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## Algebra Solutions

Our goal here is to introduce some of the equation solving techniques that may be helpful for kids in understanding Algebra.

We start here with a very simple technique.

## Let us say that there is an equation x + 3 = 9 and we need to solve it. Solving an equation means that we need to know the possible values of the variables that when put into the equation will satisfy it. We present the solution next.

Explanation:.

x + 3 = 9 (original equation ) x = 9 – 3 (take the constant term to the other side so that the variable x is isolated.) x = 6

So x = 6 is the required solution to the equation.

## You have an equation 3x = 9. Find x.

3x = 9 (original equation) x = 9/3 (take the constant term associated with x to the other side so that the variable x is isolated.) x = 3

So x = 3 is the required solution to the equation.

## We have two simultaneous equations in two variables x and y. Find x and y. x – y = 10 --------(1) x + y = 15 --------(2)

(25/2, 5/2).

In solving these equations, we use a simple Algebraic technique called " Substitution Method ". In this method, we evaluate one of the variable value in terms of the other variable using one of the two equations. And that value is put into the second equation to solve for the two unknown values.

The solution below will make the idea of Substitution clear.

Using 1 st equation,

x – y = 10 -----(1) x = 10 + y

Now we put this value of x into the 2 nd equation.

x + y = 15 -----(2) (10 + y) + y = 15 10 + 2y = 15 2y = 15 – 10 = 5 y = 5/2

Putting this value of y into any of the two equations will give us the value of x.

x + y = 15 x + 5/2 = 15 x = 15 – 5/2 x = 25/2

Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations.

## We have two simultaneous equations in two variables x and y and we need to find x and y. x – y = 10 -----(1) x + y = 15 -----(2)

(25/2, 5/2).

In Elimination Method, our aim is to "eliminate" one variable by making the coefficients of that variable equal and then adding/subtracting the two equations, depending on the case.

In this example, we see that the coefficients of all the variable are same, i.e., 1. So if we add the two equations, the –y and the +y will cancel each other giving as an equation in only x. Let me illustrate this below.

x – y = 10 x + y = 15 2x = 25 x = 25/2

Putting the value of x into any of the two equations will give y = 5/2

Elimination Method - By Equating Coefficients:

This is another very easy and useful equation solving technique that is extensively used in Algebraic calculations. We illustrate this method through an example.

## Given two equations in two variables x and y. Find the values of x and y that satisfy these equations simultaneously. 2x – y = 10 ------(1) x + 2y = 15 ------(2)

( 7, 4).

In this example, we see that neither the coefficients of x nor those of y are equal in the two equations. So simple addition and subtraction will not lead to a simplified equation in only one variable. However, we can multiply a whole equation with a coefficient (say we multiply equation (2) with 2) to equate the coefficients of either of the two variables.

After multiplication, we get

2x + 4y = 30 ------(2)'

Next we subtract this equation (2)’ from equation (1)

2x – y = 10 2x + 4y = 30 –5y = –20 y = 4

Putting this value of y into equation (1) will give us the correct value of x.

2x – y = 10 ------(1) 2x – 4 = 10 2x = 10 + 4 = 14 x = 14/2 = 7

Hence (x , y) =( 7, 4) gives the complete solution to these two equations.

## Different Equation Types

In Algebra , sometimes you may come across equations of the form Ax + B = Cx + D where x is the variable of the equation, and A,B,C,D are coefficient values (can be both positive and negative).

In the next section, we present an example of this type of equation and learn how to solve it through simple Algebraic techniques.

## How to find system of linear equations using Simple Algebric techniques

Find a value of x that satisfies this equation: 4x – 3 = 3x + 8.

We are given that

4x – 3 = 3x + 8

Separating the variables and the coefficients gives:

4x – 3x = 8 + 3

(Note: Taking a constant or a variable term to the left hand side from the right hand side (or vice versa) changes its sign as illustrated above.)

Simplifying the above equation on the L.H.S (Left Hand Side) and the R.H.S (Right Hand Side) gives

Hence x = 11 is the required solution to the above equation.

