. x × \(\frac{3}{4}\) = 9 x = 9 × \(\frac{4}{3}\) x = 3 × 4 = 12 x = 12 she can make 12 skirts from 9 yards of fabric.

Question 12. Part B Explain how you determined which operation was needed to write the equation Type below: _____________

Answer: Division operation is needed to write the equation to know how many x skirts she can make from 9 yards of fabric.

Question 13. Karen is working on her math homework. She solves the equation \(\frac{b}{8}\) = 56 and says that the solution is b = 7. Do you agree or disagree with Karen? Use words and numbers to support your answer. If her answer is incorrect, find the correct answer. Type below: _____________

Answer: Karen is working on her math homework. She solves the equation \(\frac{b}{8}\) = 56 and says that the solution is b = 7. I Disagree with Karen. b/8 = 56; multiply both sides by 8 to solve for b, and you get b = 448

Question 14. There are 70 historical fiction books in the school library. Historical fiction books make up \(\frac{1}{10}\) of the library’s collection. The equation \(\frac{1}{10}\)b = 70 can be used to find out how many books the library has. Solve the equation to find the total number of books in the library’s collection. Use numbers and words to explain how to solve \(\frac{1}{10}\)b = 70. Type below: _____________

Answer: Given Number of historical books = 70 The equation used to find the totals number of books in the library collection. \(\frac{1}{10}\)b = 70 b = 70 × 10 b = 700 Hence there are 700 books in the library collection.

Question 15. Andy drove 33 miles on Monday morning. This was \(\frac{3}{7}\) of the total number of miles he drove on Monday. Solve the equation \(\frac{3}{7}\)m = 33 to find the total number of miles Andy drove on Monday. ______ miles

Answer: 77 miles

Explanation: Andy drove 33 miles on Monday morning. This was \(\frac{3}{7}\) of the total number of miles he drove on Monday. \(\frac{3}{7}\)m = 33 3 × m = 33 × 7 3 × m = 231 m = 231/3 m = 77 miles Therefore the total number of miles Andy drove on Monday is 77 miles.

Question 16. The maximum number of players allowed on a lacrosse team is 23. The inequality t≤23 represents the total number of players t allowed on the team. Two possible solutions for the inequality are _____ and _____. Two possible solutions for the inequality are _____ and _____

Answer: The maximum number of players allowed on a lacrosse team is 23. t ≤ 23 Thus the two possible solutions for the inequality are 22 and 23.

Question 17. Mr. Charles needs to have at least 10 students sign up for homework help in order to use the computer lab. The inequality h ≥ 10 represents the number of students h who must sign up. Select possible solutions of the inequality. Mark all that apply. Options: a. 7 b. 8 c. 9 d. 10 e. 11 f. 12

Answer: D, E

Explanation: Mr. Charles needs to have at least 10 students sign up for homework help in order to use the computer lab. h ≥ 10 The number near to 10 is 10 and 11 Thus the correct answers are options D and E.

Question 18. The maximum capacity of the school auditorium is 420 people. Write an inequality for the situation. Tell what type of numbers the variable in the inequality can represent. Type below: _____________

Answer: The maximum capacity of the school auditorium is 420 people Let x be the maximum people The inequality is x is less than or equal to 420. x ≤ 420

Answer: Agree with Dylan. Because the dark circle shows that it is not the solution.

Question 20. Part B Suppose Cydney’s graph had an empty circle at 14. Write the inequality represented by this graph. Type below: _____________

Conclusion:

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Go Math Grade K Answer Key Chapter 4 Represent and Compare Numbers to 10

Teach your kid the basic math skills he might need using the Go Math Grade 4 Answer Key Represent and Compare Numbers to 10. Help them improve their math proficiency using the Grade K HMH Go Math Solution Key provided by subject experts. Kick start your preparation with the quick resources available and clear the exam with flying colors. Download the HMH Go Math Grade K Answer Key Ch 4 for free of cost and resolve your queries whenever you want and from anywhere and anytime.

Go Math Grade K Chapter 4 Answer Key Represent and Compare Numbers to 10

Go Math Grade K Ch 4 Represent and Compare Numbers to 10 Solution Key covers questions from Lessons, Homework Practice, Extra Practice, Review Tests, etc. Access the Topicwise Go Math Kindergarten Chapter 4 Answers available via quick links and clarify all your doubts. All you need to do is just tap on the relevant links and clear the doubts concerning the particular topic in no time. Score better grades in exams compared to others with consistent practice as practice is the only key to success.

Represent and Compare Numbers to 10

• Represent and Compare Numbers to 10 Show What You Know – Page 178
• Represent and Compare Numbers to 10 Vocabulary Builder – Page 179
• Represent and Compare Numbers to 10 Game Spin and Count! – Page 180
• Represent and Compare Numbers to 10 Vocabulary Game – Page(180A-180B)

Lesson: 1 Model and Count 10

• Lesson 4.1 Model and Count 10 – Page(181-186)
• Model and Count 10 Homework & Practice 4.1 – Page(185-186)

Lesson: 2 Count and Write to 10

• Lesson 4.2 Count and Write to 10 – Page(187-192)
• Problem Solving • Applications – Page 190
• Count and Write to 10 Homework & Practice 4.2 – Page(191-192)

Lesson: 3 Algebra • Ways to Make 10

• Lesson 4.3 Algebra • Ways to Make 10 – Page(193-198)
• Algebra • Ways to Make 10 Homework & Practice 4.3 – Page(197-198)

Lesson: 4 Count and Order to 10

• Lesson 4.4 Count and Order to 10 – Page(199-204)
• Count and Order to 10 Homework & Practice 4.4 – Page(203-204)

Mid-Chapter Checkpoint

• Represent and Compare Numbers to 10 Mid-Chapter Checkpoint – Page 202

Lesson: 5 Problem Solving • Compare by Matching Sets to 10

• Lesson 4.5 Problem Solving • Compare by Matching Sets to 10 – Page(205-210)
• Problem Solving • Compare by Matching Sets to 10 Homework & Practice 4.5 – Page(209-210)

Lesson: 6 Compare by Counting Sets to 10

• Lesson 4.6 Compare by Counting Sets to 10 – Page(211 – 216)
• Compare by Counting Sets to 10 Homework & Practice 4.6 – Page(215-216)

Lesson: 7 Compare Two Numbers

• Lesson 4.7 Compare Two Numbers – Page(217-222)
• Compare Two Numbers Homework & Practice 4.7 – Page(221-222)
• Represent and Compare Numbers to 10 Review/Test – Page(223-226)

Represent and Compare Numbers to 10 Show What You Know

Represent and Compare Numbers to 10 Vocabulary Builder

Explanation: In the above picture There are 3 number of carrots and the 3 number of celery sticks. There are same number of carrots and celery sticks. There are 4 trees and 2 bushes. Trees are greater. Bushes are less.

