problem solving equations with fractions lesson 8 7

Solving Equations with Fractions

Related Topics: More Lessons for Grade 8 Math Math Worksheets

Examples, solutions, videos, worksheets, stories and songs to help Grade 8 students learn how to solve equations with fractions.

In this lesson, we will learn how to solve

• one-step equations with fractions
• two-step equations with fractions
• multi-step equations with fractions

Solve One Step Equations With Fraction by Adding or Subtracting This video provides two examples of how to solve a one step linear equation by adding and subtracting.

Solve One Step Equations With Fraction by Multiplying This video provides two examples of how to solve a one step linear equation by multiplying.

Solving One Step Equations Involving Fractions

Solve One Step Equations with Fractions This video provides several examples of how to solve one step equation containing fractions.

Solving Two Step Equations Involving Fractions This video explains how to solve two step equations involving fractions.

Solving Equations with Fractions Learn to solve equations that involve fractions by either multiplying both sides of the equation by the reciprocal of the fraction, or multiplying both sides of the equation by the denominator of the fraction. The goal is to get rid of the fraction as soon as possible.

Equations with fractions in them How to solve linear equations that contain fractions.

Solving Multi-Step Equations with Fractions This lesson looks at four examples and the process that can be used to solve equations that contain fractions

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Equations involving Fractions Practice Questions

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4.7 Solve Equations with Fractions

Learning objectives.

• Determine whether a fraction is a solution of an equation
• Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality
• Solve equations using the Multiplication Property of Equality
• Translate sentences to equations and solve

Be Prepared 4.7

Before you get started, take this readiness quiz. If you miss a problem, go back to the section listed and review the material.

• Evaluate x + 4 when x = −3 x + 4 when x = −3 If you missed this problem, review Example 3.23 .
• Solve: 2 y − 3 = 9 . 2 y − 3 = 9 . If you missed this problem, review Example 3.61 .
• Solve: y − 3 = −9 y − 3 = −9 If you missed this problem, review Example 4.28 .

Determine Whether a Fraction is a Solution of an Equation

As we saw in Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality , a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. In those sections, we found whole number and integer solutions to equations. Now that we have worked with fractions, we are ready to find fraction solutions to equations.

The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.

Determine whether a number is a solution to an equation.

• Step 1. Substitute the number for the variable in the equation.
• Step 2. Simplify the expressions on both sides of the equation.
• Step 3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

Example 4.95

Determine whether each of the following is a solution of x − 3 10 = 1 2 . x − 3 10 = 1 2 .

• ⓐ x = 1 x = 1
• ⓑ x = 4 5 x = 4 5
• ⓒ x = − 4 5 x = − 4 5

Since x = 1 x = 1 does not result in a true equation, 1 1 is not a solution to the equation.

Since x = 4 5 x = 4 5 results in a true equation, 4 5 4 5 is a solution to the equation x − 3 10 = 1 2 . x − 3 10 = 1 2 .

Since x = − 4 5 x = − 4 5 does not result in a true equation, − 4 5 − 4 5 is not a solution to the equation.

Try It 4.189

Determine whether each number is a solution of the given equation.

x − 2 3 = 1 6 : x − 2 3 = 1 6 :

• ⓑ x = 5 6 x = 5 6
• ⓒ x = − 5 6 x = − 5 6

Try It 4.190

y − 1 4 = 3 8 : y − 1 4 = 3 8 :

• ⓐ y = 1 y = 1
• ⓑ y = − 5 8 y = − 5 8
• ⓒ y = 5 8 y = 5 8

Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality

In Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality , we solved equations using the Addition, Subtraction, and Division Properties of Equality. We will use these same properties to solve equations with fractions.

Addition, Subtraction, and Division Properties of Equality

For any numbers a , b , and c , a , b , and c ,

In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.

Example 4.96

Solve: y + 9 16 = 5 16 . y + 9 16 = 5 16 .

Since y = − 1 4 y = − 1 4 makes y + 9 16 = 5 16 y + 9 16 = 5 16 a true statement, we know we have found the solution to this equation.