In the equation A x + B = C x + D , the coefficients A, B, C, D may also be any decimal numbers. For example, the equation could be of this form: 4x + 3.2 = 6.1x + 5.2 -- But you are not supposed to be confused with the method. There will be no change in the equation solving strategy and once you have learnt the above method, you do not need to bother about the coefficients at all.

Next we present and try to solve the examples in a more detailed step-by-step approach. Examples given next are similar to those presented above and have been shown in a way that is more understandable for kids.

## Solve x + 4 = 11 to find the value of x.

x + 4 – 4 = 11 – 4 x = 7

Hence x = 7 is the solution to the given equation.

## How to solve system of Linear Equations by Elimination

Solve the following system using elimination. 2x + y = 15 3x – y = 10, (5, 5).

If we use the method of addition in solving these two equations, we can see that what we get is a simplified equation in one variable, as shown below.

2x + y = 15 ------(1) 3x – y = 10 ------(2) ______________ 5x = 25

What we are left with is a simplified equation in x alone. i.e., 5x = 25

5x/5 = 25/5 x = 5

2(5) + y = 15 10 + y = 15

Which is another equation in a single variable y.

10 + y – 10 = 15 – 10 y = 5

Hence the solution to the system of equations is (x , y) = (5, 5)

## Solve the following system using Elimination x + 2y = 15 x – y = 10

( 35/3 , 5/( 3 )).

With a little observation, we can conclude that if we directly add these two equations, we are not going to reach any simple equation. Let us show this below.

x + 2y = 15 ------(1) x – y = 10 ------(2) ______________ 2x + y = 25

Which is another equation in 2 variables x and y. So our problem doesn’t seem to have reduced. If instead of adding the two equations directly, I multiply the entire equation (1) with – 1, and then add the resulting equation into equation (2), the +x will be cancelled out with – x as shown next.

– ( x + 2y ) = – 15 – x – 2y = – 15 ------(1’)

– x – 2y = – 15 ------(1’) x – y = + 10 ------(2) ______________ – 3y = – 5

-3y/-3=-5/-3 y = 5/3

x – 5/3 = + 10

x – 5/3 + 5/3 = 10 + 5/3 x = (30+ 5)/3 = 35/3

Hence the solution to the given system of equations is (x , y) = ( 35/3 , 5/( 3 ))

## Solve the following system of linear equations using Elimination. 8x – 13y = 2 –4x + 6.5y = –2

Apparently, this system seems to be a bit complex and one might think that no cancellation of terms is possible. But a close observation and a simple multiplication can lead us in the right direction.

We are given two equations:

8x – 13y = 2 ------(1) –4x + 6.5y = –2 ------(2)

2(–4x + 6.5y ) = 2(–2) –8x + 13y = –4 ------(2’)

8x – 13y = 2 ------(1) –8x + 13y = –4 ------(2’) ______________ 0 = –2

But this is not true!! 0≠ –2

Hence the two equations constitute an inconsistent system of linear equations and thus do no have a solution (At no point do the two straight lines intersect => No solution!)

## How to solve system of Linear Equations by Substitution

Solve the following system of linear equations by substitution. 2x – 2y = –2 x + y = 24, 23/2 , 25/( 2 ).

In this method of equation solving, we work out on any of the given equations for one variable value, and then substitute that value in the other equation. It gives us an equation in a single variable and we can use a single variable equation solving technique to find the value of that variable (as shown in examples above). Let us solve the given system now

2x – 2y = –2 ------(1) x + y = 24 ------(2)

Now the next question is: which equation to pick up. There is no particular criteria for this choice. One can simply choose an equation that makes the calculations simpler. E.g., in this example, the equation (2) is easier to work on.

x + y = 24 ------(2)

2(24 – y) – 2y = –2 48 – 2y – 2y = –2 48 – 4y = – 2

48 – 4y – 48 = –2 –48 –4y = –50

-4y/-4 = -50/-4 y = 50/4 = 25/2

x + 25/2 = 24

x + 25/2 - 25/2 = 24 - 25/2 x = (48 - 25)/2 = 23/2

Hence the solution to the given system of equations is (x , y) = ( 23/2 , 25/( 2 ))

Note: Next we show what happens if we substitute the value of x into the same equation that we used to compute it (equation (2) in this example)

x + y = 24 24 – y + y = 24 ∵ (x = 24 – y) 24 = 24

This is the result that we are left with. There is nothing wrong with 24 being equal to 24, but then what should we do with it? Of course we have not been looking to prove this in the first place!!