Represent and Compare Numbers to 10 Game Spin and Count!

DIRECTIONS Play with a partner. Place game markers on START. Use a pencil and a paper clip to spin for a number. Take turns spinning. Each player moves his or her marker to the next space that has the same number of objects as the number on the spinner. The first player to reach END wins.

Represent and Compare Numbers to 10 Vocabulary Game

DIRECTIONS Shuffle the Word Cards. Place each card facedown on a different square above. A player turns over two cards. If they match, the player tells what they know about the word and keeps the cards. If they do not match, the player turns the cards facedown again. Players take turns. The player with more pairs wins.

The Write Way DIRECTIONS Draw to show how to compare two sets of objects. Reflect Be ready to tell about your drawing.

Lesson 4.1 Model and Count 10

Essential Question How can you show and count 10 objects?

Share and Show

Problem Solving • Applications

HOME ACTIVITY • Ask your child to show a set of nine objects. Then have him or her show one more object and tell how many objects are in the set.

Model and Count 10 Homework & Practice 4.1

Lesson 4.2 Count and Write to 10

Essential Question How can you count and write up to 10 with words and numbers?

Explanation: Given a basket of eggs with two set of 9 , 1. The set with 9 eggs is represented with 3 eggs in three rows. Adding these  two sets 9 + 1 make 10. 10 is represented.

Explanation: Given a basket of eggs with two set of 5, 4. Adding these  two sets 5 + 4 make 9. 9 is represented.

Explanation: Given a basket of eggs with two set of 5. Adding these  two sets 5 + 5 make 10. 10 is represented.

Explanation: Given a basket of eggs . The eggs are represented in ten frame format. 10 is represented.

Explanation: Given a basket of eggs with two set of 4. Adding these  two sets 4+ 4 make 8. 8 is represented.

Explanation: Given a basket of eggs with two set of 9, 1. The set with 9 eggs represents 3 eggs each in three rows. Adding these  two sets 9 + 1 make 10. 10 is represented.

Count and Write to 10 Homework & Practice 4.2

Explanation: Given a set of bowling pins. They are stacked in 1, 2, 3, 4 format. Adding these stacks 1 +2 + 3 + 4 makes 10.

Explanation: Given a set of bowling pins. They are stacked in a circular format. I have numbered the pins and numbered to 10.

Lesson 4.3 Algebra • Ways to Make 10

Essential Question How can you use a drawing to make 10 from a given number?

Algebra • Ways to Make 10 Homework & Practice 4.3

Lesson 4.4 Count and Order to 10

Essential Question How can you count forward to 10 from a given number?

Listen and Draw

HOME ACTIVITY • Write the numbers 1 to 10 in order on a piece of paper. Ask your child to point to each number as he or she counts to 10. Repeat beginning with a number other than 1 when counting.

Count and Order to 10 Homework & Practice 4.4

Represent and Compare Numbers to 10 Mid-Chapter Checkpoint

Concepts and Skills

Lesson 4.5 Problem Solving • Compare by Matching Sets to 10

Essential Question How can you solve problems using the strategy make a model?

Unlock the Problem

Try Another Problem

On Your Own

HOME ACTIVITY • Ask your child to show two sets of up to 10 objects each. Then have him or her compare the sets by matching and tell which set has more objects.

Problem Solving • Compare by Matching Sets to 10 Homework & Practice 4.5

Lesson 4.6 Compare by Counting Sets to 10

Essential Question How can you use counting strategies to compare sets of objects?

Explanation: Given two sets of counters with different number of counter values. set 1 have 7 counters. set 2 have 6 counters. The set with greater counters is circled.

Explanation: Given two sets of counters with different number of counter values. set 1 have 6 counters. set 2 have 9 counters. The set with greater counters is circled.

Explanation: Given two sets of counters with different number of counter values. set 1 have 8 counters. set 2 have 7 counters. The set with less counters is circled.

Explanation: Given two sets of counters with different number of counter values. set 1 have 7 counters. set 2 have 9 counters. The set with less counters is circled.

Explanation: Given two sets of counters with different number of counter values. set 1 have 9 counters. set 2 have 10 counters. The set with less counters is circled.

Explanation: Number of hats Megan brought are = 8 hats Number of gifts Megan brought are = 8 gifts She brought equal number of gifts and hats for the party.

Question 8. Answer:

HOME ACTIVITY • Show your child two sets of up to 10 objects. Have him or her count the objects in each set. Then have him or her compare the numbers of objects in each set, and tell what he or she knows about those numbers.

Compare by Counting Sets to 10 Homework & Practice 4.6

Lesson 4.7 Compare Two Numbers

Essential Question How can you compare two numbers between 1 and 10?

DIRECTIONS Look at the numbers. As you count forward does 7 come before or after 8? Is it greater or less than 8? Circle the words that describe the numbers when comparing them. 7 7 is less than 8 7 is greater than 8

Explanation: Given two numbers. 10, 5. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the given numbers 10 and 5. 10 is 5 counts more than 5. The greater number is circled.

Explanation: Given two numbers. 6, 4. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the given numbers 4 and 6. 6 is 2 counts more than 4. The greater number is circled.

Explanation: Given two numbers. 7 , 9. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 7 and 9. 9 is 2 counts more than 7. The greater number is circled.

Explanation: Given two numbers. 10 , 8. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 10 and 8. 10 is 2 counts more than 8. The greater number is circled.

Explanation: Given two numbers. 2 , 4. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 2 and 4. 2 is 2 counts less than 4.The number that is less is circled.

Explanation: Given two numbers. 5, 3. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 5 and 3. 3 is 2 counts less than 5. The number that is less is circled.

Explanation: Given two numbers. 8, 9. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 8 and 9. 8 is 1 counts less than 9.The number that is less is circled.

Explanation: Given two numbers. 6, 8 The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 6 and 8. 6 is 2 counts less than 8.The number that is less is circled.

HOME ACTIVITY • Write the numbers 1 to 10 on individual pieces of paper. Select two numbers and ask your child to compare the numbers and tell which number is greater and which number is less.

Compare Two Numbers Homework & Practice 4.7

Explanation: Given two numbers. 8, 5 The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 8 and 5. 8 is 3 counts more than 5. The greater number is circled.

Explanation: Given two numbers. 10 , 7. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 10 and 7. 10 is 3 counts more than 7. The greater number is circled.

Explanation: Given two numbers. 6, 9. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the given numbers 6 and 9. 9 is 3 counts more than 6. The greater number is circled.

Explanation: Given two numbers. 6, 4 The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 6 and 4. 4 is 2 counts less than 6.The number that is less is circled.

Explanation: Given two numbers. 8,7 The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 8 and 7. 7 is 1 counts less than 8.The number that is less is circled.

Explanation: Given two numbers. 5, 3 The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 5 and 3. 3 is 2 counts less than 5.The number that is less is circled.