Try It 4.191

Solve: y + 11 12 = 5 12 . y + 11 12 = 5 12 .

Try It 4.192

Solve: y + 8 15 = 4 15 . y + 8 15 = 4 15 .

We used the Subtraction Property of Equality in Example 4.96 . Now we’ll use the Addition Property of Equality.

Example 4.97

Solve: a − 5 9 = − 8 9 . a − 5 9 = − 8 9 .

Since a = − 1 3 a = − 1 3 makes the equation true, we know that a = − 1 3 a = − 1 3 is the solution to the equation.

Try It 4.193

Solve: a − 3 5 = − 8 5 . a − 3 5 = − 8 5 .

Try It 4.194

Solve: n − 3 7 = − 9 7 . n − 3 7 = − 9 7 .

The next example may not seem to have a fraction, but let’s see what happens when we solve it.

Example 4.98

Solve: 10 q = 44 . 10 q = 44 .

The solution to the equation was the fraction 22 5 . 22 5 . We leave it as an improper fraction.

Try It 4.195

Solve: 12 u = −76 . 12 u = −76 .

Try It 4.196

Solve: 8 m = 92 . 8 m = 92 .

Solve Equations with Fractions Using the Multiplication Property of Equality

Consider the equation x 4 = 3 . x 4 = 3 . We want to know what number divided by 4 4 gives 3 . 3 . So to “undo” the division, we will need to multiply by 4 . 4 . The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.

The Multiplication Property of Equality

For any numbers a , b , a , b , and c , c ,

If you multiply both sides of an equation by the same quantity, you still have equality.

Let’s use the Multiplication Property of Equality to solve the equation x 7 = −9 . x 7 = −9 .

Example 4.99

Solve: x 7 = −9 . x 7 = −9 .

Try It 4.197

Solve: f 5 = −25 . f 5 = −25 .

Try It 4.198

Solve: h 9 = −27 . h 9 = −27 .

Example 4.100

Solve: p −8 = −40 . p −8 = −40 .

Here, p p is divided by −8 . −8 . We must multiply by −8 −8 to isolate p . p .

Try It 4.199

Solve: c −7 = −35 . c −7 = −35 .

Try It 4.200

Solve: x −11 = −12 . x −11 = −12 .

Solve Equations with a Coefficient of −1 −1

Look at the equation − y = 15 . − y = 15 . Does it look as if y y is already isolated? But there is a negative sign in front of y , y , so it is not isolated.

There are three different ways to isolate the variable in this type of equation. We will show all three ways in Example 4.101 .

Example 4.101

Solve: − y = 15 . − y = 15 .

One way to solve the equation is to rewrite − y − y as −1 y , −1 y , and then use the Division Property of Equality to isolate y . y .

Another way to solve this equation is to multiply both sides of the equation by −1 . −1 .

The third way to solve the equation is to read − y − y as “the opposite of y .” y .” What number has 15 15 as its opposite? The opposite of 15 15 is −15 . −15 . So y = −15 . y = −15 .

For all three methods, we isolated y y is isolated and solved the equation.

Try It 4.201

Solve: − y = 48 . − y = 48 .

Try It 4.202

Solve: − c = −23 . − c = −23 .

Solve Equations with a Fraction Coefficient

When we have an equation with a fraction coefficient we can use the Multiplication Property of Equality to make the coefficient equal to 1 . 1 .

For example, in the equation:

The coefficient of x x is 3 4 . 3 4 . To solve for x , x , we need its coefficient to be 1 . 1 . Since the product of a number and its reciprocal is 1 , 1 , our strategy here will be to isolate x x by multiplying by the reciprocal of 3 4 . 3 4 . We will do this in Example 4.102 .

Example 4.102

Solve: 3 4 x = 24 . 3 4 x = 24 .

Notice that in the equation 3 4 x = 24 , 3 4 x = 24 , we could have divided both sides by 3 4 3 4 to get x x by itself. Dividing is the same as multiplying by the reciprocal, so we would get the same result. But most people agree that multiplying by the reciprocal is easier.