Hence we conclude that there is no point in substituting the computed value into the same equation that was used for its computation. Always use the other equation!

## Solve the following system of linear equations by method of substitution. y = 24 – 4x 2x + y/2 = 12

y = 24 – 4x.

As shown in the above example, we compute the variable value from one equation and substitute it into the other.

y = 24 – 4x ------(1) 2x + y/2 = 12 ------(2)

Here we choose equation (1) to compute the value of x. Since equation (1) is already in its most simplified form:

(Putting this value of y into equation (2) and then solving for x gives)

2x + (24-4x)/2 = 12 ------(2) (∵ y = 24 – 4x) 2x + 24/2- 4x/2 = 12 2x + 12 – 2x = 12 12 = 12

You might feel that this is the same scenario as discussed above (that of 24 = 24). But wait! You are trying to jump at a conclusion a bit too early. In the previous scenario, the result 24 = 24 had resulted because we put the variable value into the same equation that we used for its computation. Here we have not done that.

The result 12 = 12 has got something to do with the nature of the system of equations that we are given. No matter what solving technique you might be using, a solution to a system of linear equations lies at a single point where their lines intersect. In this scenario, the two lines are basically the same (one line over the other. The following figure shows this scenario.

Such a system is called a dependent system of equations. And solution to such a system is the entire line (every point on the line is a point of intersection of the two lines)

Hence the solution to the given system of equations is the entire line: y = 24 – 4x

Another possible Scenario:

Similar to this example, there exists another scenario where substitution of one variable into the 2 nd equation leads to a result similar to one shown below:

Such a scenario arises when there exists no solution to the given system of equations. I.e., when the two lines do not intersect at any point at all.

Hence in case of such a result, where your basic Math rules seem to fail, a simple conclusion is that no solution to the given system exists. Such a system of equations is called an Inconsistent system.

## Popular Algebra Links

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## How to Solve an Algebra Problem

## Introduction: How to Solve an Algebra Problem

This instructable is about how to solve hard algebra problems. You may have to look closely on some of the steps, though. Sorry about that.

## Step 1: Write Down the Problem

The first thing you do when solving an algebra problem is write the problem and the steps down. Teachers love seeing the steps and it makes it easier to catch mistakes. FYI, I didn't get this problem out of a textbook. I just randomly came up with it.

## Step 2: PEMDAS

The acronym PEMDAS is something you will use every time you work with equations. It stands for Parenthesis, Exponents, Multiply, Divide, Add, and Subtract. Before we begin solving, I would like to say that we are leaving the 180x for later. Don't do anything to it yet.

## Step 3: Solve the Parenthesis

There are 2 ways to solve parenthesis: the Distribution Property, or just solving the equation in the parenthesis. During Distribution, you multiply the number immediately outside the parenthesis by each of the integers on the inside. Either method works, so do whichever you like most. I stuck with just solving the equation that was in the parenthesis: 30/3=10.

## Step 4: Handle the Exponents/ Square Roots

There are no exponents in this problem, but I drew one so you could see what one looks like. When dealing with exponents, you do not multiply the big number by the little one. The exponent (little number) tells you how many times the big number is multiplied by itself . In this case; 3*3=9

## Step 5: Multiply

Multiplication can be viewed in many forms, such as 2(10), 2*10, or 2x10. All of these mean the exact same thing: 2 times 10. In algebra, however, the variable "x" is quite common, so you may want to stop using that form. Multiply 2(10)=20, and 5*11=55.

## Step 6: Divide

Now we divide. Multiplying and dividing can be done in the same step, but I just chose not to. 2/1=2.