Explanation: Given two numbers. 7, 8. The counting order is represented as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So, the  given numbers 8 and 7. 8 is 1 counts more than 7. The greater number is circled.

Represent and Compare Numbers to 10 Review/Test

Hope the information shared regarding the Go Math Grade K Chapter 4 has helped you to a possible extent. In Case of any queries feel free to reach us via the comment section so that our experts can guide you with a possible solution. Bookmark our site for more updates on Gradewise HMH Go Math Answers in a matter of seconds.

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Mom and More

10 Helpful Worksheet Ideas for Primary School Math Lessons

Posted: October 30, 2023 | Last updated: December 7, 2023

Mathematics is a fundamental subject that shapes the way children think and analyze the world. At the primary school level, laying a strong foundation is crucial. While hands-on activities, digital tools, and interactive discussions play significant roles in learning, worksheets remain an essential tool for reinforcing concepts, practicing skills, and assessing understanding. Here’s a look at some helpful worksheets for primary school math lessons.

Comparison Chart Worksheets

Comparison charts provide a visual means for primary school students to grasp relationships between numbers or concepts. They are easy to make at www.storyboardthat.com/create/comparison-chart-template , and here is how they can be used:

• Quantity Comparison: Charts might display two sets, like apples vs. bananas, prompting students to determine which set is larger.
• Attribute Comparison: These compare attributes, such as different shapes detailing their number of sides and characteristics.
• Number Line Comparisons: These help students understand number magnitude by placing numbers on a line to visualize their relative sizes.
• Venn Diagrams: Introduced in later primary grades, these diagrams help students compare and contrast two sets of items or concepts.
• Weather Charts: By comparing weather on different days, students can learn about temperature fluctuations and patterns.

Number Recognition and Counting Worksheets

For young learners, recognizing numbers and counting is the first step into the world of mathematics. Worksheets can offer:

• Number Tracing: Allows students to familiarize themselves with how each number is formed.
• Count and Circle: Images are presented, and students have to count and circle the correct number.
• Missing Numbers: Sequences with missing numbers that students must fill in to practice counting forward and backward.

Basic Arithmetic Worksheets

Once students are familiar with numbers, they can start simple arithmetic. 

• Addition and Subtraction within 10 or 20: Using visual aids like number lines, counters, or pictures can be beneficial.
• Word Problems: Simple real-life scenarios can help students relate math to their daily lives.
• Skip Counting: Worksheets focused on counting by 2s, 5s, or 10s.

Geometry and Shape Worksheets

Geometry offers a wonderful opportunity to relate math to the tangible world.

• Shape Identification: Recognizing and naming basic shapes such as squares, circles, triangles, etc.
• Comparing Shapes: Worksheets that help students identify differences and similarities between shapes.
• Pattern Recognition: Repeating shapes in patterns and asking students to determine the next shape in the sequence.

Measurement Worksheets

Measurement is another area where real-life application and math converge.

• Length and Height: Comparing two or more objects and determining which is longer or shorter.
• Weight: Lighter vs. heavier worksheets using balancing scales as visuals.
• Time: Reading clocks, days of the week, and understanding the calendar.

Data Handling Worksheets

Even at a primary level, students can start to understand basic data representation.

• Tally Marks: Using tally marks to represent data and counting them.
• Simple Bar Graphs: Interpreting and drawing bar graphs based on given data.
• Pictographs: Using pictures to represent data, which can be both fun and informative.

Place Value Worksheets

Understanding the value of each digit in a number is fundamental in primary math.

• Identifying Place Values: Recognizing units, tens, hundreds, etc., in a given number.
• Expanding Numbers: Breaking down numbers into their place value components, such as understanding 243 as 200 + 40 + 3.
• Comparing Numbers: Using greater than, less than, or equal to symbols to compare two numbers based on their place values.

Fraction Worksheets

Simple fraction concepts can be introduced at the primary level.

• Identifying Fractions: Recognizing half, quarter, third, etc., of shapes or sets.
• Comparing Fractions: Using visual aids like pie charts or shaded drawings to compare fractions.
• Simple Fraction Addition: Adding fractions with the same denominator using visual aids.

Money and Real-Life Application Worksheets

Understanding money is both practical and a great way to apply arithmetic.

• Identifying Coins and Notes: Recognizing different denominations.
• Simple Transactions: Calculating change, adding up costs, or determining if there’s enough money to buy certain items.
• Word Problems with Money: Real-life scenarios involving buying, selling, and saving.

Logic and Problem-Solving Worksheets

Even young students can hone their problem-solving skills with appropriate challenges.

• Sequences and Patterns: Predicting the next item in a sequence or recognizing a pattern.
• Logical Reasoning: Simple puzzles or riddles that require students to think critically.
• Story Problems: Reading a short story and solving a math-related problem based on the context.

Worksheets allow students to practice at their own pace, offer teachers a tool for assessment, and provide parents with a glimpse into their child’s learning progression. While digital tools and interactive activities are gaining prominence in education, the significance of worksheets remains undiminished. They are versatile and accessible and, when designed creatively, can make math engaging and fun for young learners.

The post 10 Helpful Worksheet Ideas for Primary School Math Lessons appeared first on Mom and More .

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The staff development committee of Information Technology department organized a workshop entitled " Theory of Inventive Problem Solving TRIZ (Math Application)” on Tuesday, 21-11-2023 between 12:00 noon to 01.00 pm@ IT206.

The resource person was Mr. Qasim mohammed Alkulaifeen, Lecturer-Math, UTAS – IBRA. The workshop was very interactive, informative, and practically covered broad areas, about the theory of inventive and problem solving using the Mathematical Application. The speaker was honored by providing appreciation certificate from the department by HOS-Math.

HOS-Math, has given the vote of thanks for giving us the eye opening session to all our Staff members.