Try It 4.203

Solve: 2 5 n = 14 . 2 5 n = 14 .

Try It 4.204

Solve: 5 6 y = 15 . 5 6 y = 15 .

Example 4.103

Solve: − 3 8 w = 72 . − 3 8 w = 72 .

The coefficient is a negative fraction. Remember that a number and its reciprocal have the same sign, so the reciprocal of the coefficient must also be negative.

Try It 4.205

Solve: − 4 7 a = 52 . − 4 7 a = 52 .

Try It 4.206

Solve: − 7 9 w = 84 . − 7 9 w = 84 .

Translate Sentences to Equations and Solve

Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.

When you add, subtract, multiply or divide the same quantity from both sides of an equation, you still have equality.

In the next few examples, we’ll translate sentences into equations and then solve the equations. It might be helpful to review the translation table in Evaluate, Simplify, and Translate Expressions .

Example 4.104

Translate and solve: n n divided by 6 6 is −24 . −24 .

Try It 4.207

Translate and solve: n n divided by 7 7 is equal to −21 . −21 .

Try It 4.208

Translate and solve: n n divided by 8 8 is equal to −56 . −56 .

Example 4.105

Translate and solve: The quotient of q q and −5 −5 is 70 . 70 .

Try It 4.209

Translate and solve: The quotient of q q and −8 −8 is 72 . 72 .

Try It 4.210

Translate and solve: The quotient of p p and −9 −9 is 81 . 81 .

Example 4.106

Translate and solve: Two-thirds of f f is 18 . 18 .

Try It 4.211

Translate and solve: Two-fifths of f f is 16 . 16 .

Try It 4.212

Translate and solve: Three-fourths of f f is 21 . 21 .

Example 4.107

Translate and solve: The quotient of m m and 5 6 5 6 is 3 4 . 3 4 .

Our solution checks.

Try It 4.213

Translate and solve. The quotient of n n and 2 3 2 3 is 5 12 . 5 12 .

Try It 4.214

Translate and solve The quotient of c c and 3 8 3 8 is 4 9 . 4 9 .

Example 4.108

Translate and solve: The sum of three-eighths and x x is three and one-half.

We write the answer as a mixed number because the original problem used a mixed number.

Is the sum of three-eighths and 3 1 8 3 1 8 equal to three and one-half?

The solution checks.

Try It 4.215

Translate and solve: The sum of five-eighths and x x is one-fourth.

Try It 4.216

Translate and solve: The difference of one-and-three-fourths and x x is five-sixths.

ACCESS ADDITIONAL ONLINE RESOURCES

• Solve One Step Equations With Fractions
• Solve One Step Equations With Fractions by Adding or Subtracting
• Solve One Step Equations With Fraction by Multiplying

Section 4.7 Exercises

Practice makes perfect.

In the following exercises, determine whether each number is a solution of the given equation.

x − 2 5 = 1 10 : x − 2 5 = 1 10 :

• ⓑ x = 1 2 x = 1 2
• ⓒ x = − 1 2 x = − 1 2

y − 1 3 = 5 12 : y − 1 3 = 5 12 :

• ⓑ y = 3 4 y = 3 4
• ⓒ y = − 3 4 y = − 3 4

h + 3 4 = 2 5 : h + 3 4 = 2 5 :

• ⓐ h = 1 h = 1
• ⓑ h = 7 20 h = 7 20
• ⓒ h = − 7 20 h = − 7 20

k + 2 5 = 5 6 : k + 2 5 = 5 6 :

• ⓐ k = 1 k = 1
• ⓑ k = 13 30 k = 13 30
• ⓒ k = − 13 30 k = − 13 30

In the following exercises, solve.