## Step 7: Add/ Subtract (aka, Combine Like Terms)

Now you add and subtract. In algebra, we do this thing called "combining like terms." This means that we see 17-55 as a 17 and -55. We just add those, and the rest of the numbers up in a calculator to get the next answer. I circled the positive numbers and boxed the negative number to help you see my point. 20+17-55+2=-16.

## Step 8: Find X by Division

Next, we solve for x. We do this by dividing both side of the equal sign by 180. The slash shows that the 2 "180's" cancel each other out, leaving x behind (180x means 180*x, by the way). X= -16/180.

## Step 9: The Final Answer

Lastly, divide -16 by 180, and you have your final answer!

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Question 1 year ago on Step 8

what makes it -16?

Question 1 year ago on Step 9

when I divide the last step I get -0.08888889 Why go with 0.08 and not 0.08888889?

Tip 2 years ago

I LIKE YA CUT G

2 years ago

HI I LIKE YA CUT G

Question 2 years ago on Step 5

How did you multipled your 5.11 which gave you 55

10 years ago on Step 9

You can leave the final answer as the fraction -16/180, but I don't like answers that end in fractions, so I made it a decimal by dividing -16 by 180.

Reply 4 years ago

correct me if im wrong but, 180x=2(30 / 3) + 17- 5 X 11+ 2 / 11 = 49/495 proof, 180x = 2 X 30/3 + 17- 5 X 11+ 2/11 180X = 60/3+17- 5 x 11 + 2/11 180x= 20 + 17- 55+ 2/11 180x = 196/11 x= 196/11/180 x =196/ 11 x 180 x = 49/495 srry if im incorrect im in 5th grade

Reply 7 years ago

If you were to leave your answer as a factor you should factor it though.

-16/180 is -4/45

8 years ago on Introduction

Hey, nice ible, although there is one mistake in the second pic in step 3. You can't just multiply both integers in the parenthesis. You have to imagine it as a fraction and multiply just the first number, so it's like 2(30/3) = (2*30)/3 = 60/3 = 20, not 60/6 ;)

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## What can QuickMath do?

QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students.

- The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and cancelling common factors within a fraction.
- The equations section lets you solve an equation or system of equations. You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require.
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## Praxis Core Math

Course: praxis core math > unit 1.

- Algebraic properties | Lesson
- Algebraic properties | Worked example
- Solution procedures | Lesson
- Solution procedures | Worked example
- Equivalent expressions | Lesson
- Equivalent expressions | Worked example
- Creating expressions and equations | Lesson
- Creating expressions and equations | Worked example

## Algebraic word problems | Lesson

- Algebraic word problems | Worked example
- Linear equations | Lesson
- Linear equations | Worked example
- Quadratic equations | Lesson
- Quadratic equations | Worked example

## What are algebraic word problems?

What skills are needed.

- Translating sentences to equations
- Solving linear equations with one variable
- Evaluating algebraic expressions
- Solving problems using Venn diagrams

## How do we solve algebraic word problems?

- Define a variable.
- Write an equation using the variable.
- Solve the equation.
- If the variable is not the answer to the word problem, use the variable to calculate the answer.

## What's a Venn diagram?

- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text
- (Choice A) $ 4 \$4 $ 4 dollar sign, 4 A $ 4 \$4 $ 4 dollar sign, 4
- (Choice B) $ 5 \$5 $ 5 dollar sign, 5 B $ 5 \$5 $ 5 dollar sign, 5
- (Choice C) $ 9 \$9 $ 9 dollar sign, 9 C $ 9 \$9 $ 9 dollar sign, 9
- (Choice D) $ 14 \$14 $ 1 4 dollar sign, 14 D $ 14 \$14 $ 1 4 dollar sign, 14
- (Choice E) $ 20 \$20 $ 2 0 dollar sign, 20 E $ 20 \$20 $ 2 0 dollar sign, 20
- (Choice A) 10 10 1 0 10 A 10 10 1 0 10
- (Choice B) 12 12 1 2 12 B 12 12 1 2 12
- (Choice C) 24 24 2 4 24 C 24 24 2 4 24
- (Choice D) 30 30 3 0 30 D 30 30 3 0 30
- (Choice E) 32 32 3 2 32 E 32 32 3 2 32
- (Choice A) 4 4 4 4 A 4 4 4 4
- (Choice B) 10 10 1 0 10 B 10 10 1 0 10
- (Choice C) 14 14 1 4 14 C 14 14 1 4 14
- (Choice D) 18 18 1 8 18 D 18 18 1 8 18
- (Choice E) 22 22 2 2 22 E 22 22 2 2 22