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• Senior Project Orientation by Business Studies
• Renewable Energy System Modelling and Installation Online Workshop by Engineering
• Recon: Student-Centered Learning Workshop for Engineering Staff by Engineering
• College Policies Awareness Webinar by Engineering
• Exploring Python Programming for Mathematical Application by Information Technology
• Community Outreach Program - Workshop on Computer Hardware Servicing by ICEC
• Introduction to Robotics, Mechanical Assembly and Robot C Programming by Engineering Department
• Community Outreach Program – Solar-Powered Pumping System Webinar by Engineering Department
• Basic Arduino Programming and Interfacing Online Workshop for Students by Engineering Department
• Approaches to Increase Scientific Publications in Highly Reputed Journals by Business Studies Department
• Workshop on PV System Design & Implementation to Domestic Load by Engineering Department
• Awareness seminar on General Electrical Safety by Engineering Department
• English Society Celebrate Oman's National Day by English Language Centre
• Workshop on Developing Practical Skills Using DBMS Software for Computer Engineering Students by Engineering Department
• UTAS - Ibra Student Attended Huawei ICT Skill National Final Examination by IT Department
• Specialization Orientation Program by BS Department
• "Ascend Yourself, the Bottom is Crowded" by IT Department
• تمويل 13 بحثاً علمياً في جامعة التقنية والعلوم التطبيقية بإبراء by Administration
• UTAS - Ibra Student Qualified for the Huawei ICT Skill Competition by Information Technology
• UTAS - Ibra students attended the Oman Collegiate Programming Competition by Information Technology
• Health Safety Environment Committee by Information Technology
• UTAS - Ibra Students took the Huawei ICT Skill Exam by Information Technology
• English Language Centre Writing Centre Commencement and Invitation to Teach by English Language Centre
• IEEE Online Campaign for AY 2021-2022 by Engineering
• "Be Faster and Smarter at Work" by Business Studies
• Data Cleaning and Validation by Business Studies
• Bridging the Gap between Academia and Industry by Business Studies
• HSE Induction Program for New Engineering Staff by Engineering
• New Students Induction Program by Business Studies
• Live Webinar on Introduction to Materials by Engineering
• Project Orientation on Senior and Graduating Students by Business Studies
• Workshop on Writing Research Proposal by Business Studies
• A Webinar on "AdTrac - Counselling - Guidelines to Advisors" to Engineering Advisors by Engineering
• Transformation of Engineering Education by Engineering
• Staff Webinar on Stress - A Powerful Driving Force, Not an Obstacle by Engineering
• Engineering Hosted Research Methodologies Webinar by Engineering
• Online Course Project Orientation by Information Technology
• ELC celebrated Semester 1- Opening Lunch by English Language Centre
• البرنامج التعريفي للطلبة الجدد 2021-2022 by Administration
• Engineering Online Discussion Forum Based on Synergizing Clean Energy and Green Transportation for Smart Cities by Engineering
• Awareness Program - Advising and Registration by Business Studies
• Engineering's Dr. Shamganth K Joins the Elite Rank of IEEE Senior Member by Engineering

AY2020-2021

• Webinar on Building Mobile Robots and Arduino Based Robot Assembling and Programming by Engineering
• IT Department Awarding Day 2021 by IT
• Webinar on 'Scalability Challenges in IoT - Edge Computing, Block Chain and Industry Revolution 4.0' by IT
• Entrepreneurial Opportunities in Omani Market Webinar by Business Studies
• New Students' Induction Program by Business Studies
• مؤتمر علمي يستعرض تقنيات الثورة الصناعية الرابعة بإبراء by Administration
• IEEE UTAS - Ibra Student Branch Conducts IEEE Membership Campaign Online by Engineering
• Engineering OJT Trainees' Final Presentation: Appraised by External Examiners by Engineering
• HSEC at ELC conducted talk series for staff and students in the month of March, 2021. by English Language Centre
• Engineering Students Virtual Training on Network Modeling and Simulation using OPNET Riverbed Modeler by Engineering
• Engineering QA Awareness and Active Participation in the Committee Work Webinar by Engineering
• Entrepreneurship Unit Hosted Webinar entitled 'Entrepreneurship and Innovation Camp' by Engineering
• Two-Day Electrical Skills Enhancement Workshop for Engineering Students by Engineering
• Training in Computer Network Configuration and Installation by Engineering
• A Virtual Industrial Tour to Telecommunication Regulatory Authority by Engineering
• Introduction to Raspberry Pi Webinar for Engineering Staff by Engineering
• Effective Use of Autodesk Inventor Professional Engineering Program in Online Teaching by Engineering
• Diabetes Mellitus Awareness Program Conducted for Engineering Staff by Engineering
• Engineering Alumna Speaks on a Webinar entitled 'Passion of Artificial Intelligence of Things' by Engineering
• 2nd INNOVATIA 2021 by Business Studies
• IT Department’s Workshop with PDO by Information Technology
• Engineering HSE Committee Went Virtual on Fire Extinguisher Training by Engineering
• Engineering Department Hosted Webinar on Project Management Skills Required for Administrative & Technical Professionals by Engineering
• Workshop on Photovoltaic System Design and Installation by Engineering
• Hands On Training on Packet Tracer by Engineering
• Engineering Department hosted a Webinar on "Latest Mobile Technologies" for OJT & Prospective Graduates by Engineering
• Engineering Hosted CNC Programming and Fusion 360 Manufacturing Webinar for Mechanical Engineering Staff by Engineering
• IT STAFF AND STUDENTS PRESENTED PAPERS IN INTERNATIONAL CONFERENCE by Information Technology
• Google Hash Code Programming Competition by Information Technology
• A Talk on MS-Project Application in Course Project by Information Technology
• Webinar on "Arduino Programming and Interfacing" by Information Technology
• IEEE- UTAS-Ibra Student Branch Hosted a National Wide Webinar on "Introduction to Deep Learning" by Engineering
• Workshop on Literature Review by Business Studies
• WORKSHOP ON HOW TO FIND INNOVATIVE IDEA by Business Studies
• Webinar on "Review of Literature & Referencing" by Information Technology
• BlockChain Technology by Information Technology
• Report on Senior/Graduation Project Orientation by Business Studies
• Management Research using the Analytic Hierarchy Process (AHP) Technique by Business Studies
• Case Development and Writing by Business Studies
• Risk Management Information System Awareness Program for Engineering Staff by Engineering
• Staff Lunch Treat by English Languge Centre
• External Examiners from Various Industries for Engineering Final Project Evaluation by Engineering
• Building a Culture of Participation by ETC
• Awareness on Common Password Threats by ETC
• National Speech Contest by ELC
• ‘Teaching Writing Using Internet Tools’- Webinar by ELC
• Webinar on Occupational Health and Safety by ETC
• Staff of the Month Awarding Ceremony by ELC
• IT Students Graduation Project in Final Round of Upgrade by IT
• UTAS-Ibra Students compete in Speech Contest by ELC
• Awareness on Staff Related Policies for ETC Staff by ETC
• Engineering hosted a 'Network Modeling and Simulation using OPNET' Online Workshop by Engineering
• Higher Education & Industry Symposium – Bolstering Connection Towards Oman 2040 by Engineering
• IEEE – Ibra bagged the MOST ACTIVE STUDENT BRANCH AWARD in Oman by Engineering
• Induction Program for New Intake Students by IT
• Inspiring Students Through Alumni Experience by IT
• Policy Awareness Webinar by Engineering
• Engineering Students Take Part in 'HUAWEI ICT COMPETITION MIDDLE EAST 2020' by Engineering
• Incentive Methods by Business Studies
• Methods in Training and Development by Business Studies
• National Day Celebration by Engineering
• نظرا لأهمية غرس قيم المواطنة الحقيقية و تنمية حس المسؤولية المجتمعية بين by Admin
• Specialization Orientation program by Business Studies
• Business Quiz by Business Studies
• Research Questionnaire Preparation by Business Studies
• WORKSHOP ON RESEARCH METHODOLOGY FOR GRADUATE PROJECT by Business Studies
• Engineering Alumna Speaks on a Webinar Entitled "How Course Projects and Internship Programs Support the Recruitment Procedure of Scholars" by Engineering
• Workshop on Curriculum Development and Review Framework by Business Studies
• Webinar - New Features of ADTRAC by Engineering
• Financial Modeling Using Excel Workshop by Business Studies
• ETC organizes Webinar on ProQuest Databases and Turnitin by ETC
• UTAS-Ibra Conducted Its First Ever Online Examination with ETC Technical Team by ETC
• Live Webinar: Importance of Soft Skills & Personal Branding by Engineering
• Live Webinar: The significance of Omani Graduate achievements in the reality of work by Engineering
• Induction Program to New Students by Business Studies
• Intelligent Electric Vehicle Charging System by Engineering
• Smart City Development through IoT by Engineering Department
• Role of the Smart Grid in Facilitating the Integration of Renewables by Engineering
• "EMERGING DESIGN AND TECHNOLOGIES IN ENGINEERING Internet of Things (IoT) & MICROSTRIP ANTENNA" by Engineering
• Artificial Intelligence for Business Operations by Engineering Department
• 5G Live: Webinar – "Get in-depth knowledge about future Generation Technology and designing of 5G business strategy" by Engineering
• An Online Live Seminar on 'Adopting Green Marketing Strategy and its relation to Production Cost in Industries' by Engineering
• Exploring the Rule-based Expert Systems for Smart Healthcare in 21st century by Engineering
• Making an effective Video lecture - Use of Different Tools and Software by Engineering
• Webinar on the Challenged Experienced by Young Entrepreneurs: Strategy to Way Forward by OJT
• Audit Committee Overlapping and Forward-looking Disclosure: An Untold Story of Omani Financial Institutions by Business Studies
• Online Seminar on EMI/EMC technologies for Industrial IOT Platform by OJT
• Webinar on Smart Grid by Engineering
• Webinar on Strategies for an Entrepreneurial Career by Business Studies
• Automotive Safety Technologies by Engineering
• Condition Monitoring of Transformers Using Dissolved Gas Analysis Techniques by Engineering