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- Math Article

## Algebra Problems

Algebra problems are not only based on algebraic expressions but also on various types of equations in Maths where a quantity or variable is unknown to us. Many of us are familiar with the word problem, but are we aware of the fact and problems related to variables and constants? When we say 5 it means a number but what if we say x=5 or 5y or something like that?

This is where algebra came into existence algebra is that branch of mathematics which not only deals with numbers but also variable and alphabets. The versatility of Algebra is very deep and very conceptual, all the non-numeric character represents variable and numeric as constants. Let us solve some problems based algebra with solutions which will cover the syllabus for class 6, 7, 8. Below are some of the examples of algebraic expressions .

For example.

Algebra Word Problems deal with real-time situations and solutions which can be solved using algebra.

## Basic Algebra Identities

- (a + b) 2 = a 2 + b 2 + 2ab
- (a – b) 2 = a 2 + b 2 – 2ab
- a 2 – b 2 = (a + b)(a – b)
- a 2 + b 2 = (a + b) 2 – 2ab = (a – b) 2 + 2ab
- a 3 + b 3 = (a + b)(a 2 – ab + b 2 )
- a 3 – b 3 = (a – b)(a 2 + ab + b 2 )
- (a + b) 3 = a 3 + 3ab(a + b) + b 3
- (a – b) 3 = a 3 – 3ab(a – b) – b 3

## Algebra problems With Solutions

Example 1: Solve, (x-1) 2 = [4√(x-4)] 2 Solution: x 2 -2x+1 = 16(x-4)

x 2 -2x+1 = 16x-64

x 2 -18x+65 = 0

(x-13) (x-5) = 0

Hence, x = 13 and x = 5.

## Algebra Problems for Class 6

In class 6, students will be introduced with an algebra concept. Here, you will learn how the unknown values are represented in terms of variables. The given expression can be solved only if we know the value of unknown variable. Let us see some examples.

Example: Solve, 4x + 5 when, x = 3.

Solution: Given, 4x + 5

Now putting the value of x=3, we get;

4 (3) + 5 = 12 + 5 = 17.

Example: Give expressions for the following cases:

(i) 12 added to 2x

(ii) 6 multiplied by y

(iii) 25 subtracted from z

(iv) 17 times of m

(i) 12 + 2x

## Algebra Problems for Class 7

In class 7, students will deal with algebraic expressions like x+y, xy, 32x 2 -12y 2 , etc. There are different kinds of the terminology used in case algebraic equations such as;

- Coefficient

Let us understand these terms with an example. Suppose 4x + 5y is an algebraic expression, then 4x and 5y are the terms. Since here the variables used are x and y, therefore, x and y are the factors of 4x + 5y. And the numerical factor attached to the variables are the coefficient such as 4 and 5 are the coefficient of x and y in the given expression.

Any expression with one or more terms is called a polynomial. Specifically, a one-term expression is called a monomial; a two-term expression is called a binomial, and a three-term expression is called a trinomial.

Terms which have the same algebraic factors are like terms . Terms which have different algebraic factors are unlike terms . Thus, terms 4xy and – 3xy are like terms; but terms 4xy and – 3x are not like terms.

Example: Add 3x + 5x

Solution: Since 3x and 5x have the same algebraic factors, hence, they are like terms and can be added by their coefficient.

3x + 5x = 8x

Example: Collect like terms and simplify the expression: 12x 2 – 9x + 5x – 4x 2 – 7x + 10.