AY2019-2020

• Webinar on "Augmented Reality Using Unity Engine & Vuforia" by IT
• Webinar on Thinking Skills and Problem Solving by IT
• Workshop on Data Analysis vs Data Analytics by IT
• Workshop on Quality Awareness by IT
• IT Students Presented Papers in SNSIR4.0 Webinar, HCT, Muscat by IT
• Math Lessons in GeoGebra with Moodle Integration by IT
• External Examiners from various HEI’s for Engineering Student’s Course Project Evaluations by Engineering
• ETC Goes Online Training/Courses by ETC
• Development of Practical Skills in the use of DBMS Software by Engineering
• 8th Annual Programming Competition by Engineering
• Hands on Training in Packet Tracer by Engineering
• Workshop on Mathematics using Apps, Software and IT Technology by Information Technology
• IT Students Awarding Day by IT
• Engineering students Won Second place in Robotics Challenge by Engineering
• Parents’ Perceptions on Graduate-Children Choosing Entrepreneurship as their Career by Business Studies
• Teachers’ Day Celebration at ELC by ELC
• Poster presentation for 16th International Gas Research Conference (IGRC 2020) hosted by Oman LNG by Engineering
• Questionnaire Preparation Workshop for Senior/Graduation Project Students by Business
• Workshop on Literature Review and Referencing for the Senior/Graduate Project by Business
• Ericsson Oman L.L.C, Muscat – Guest Lecture by Engineering
• Dr. CK Chairs a Session and Presents a Paper at SQU’s 4th International Conference on Language, Linguistics, Literature, and Translation by IT
• Configuring Cisco IOS using Python by IT
• Al Omran Private School Community Outreach Program by IT
• Guest Lecture on Generator and it's Protection by Engineering
• Innovative Idea Poster Presentation by Business Studies
• Does Gamification Enhance Learning Experience of Accounting Courses? – An Experimental study at Ibra College of Technology by Business Studies
• Engineering students educational field trip to Sultan Qaboos Port,Muscat by Engineering
• Strategic Plan 2019-2024 Awareness by ELC
• Workshop on SCL Tools by Business
• Risk Management Approach in Curriculum and Program Quality Improvement by IT
• Business Department Inducts New Students by Business
• Business Department Conducts Senior/Graduation Project Orientation by Business
• Senior/Graduation Project Workshop on Microsoft Word Preparation by Business
• Toastmasters Speaking Workshop by ELC
• Workshop on Innovative Idea Generation by Business
• Strategic Plan- 2019-2024 Awareness Program by Business
• Course Project Orientation by IT
• IQAC-Strategic Plan 2019-2024 Awareness Workshop by IT
• GFP Common Quality Audit Portfolio And Quality Audit Visit Awareness Workshop by ELC
• Academic Advising and Improvement Action based on Student’s Registration Feedback by IT
• Reviewing and Developing Curriculum of Electrical and Electronics Engineering Workshop by Engineering
• IEEE Annual Gathering in Oman by Engineering
• ELC Held A Semester End Buffet Lunch by English Language Centre
• Electrical Machine Testing and Wiring Design Workshop by Engineering
• ICT Staff Attend 'ACT IP DAY' by IT
• Dean's Honor List by ELC
• ICT SKILLS COMPETITION 2019 CONCLUDED by IT
• Technology Integrated in English Language Teaching by ELC
• Omanisation Lecture by Business Studies
• Industry Visit HIGH TOWER GROUP OF COMPANIES by Engineering
• Industry Visit to ENGIE STOMO by Engineering
• Field Study Visit Munisifeh Ruins by Engineering
• Mathematics Challenge Battle 1 by Information Technology
• Training on Total Station by Engineering
• Network Modeling and Simulation by Engineering
• Use of Microsoft Office Workshop by ELC
• Speech Competition Day by ELC
• Peer Tutoring Workshop by ELC
• ELC Celebrated 49th National Day by ELC
• Integrity Day at ELC by ELC
• Industry Visit to Omani Securities Association by Business Studies
• Workshop on SPSS by Business Studies
• Field visit to The Chedi Hotel by Engineering
• تقرير محاضرة ( قانون العمل العماني ) Lecture Report (Oman Labor Law) by Business Studies
• Industrial visit Haya Water Muscat by Business Studies
• HVAC and Building Services study tour at Mohammed Al Ameen Masjid, Muscat by Engineering
• ICT Skills Competition 2019 Elimination Round by IT
• Engineering students selected as Panelists for Robotic Competition for School students by Engineering
• 49th National Day Celebration of Engineering Department by Engineering
• Community Engagement Program - National Day Celebration at Al-Kamil Power Station by Engineering
• Community Service at Al-Wafa Social Center by IT
• Academic Advising Workshop by Business
• Development of Practical Skills in the use of DBMS software by Engineering
• CNC Programming On Fanuc Control by Engineering
• Research Through Student Projects by Engineering
• Academic Counselling for Students by Business Studies
• WORKSHOP ON BEARING MAINTENANCE AND LUBRICATION by Engineering
• Specialization choice induction program for SPRING 2020 by Engineering
• Workshop on ‘Microsoft Excel’ by Business Studies
• Engineering students Industrial visit to Power Station, Alkamil by Engineering
• Employers Recommended Training Program on PSTN Modeling and Simulation to Telecommunication and computer students by Engineering
• Workshop on Entrepreneurship by Business
• Health and Safety Awareness by Business
• Regional Final Exam for Huawei ICT Skills by IT
• Omani Women’s Day Celebration by Business
• Hands on Training in Packet tracer by Engineering
• Cooperation Agreement between Ibra College of Technology and Omani Society for Education Technologies by Administration
• IT Students won 3rd Prize in SITAM 2019 by IT
• ICT Graduation Ceremony 2019 by Administration
• Microsoft Word Presentation Workshop by Business
• Questionnaire Preparation Workshop by Business
• IT Students attend Huawei ICT Skill Workshop by IT
• Global Careers In Cyber Security – The Importance Of Skills And Education by IT
• Health and Safety Awareness Program by Engineering
• Policy Management Portal Workshop by IT
• Enhancing Blended Learning Workshop by IT
• Academic Advising by IT
• Robotics Workshop by Engineering
• Community Outreach at Al Omran Private School by IT
• RMC conducts Risk Management Planning for ICT Staff by QA
• Industrial Visit to OMANTEL Ibra by Engineering
• Engineering students Industrial visit to Oman Fiber Optic Company SAOC Muscat by Engineering
• Training Program on 'Computer Hardware Servicing' by Engineering
• 3D Modeling Workshop by Engineering
• Training program on Computer Network Administration by Engineering
• A Talk on 'Hands on Security' by IT
• Basic Electronics Skill Development Workshop by Engineering
• Review of Literature & Referencing by Business
• ICT Alumni View On Exploring Career in Electrical Power Plant Engineering by Engineering
• Workshop on Writing Research Proposal by Business
• IP Addressing and Subnetting Student Workshop by IT
• Cooperation Agreement between the Ibra College of Technology and the Oman Chamber of Commerce and Industry, Ibra by Administration
• IEEE Campaign by Engineering
• ELC conducts Policy Awareness Program for staff by ELC
• Job Search Technique Presentation and Training Orientation by OJT
• Teaching and Learning Workshop by IT
• ELC conducts an 'Induction Program' for New Intake Students by ELC
• Memorandum of cooperation was signed between Ibra Technical College and Muscat College by Administration
• ELC Staff Gathering by ELC
• ELC conducts Flipped Classroom Using Edpuzzle Training by ELC
• Induction Program for new intake by IT
• Workshop on 'ISA Report Writing' by IT
• A talk on 'CLASSROOM ASSESSMENT TECHNIQUES' by IT
• Academic Advising Workshop by IT
• A talk on 'Augmented Reality' by IT

AY2018-2019

• ELC Year-End Get-to-Gather by ELC
• Alumni Workshop on CV Writing by Business
• Report for the workshop "Recent Trends in Accounting" by Business
• Alumni workshop on "Fundamental Analysis" by Business
• Student Appreciation Day 2019 by Business
• ICT Business Studies’ Benchmarking with Muscat College by Business
• Academic Counselling for Students by Business
• Engineering Students Awarding Day 2019 by Engineering
• Reimagining Teaching and Learning in Higher Education by Business
• IT Student Awarding Day 2019 by IT
• Center for career guidance Department in coordination with the Department of Engineering Organized Mock Interview & Motivational Talk to Prospective Graduates by Engineering
• How to Use the Voice Recorder for the Speaking Exam by English Language Centre
• Report on Guest Lecture – Functions of Human Resource Management by Business Studies
• Specialization choice induction program by Engineering
• Specialization Orientation Program by Business Studies
• HCT benchmarks ICT’s Staff Performance Appraisal System by Educational Technology Centre
• Alumni Motivational Talk Series by Engineering Department
• Review of Literature & Referencing by IT Department
• Benchmarking between Engineering Department of Ibra College of Technology and Nizwa College of Technology by Engineering Department
• IT Students at Oman Data Park by IT Department
• Memorandum of Cooperation with Bahwan Veolia Water Desalination, Sur by IT Department
• Students Participation in an International Conference by Business Studies Department
• Community Outreach Program by IT Department
• Accounting Students Participation in a Research Conference Organized by Arab Open University, Oman by Business Studies Department
• Session Chair in an International Research Conference by Business Studies Department
• Presentation on Student Request System by IT
• IT Staff Attended 2nd Academia And Industry Meeting Workshop by IT
• IT Students at Oman Netacad Student Day 2019 by IT
• IT Students Presented Papers at AOU-ISRC 2019 by IT
• IT Students Presented Papers at CICTAB 2019 by IT
• English Language Teaching Professional Development Workshop by ELC
• Workshop on Digital Marketing by IT
• Engineering-Alumni-as-Industry-Expert-Panelists by Engineering
• ICT Risk Management Committee meets SQU Risk Management Team by RMC
• OJT Department Benchmarking with Nizwa College of Applied Science by OJT Department
• Guest Lecture – Management Accounting II by Business
• Telecommunication Students Final OJT Presentation at Omantel by Engineering
• Ibra College of Technology in Observance of Earth Hour by ETC
• College OJT Orientation Program by OJT Department
• Art of Model Making Competition by Engineering
• Advanced Training on CNC by Engineering
• ELC Staff attend the 2nd Sohar University TESOL Symposium by ELC
• ELC students attend the Open Day at Shinas College by ELC
• Critical Thinking Workshop by ELC
• Mathematics Challenge 2019 by IT
• Practicals on Fundamentals of Computer Security by IT
• ICT Innovatia 2019 by Business
• Guest lecture on "Digital forensics" by IT
• Familiarization of Practical Antenna Systems by Engineering
• Innovatia 2019 by Business Studies
• Engineering Robotics Team Wins 2nd Place by Engineering
• Health and Safety Exhibition by Engineering
• Integrated Petroleum Services Plant Visit by Engineering
• Employee Recruitment And Compensation Practices by Business
• ICT Networking Competition–2019 by IT
• A Talk on Free & Open Source Software (FOSS) by IT
• Using Turnitin in Moodle by IT
• Fiesta Mathematica 2019 by IT
• Community Outreach Program by Engineering
• 7th Annual Programming Competition by Engineering
• Students Specialization Meet by IT
• Educational Tour at Oman Convention and Exhibition Centre by Engineering
• IEEE-ICT Students Shine in IEEE Oman Congress 2019 by Engineering
• ELC Celebrates Teachers' Day by ELC
• Industry Exposure for Bachelor of Computer Engineering Students by Engineering
• Collaboration with Public Authority for Consumer Protection by Business
• An Industrial Visit to OMANTEL, Ibra by Engineering
• PLANT VISIT in ENGIE STOMO Company by Engineering
• National Museum Visit by ELC
• Engineering Students Visit Oman Fiber Optic Company by Engineering
• Towards Spreading the Culture of Research and Innovation by Business Studies
• Workshop on Questionnaire Preparation by Business
• ELC Conducts Policy Awareness Meeting by ELC
• ELC (IQAC) Reviews the Spot Audit Checklists by ELC
• MEMS Energy Harvester (Engineering Lecture Series) by Engineering
• Industrial Visit to Sohar Aluminum by Engineering
• Senior Project Workshop on Research Methodology by Business Studies
• Senior Project Workshop on Manchester Phrase Bank and Literature Review by Business Studies
• Senior Project Workshop on Writing Research Proposal by Business Studies
• Seminar on Visualization of Temperature Distribution during Real Time Multiphase Flow Process in Various Applications Docs by Engineering
• Induction Program to New Students Docs by Business Studies
• Workshop on Google Docs by ETC
• Induction Program for new intake IT students by IT
• Seminar on Project Based Learning by IT
• A talk on "Network OS Administration & Security" by IT
• Workshop on Entrepreneur by Business Studies
• ETC Participated the Seminar on Legal Legislation for Women Working in the Government Sector by ETC
• Engineering Induction Program for New Intake by Engineering
• IT STUDENTS VISIT OMANTEL - Semester II by IT
• ProQuest Training by ETC
• Business Students’ participation on Agriculture, Fisheries and Food Investment Forum by Business
• IT Course Project Orientation by IT
• المكتب الوطني لنقل التكنولوجيا by Admin
• A Team of ETC System Developers Attended the Oracle Seminar at Kempinski by ETC
• مشاركة الكلية التقنية في اجتماع غرفة تجارة وصناعة عمان لمناقشة أهمية قطاع المعارض والمؤتمرات by Student Affairs Office
• Student Centered Learning using Technology by Business Department
• Students’ Project Evaluation by Experts from Industries by Engineering Dept
• Plagiarism Awareness Workshop by ELC
• Centre Staff General Meeting with New HoC by ETC
• Academic Advising - Tips & Discussion by IT Department
• Internal Project Presentation by IT Department
• Staff Gathering by IT Department
• Workshop on Vocabulary Development by English Language Centre
• IEEE-ICT Student Branch Conducts Short-Term Course on 4G Network Essentials by Engineering
• Community Engagement Activity by Business Studies
• GREEN SCARF Project Won Second Place in NAMA Architectural Design Competition by Engineering Dept
• An Industrial Visit to Al Kamil Power Station by Engineering Dept
• Engineering Department Celebrates 48th National Day by Engineering Dept
• Engineering Students Win Second Place and Fourth Place in Al Roya Youth Initiative Awards by Engineering Dept
• Probation Students Meeting by Business Studies
• MONOLITH 2018 by Business Studies
• ETC Participated in ServiceDesk Plus Workshop by ETC
• Awareness Program on Policy Management and Approved Policies for ETC Staff by ETC
• ETC Participates in ICT Open Day 2019 by ETC
• ETC Lab in-charge attend Installation and Use of Electronic Whiteboard Workshop by ETC
• ETC holds Workshop on ProQuest by ETC
• Senior Project Workshop on SPSS by Business Studies
• Engineering Students Got 1st and 3rd Prize in Invasion 2018 Competition by Engineering
• ICT Celebrates 48th Glorious National Day by Admnistration
• ETC holds Risk Management Awareness Program by ETC
• ETC staff attends Awareness Program on ISA Requirements by ETC
• Strengthening the Relationship with Alumni through Personality Development and Career Guidance Seminar by Engineering Dept
• Two Day Event - Muscat Securities Market Awareness Program by Business Dept
• Workshop on Flip Activity by Business Dept
• IT Students in Community Out Reach Program by IT Dept
• First Aid Skills Awareness Program by Engineering Dept
• CIMS Awareness Program for IT students by IT Dept
• سبع مائة و اربعة و خمسون خريج و خريجة ترفدهم تقنية ابراء الى سوق العمل by Admin
• Electrical and Electronics Measuring Instruments and Equipment Workshop by Engineering Dept
• Workshop on 'Project Poster Preparation' by IT Dept
• Induction Program for New Intake Diploma 1st Year Students by IT Dept
• Workshop on 'Assessment of Operational Plan - AY2017-2018' by IT Dept
• Industrial Visit to Bin Abdan International Consultancy Group by Business Dept
• Preparation for Examination and Concentration in Studies by Business Dept
• Community Engagement Workshop by Business Dept
• Workshop on Policy Review by IT Dept
• Linkage with Industry Promotes Partnership with Academe Through Guest Lecture on GSM NETWORK by Engineering Dept
• Staff Awareness Program on Reviewed Policies by Engineering Dept
• Training on Basic Computer Network Administration by Engineering Dept
• IT Society Students in Community Service at Al-Wafa Social Center by IT Dept
• IT Students Visit Omantel Ibra by IT Dept
• Workshop on MATLAB programming and SIMULINK modeling by Engineering Dept
• Workshop on Website Development using Wix by Engineering Dept
• Workshop on PCB Development by Engineering Dept
• IT Students Visit Oman Fiber Optics by IT Dept
• Al Roya Youth Initiative Award Awareness Program by Engineering Dept
• Report on Workshop held on 12th September 2018 by Business Studies
• A talk on "Software Methodologies and System requirement specifications" by IT
• A talk on "Course Project Orientation" by IT
• Workshop on “Review of Literature & Referencing” by IT
• Business Induction Program Semester 1 by Business Studies
• Staff Students Gathering by Engineering
• 4th Teaching and Learning Conference by ELC
• Engineering Student Gathering by Engineering
• Writing Research Proposal Workshop by Admininistration
• Workshop on Effective Project Guidance and Assessment by QA Dept