Solution: 12x 2 – 9x + 5x – 4x 2 – 7x + 10

= (12 – 4)x 2 – 9x + 5x – 7x + 10

= 8x 2 – 11x + 10

## Algebra Problems for Class 8

Here, students will deal with algebraic identities. See the examples.

Example: Solve (2x+y) 2

Solution: Using the identity: (a+b) 2 = a 2 + b 2 + 2 ab, we get;

(2x+y) = (2x) 2 + y 2 + 2.2x.y = 4x 2 + y 2 + 4xy

Example: Solve (99) 2 using algebraic identity.

Solution: We can write, 99 = 100 -1

Therefore, (100 – 1 ) 2

= 100 2 + 1 2 – 2 x 100 x 1 [By identity: (a -b) 2 = a 2 + b 2 – 2ab

= 10000 + 1 – 200

## Algebra Word Problems

Question 1: There are 47 boys in the class. This is three more than four times the number of girls. How many girls are there in the class?

Solution: Let the number of girls be x

As per the given statement,

4 x + 3 = 47

4x = 47 – 3

Question 2: The sum of two consecutive numbers is 41. What are the numbers?

Solution: Let one of the numbers be x.

Then the other number will x+1

Now, as per the given questions,

x + x + 1 = 41

2x + 1 = 41

So, the first number is 20 and second number is 20+1 = 21

## Linear Algebra Problems

There are various methods For Solving the Linear Equations

- Cross multiplication method
- Replacement method or Substitution method
- Hit and trial method

There are Variety of different Algebra problem present and are solved depending upon their functionality and state. For example, a linear equation problem can’t be solved using a quadratic equation formula and vice verse for, e.g., x+x/2=7 then solve for x is an equation in one variable for x which can be satisfied by only one value of x. Whereas x 2 +5x+6 is a quadratic equation which is satisfied for two values of x the domain of algebra is huge and vast so for more information. Visit BYJU’S. where different techniques are explained different algebra problem.

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## Algebra Problems

You may solve a set of 10 questions with their detailed solutions and also a set of 50 questions, with their answers, in the applet to self test you background on how to

- Solve linear equations.
- Simplify algebraic expressions.
- Simplify absolute value expressions.
- Find the distance between two points.
- Find x intercept(s) of the graph of an equation.
- Evaluate functions.
- Find the slope of a line passing through two points.
- Find slope of a line from its equation.
- Find equation of a line.
- Solve equation with absolute value

## Algebra Problems with Detailed Solutions

Problem 1: Solve the equation

Detailed Solution.

Problem 2: Simplify the expression

Problem 3: If x <2, simplify

Problem 4: Find the distance between the points (-4 , -5) and (-1 , -1).

Problem 5: Find the x intercept of the graph of the equation .

Problem 6: Evaluate f(2) - f(1)

Problem 7: Find the slope of the line passing through the points (-1, -1) and (2 , 2).

Problem 8: Find the slope of the line

Problem 9: Find the equation of the line that passes through the points (-1 , -1) and (-1 , 2).

Problem 10: Solve the equation

## More References and links

- Step by Step Math Worksheets Solvers New !
- Math Problems and Online Self Tests.
- Basic Rules and Properties of Algebra.
- More Intermediate and College Algebra Questions and Problems with Answers .

## Solving Word Questions

With LOTS of examples!

In Algebra we often have word questions like:

## Example: Sam and Alex play tennis.

On the weekend Sam played 4 more games than Alex did, and together they played 12 games.

How many games did Alex play?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

## Turning English into Algebra

To turn the English into Algebra it helps to:

- Read the whole thing first
- Do a sketch if possible
- Assign letters for the values
- Find or work out formulas

You should also write down what is actually being asked for , so you know where you are going and when you have arrived!

Also look for key words:

## Thinking Clearly

Some wording can be tricky, making it hard to think "the right way around", such as:

## Example: Sam has 2 dollars less than Alex. How do we write this as an equation?

- Let S = dollars Sam has
- Let A = dollars Alex has

Now ... is that: S − 2 = A

or should it be: S = A − 2

or should it be: S = 2 − A

The correct answer is S = A − 2

( S − 2 = A is a common mistake, as the question is written "Sam ... 2 less ... Alex")

## Example: on our street there are twice as many dogs as cats. How do we write this as an equation?

- Let D = number of dogs
- Let C = number of cats

Now ... is that: 2D = C

or should it be: D = 2C

Think carefully now!

The correct answer is D = 2C

( 2D = C is a common mistake, as the question is written "twice ... dogs ... cats")

Let's start with a really simple example so we see how it's done:

## Example: A rectangular garden is 12m by 5m, what is its area ?

Turn the English into Algebra:

- Use w for width of rectangle: w = 12m
- Use h for height of rectangle: h = 5m

Formula for Area of a Rectangle : A = w × h

We are being asked for the Area.

A = w × h = 12 × 5 = 60 m 2

The area is 60 square meters .

Now let's try the example from the top of the page:

## Example: Sam and Alex play Tennis. On the weekend Sam played 4 more games than Alex did, and together they played 12 games. How many games did Alex play?

- Use S for how many games Sam played
- Use A for how many games Alex played

We know that Sam played 4 more games than Alex, so: S = A + 4

And we know that together they played 12 games: S + A = 12

We are being asked for how many games Alex played: A

Which means that Alex played 4 games of tennis.

Check: Sam played 4 more games than Alex, so Sam played 8 games. Together they played 8 + 4 = 12 games. Yes!

A slightly harder example:

## Example: Alex and Sam also build tables. Together they make 10 tables in 12 days. Alex working alone can make 10 in 30 days. How long would it take Sam working alone to make 10 tables?

- Use a for Alex's work rate
- Use s for Sam's work rate

12 days of Alex and Sam is 10 tables, so: 12a + 12s = 10

30 days of Alex alone is also 10 tables: 30a = 10

We are being asked how long it would take Sam to make 10 tables.

30a = 10 , so Alex's rate (tables per day) is: a = 10/30 = 1/3

Which means that Sam's rate is half a table a day (faster than Alex!)

So 10 tables would take Sam just 20 days.

Should Sam be paid more I wonder?

And another "substitution" example:

## Example: Jenna is training hard to qualify for the National Games. She has a regular weekly routine, training for five hours a day on some days and 3 hours a day on the other days. She trains altogether 27 hours in a seven day week. On how many days does she train for five hours?

- The number of "5 hour" days: d
- The number of "3 hour" days: e

We know there are seven days in the week, so: d + e = 7

And she trains 27 hours in a week, with d 5 hour days and e 3 hour days: 5d + 3e = 27

We are being asked for how many days she trains for 5 hours: d

The number of "5 hour" days is 3

Check : She trains for 5 hours on 3 days a week, so she must train for 3 hours a day on the other 4 days of the week.

3 × 5 hours = 15 hours, plus 4 × 3 hours = 12 hours gives a total of 27 hours

Some examples from Geometry:

## Example: A circle has an area of 12 mm 2 , what is its radius?

- Use A for Area: A = 12 mm 2
- Use r for radius

And the formula for Area is: A = π r 2

We are being asked for the radius.

We need to rearrange the formula to find the area

## Example: A cube has a volume of 125 mm 3 , what is its surface area?

Make a quick sketch:

- Use V for Volume
- Use A for Area
- Use s for side length of cube
- Volume of a cube: V = s 3
- Surface area of a cube: A = 6s 2

We are being asked for the surface area.

First work out s using the volume formula:

Now we can calculate surface area:

An example about Money:

## Example: Joel works at the local pizza parlor. When he works overtime he earns 1¼ times the normal rate. One week Joel worked for 40 hours at the normal rate of pay and also worked 12 hours overtime. If Joel earned $660 altogether in that week, what is his normal rate of pay?

- Joel's normal rate of pay: $N per hour
- Joel works for 40 hours at $N per hour = $40N
- When Joel does overtime he earns 1¼ times the normal rate = $1.25N per hour
- Joel works for 12 hours at $1.25N per hour = $(12 × 1¼N) = $15N
- And together he earned $660, so:

$40N + $(12 × 1¼N) = $660

We are being asked for Joel's normal rate of pay $N.