AY2017-2018

• " فقط ابتسم" حملة ينظمها طلبة الكلية التقنية بإبراء لتوزيع سلات غذائية للطلبة و الأسر المحتاجة
• الكلیة التقنیة بإبراء /حلقة عمل
• Smart Education and Technology Symposium 2018
• RMC spearheads Workshop on Incident Response Planning
• الملتقي الطلابي السابع بالكلية التقنية بعبري
• Engineering Students Got Top Awards In the Al-Roya Project Competition
• Graduation Ceremony 2017
• ICT student bagged second place in a 10-km run for higher education students
• Engineering Project Selected to Compete in the National Level of OCCI Innovation Award
• Engineering Projects Gets Nod for FURAP
• ICT purchased new PC Hardware Trainer
• An Interview with the Captain of the historic ‘Jewel of Muscat’
• The Interaction with Entrepreneurs at OCCI
• Accounting Students Participation in Research Conference
• An Interview with Mr.Hatim Al Abdissalam
• Let’s Read- The Stories of Our Ancestors
• Open Day 2017
• An Interview with the Conqueror of Mount Everest and Pumori
• 2nd Alumni Meet
• Health and Safety Workshop
• ICT students won First Place in Network Troubleshooting and Configuration
• Engineering Student Papers got accepted during 8th National Symposium on Engineering Final Year Projects
• Developing and Strengthening of On-The-Job Training Workshop
• Flipped Classroom Presentation by Dr. Azzah Ahmed Said Al Maskari
• An Interview with the Asian games medal winner Mr.Barkat Al-Harthi
• IT Networking Students Won First Place in INVENT2017
• OJT Training Orientation Seminar
• Fiesta Mathematica 2017
• Workshop on Project Development NAMA AMBASSADORS
• ICT Team visit Special Economic Zone in Duqm
• A Talk on "Smart Technologies, Security and Social Media"
• ICT Morning Walk 2017
• Guest Lecture on "Power of Mind and Self Esteem"
• ETC Team Building Activity 2017
• Business Studies student got the 2nd prize in an Intercollegiate Management Meet (Synergy 2K’17)
• ICT Conducted a Special Training Program for Young Students
• Engineering Final Project Evaluation by Industry Experts
• Engineering Robotics Team wins Second Place
• Engineering 6th Annual C Programming Competition
• Orientation Program on Professional Bodies
• Industrial Visit to OMAN FIBER OPTIC COMPANY
• Project Collaboration Meeting with Zubari Automotive Group Ibra
• Symposium on IoT and Big Data by IEEE-ICT Students and Staff Members
• Engineering Student and Staff attend Robotics Training
• Engineering Staff attends IEEE Symposium of Engineering Education (ISEEE’18)
• Engineering Department Recipient of the Renewable Energy System under Mustadeem Program
• Mechanical Engineering Students visit Desalination Plant in Sur
• Engineering conducts seminar on Student’s Centered Learning
• Engineering Department Inaugurates Professional Team
• Engineering Staff Attends "Training Program on Usage of First Aid Kit"
• Engineering Students Industrial Visit to OMANTEL
• Engineering Department Project Collaboration meeting with Mazoon Ibra
• IEEE-ICT Students Field Test Activity with TRA
• Public Safety and Evacuation Training Program
• Mechanical Students Industrial Visit to United Engineering Services (UES)
• Engineering Department Initiates MoC with Voltamp Transformer Company
• On-The-Job-Training Orientation Programme
• Engineering Staff visit Al Kamil Power Plant
• Computer Hardware Servicing and Network Servicing Workshop
• Engineering conducts workshop on 'Entrepreneurship Skills and Strategies'
• Engineering Conducts Workshop on 'Microsoft Virtual Academy'
• Architecture students visit STFA-HLG Construction Company
• Engineering-Students-Visit-Oman-Fiber-Optic-Company
• Industrial Visit to OMANTEL
• Seminar on OMANTEL NETWORK
• ELC holds "Document Management System Workshop"
• ELC RCC in ELTPDW, Ibri College of Technology
• ICT Engineering Students Project Got Among the Top Two from Sharqiyya Region
• Staff Development Program on Flipped Classroom Best Practices
• Guest Lecture to Marketing Students
• Presentation on Document Management System
• Industrial Visit to Muscat Security Market
• Workshop on Entrepreneurship
• Industrial Visit to Seeb Waste Water Treatment Plant Project
• Guest Lecture on Students Studying 2nd Year Diploma
• Senior Project Evaluation by External Examiners
• Business Studies Students Visit Central Bank of Oman
• SYNERGY-2018
• Workshop on Plagiarism Policy and Turnitin
• MSM Gathering
• Student development program on 'Co-operative and Collaborative Learning'
• Business Department Induction Program to New Students
• Business Department Probation Students Meeting
• Workshop on Final exam question paper review
• Specialization Orientation Program
• Advising and Registration Workshop
• Business Department Staff Picnic
• 47th National Day Celebration
• Industry Visit to Secretariat General for Taxation
• Industrial Visit to Central Bank of Oman
• Industrial Visit to Nabil Biscuits
• Industrial Visit to Muscat Securities Market
• Omani Women's Day
• Senior Project Report Orientation
• Tracking Flipped Classrom
• Induction Program to New Students
• WORKSHOP ON 'EMPLOYEE RELATIONS - FILING MANAGEMENT'
• Oman Skill Competition Induction held in ICT
• IT Students visit Oman Fiber Optic (OFO) Company
• Report on student development program on "Learning Styles"
• A Talk on "Programming Skills"
• Awareness Session on Planning Mechanism
• IT Students Visit Omantel Ibra
• IT Department Benchmarking Team visited Nizwa College of Technology
• IT Students Project shortlisted in Dell - EMC Envision the Future Competition
• IT Networking Advanced Diploma Students Visited Omantel Ibra
• ITPC 2018 Elimination Round
• Mathematics Challenge, Battle-2, 2017-2018
• New Students Induction Program
• Oral Health
• Report on student development program on "Motivational Speech"
• Student development program on “Co-operative and Collaborative Learning”
• Student Development program on "Motivational Speech"
• Training course on ADOBE FLASH for school students
• Workshop on Electronic Extortion
• Workshop on 'Software Testing and Plagiarism'
• Counselling Program on IT Probation Students
• Course Project with Innovative Ideas
• Cultural Event at Al Wafa Social Centre, Ibra
• Entrepreneurial Ventures and Opportunities
• IT students take part in the Huawei ICT Skill Preliminary Exam
• ICT Initiative To 3rd Riyada Award Evaluation
• ICT Students Attended the Huawei ICT Skill Workshop
• ICT Students took the Huawei ICT Skill Final Exam
• Innovation and Entrepreneurship
• IT Society – National Day Celebration
• IT Society Students visit Sharqiya University
• IT Students Participate in the Oman Collegiate Programming Contest (OCPC)
• IT Student Bagged the 2nd Best Paper Award At NSETEM 2017
• "Mathematics Challenge, Battle-1, 2017-2018"
• Memory Techniques
• OCPC Awareness Conducted in Ibra College of Technology
• Paper Presentation by IT Staff
• Project Peer Group Discussions
• Workshop on Virtualization using VMWARE
• SQU invited Mr. CK, Editor-Ibra Journal of ELT, to join a panel discussion on "Research and Publishing"

AY2016-2017

• ICT 2nd Alumni Meet