So Joel’s normal rate of pay is $12 per hour

Joel’s normal rate of pay is $12 per hour, so his overtime rate is 1¼ × $12 per hour = $15 per hour. So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660

More about Money, with these two examples involving Compound Interest

## Example: Alex puts $2000 in the bank at an annual compound interest of 11%. How much will it be worth in 3 years?

This is the compound interest formula:

So we will use these letters:

- Present Value PV = $2,000
- Interest Rate (as a decimal): r = 0.11
- Number of Periods: n = 3
- Future Value (the value we want): FV

We are being asked for the Future Value: FV

## Example: Roger deposited $1,000 into a savings account. The money earned interest compounded annually at the same rate. After nine years Roger's deposit has grown to $1,551.33 What was the annual rate of interest for the savings account?

The compound interest formula:

- Present Value PV = $1,000
- Interest Rate (the value we want): r
- Number of Periods: n = 9
- Future Value: FV = $1,551.33

We are being asked for the Interest Rate: r

So the annual rate of interest is 5%

Check : $1,000 × (1.05) 9 = $1,000 × 1.55133 = $1,551.33

And an example of a Ratio question:

## Example: At the start of the year the ratio of boys to girls in a class is 2 : 1 But now, half a year later, four boys have left the class and there are two new girls. The ratio of boys to girls is now 4 : 3 How many students are there altogether now?

- Number of boys now: b
- Number of girls now: g

The current ratio is 4 : 3

Which can be rearranged to 3b = 4g

At the start of the year there was (b + 4) boys and (g − 2) girls, and the ratio was 2 : 1

b + 4 g − 2 = 2 1

Which can be rearranged to b + 4 = 2(g − 2)

We are being asked for how many students there are altogether now: b + g

There are 12 girls !

And 3b = 4g , so b = 4g/3 = 4 × 12 / 3 = 16 , so there are 16 boys

So there are now 12 girls and 16 boys in the class, making 28 students altogether .

There are now 16 boys and 12 girls, so the ratio of boys to girls is 16 : 12 = 4 : 3 At the start of the year there were 20 boys and 10 girls, so the ratio was 20 : 10 = 2 : 1

And now for some Quadratic Equations :

## Example: The product of two consecutive even integers is 168. What are the integers?

Consecutive means one after the other. And they are even , so they could be 2 and 4, or 4 and 6, etc.

We will call the smaller integer n , and so the larger integer must be n+2

And we are told the product (what we get after multiplying) is 168, so we know:

n(n + 2) = 168

We are being asked for the integers

That is a Quadratic Equation , and there are many ways to solve it. Using the Quadratic Equation Solver we get −14 and 12.

Check −14: −14(−14 + 2) = (−14)×(−12) = 168 YES

Check 12: 12(12 + 2) = 12×14 = 168 YES

So there are two solutions: −14 and −12 is one, 12 and 14 is the other.

Note: we could have also tried "guess and check":

- We could try, say, n=10: 10(12) = 120 NO (too small)
- Next we could try n=12: 12(14) = 168 YES

But unless we remember that multiplying two negatives make a positive we might overlook the other solution of (−14)×(−12).

## Example: You are an Architect. Your client wants a room twice as long as it is wide. They also want a 3m wide veranda along the long side. Your client has 56 square meters of beautiful marble tiles to cover the whole area. What should the length of the room be?

Let's first make a sketch so we get things right!:

- the length of the room: L
- the width of the room: W
- the total Area including veranda: A
- the width of the room is half its length: W = ½L
- the total area is the (room width + 3) times the length: A = (W+3) × L = 56

We are being asked for the length of the room: L

This is a quadratic equation , there are many ways to solve it, this time let's use factoring :

And so L = 8 or −14

There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!

So the length of the room is 8 m

L = 8, so W = ½L = 4

So the area of the rectangle = (W+3) × L = 7 × 8 = 56

There we are ...

... I hope these examples will help you get the idea of how to handle word questions. Now how about some practice?

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