y + 1 3 = 4 3 y + 1 3 = 4 3

m + 3 8 = 7 8 m + 3 8 = 7 8

f + 9 10 = 2 5 f + 9 10 = 2 5

h + 5 6 = 1 6 h + 5 6 = 1 6

a − 5 8 = − 7 8 a − 5 8 = − 7 8

c − 1 4 = − 5 4 c − 1 4 = − 5 4

x − ( − 3 20 ) = − 11 20 x − ( − 3 20 ) = − 11 20

z − ( − 5 12 ) = − 7 12 z − ( − 5 12 ) = − 7 12

n − 1 6 = 3 4 n − 1 6 = 3 4

p − 3 10 = 5 8 p − 3 10 = 5 8

s + ( − 1 2 ) = − 8 9 s + ( − 1 2 ) = − 8 9

k + ( − 1 3 ) = − 4 5 k + ( − 1 3 ) = − 4 5

5 j = 17 5 j = 17

7 k = 18 7 k = 18

−4 w = 26 −4 w = 26

−9 v = 33 −9 v = 33

f 4 = −20 f 4 = −20

b 3 = −9 b 3 = −9

y 7 = −21 y 7 = −21

x 8 = −32 x 8 = −32

p −5 = −40 p −5 = −40

q −4 = −40 q −4 = −40

r −12 = −6 r −12 = −6

s −15 = −3 s −15 = −3

− x = 23 − x = 23

− y = 42 − y = 42

− h = − 5 12 − h = − 5 12

− k = − 17 20 − k = − 17 20

4 5 n = 20 4 5 n = 20

3 10 p = 30 3 10 p = 30

3 8 q = −48 3 8 q = −48

5 2 m = −40 5 2 m = −40

− 2 9 a = 16 − 2 9 a = 16

− 3 7 b = 9 − 3 7 b = 9

− 6 11 u = −24 − 6 11 u = −24

− 5 12 v = −15 − 5 12 v = −15

Mixed Practice

3 x = 0 3 x = 0

8 y = 0 8 y = 0

4 f = 4 5 4 f = 4 5

7 g = 7 9 7 g = 7 9

p + 2 3 = 1 12 p + 2 3 = 1 12

q + 5 6 = 1 12 q + 5 6 = 1 12

7 8 m = 1 10 7 8 m = 1 10

1 4 n = 7 10 1 4 n = 7 10

− 2 5 = x + 3 4 − 2 5 = x + 3 4

− 2 3 = y + 3 8 − 2 3 = y + 3 8

11 20 = − f 11 20 = − f

8 15 = − d 8 15 = − d

In the following exercises, translate to an algebraic equation and solve.

n n divided by eight is −16 . −16 .

n n divided by six is −24 . −24 .

m m divided by −9 −9 is −7 . −7 .

m m divided by −7 −7 is −8 . −8 .

The quotient of f f and −3 −3 is −18 . −18 .

The quotient of f f and −4 −4 is −20 . −20 .

The quotient of g g and twelve is 8 . 8 .

The quotient of g g and nine is 14 . 14 .

Three-fourths of q q is 12 . 12 .

Two-fifths of q q is 20 . 20 .

Seven-tenths of p p is −63 . −63 .

Four-ninths of p p is −28 . −28 .

m m divided by 4 4 equals negative 6 . 6 .

The quotient of h h and 2 2 is 43 . 43 .

Three-fourths of z z is the same as 15 . 15 .

The quotient of a a and 2 3 2 3 is 3 4 . 3 4 .

The sum of five-sixths and x x is 1 2 . 1 2 .

The sum of three-fourths and x x is 1 8 . 1 8 .

The difference of y y and one-fourth is − 1 8 . − 1 8 .

The difference of y y and one-third is − 1 6 . − 1 6 .

Everyday Math

Shopping Teresa bought a pair of shoes on sale for $48 . $48 . The sale price was 2 3 2 3 of the regular price. Find the regular price of the shoes by solving the equation 2 3 p = 48 2 3 p = 48

Playhouse The table in a child’s playhouse is 3 5 3 5 of an adult-size table. The playhouse table is 18 18 inches high. Find the height of an adult-size table by solving the equation 3 5 h = 18 . 3 5 h = 18 .

Writing Exercises

Example 4.100 describes three methods to solve the equation − y = 15 . − y = 15 . Which method do you prefer? Why?

Richard thinks the solution to the equation 3 4 x = 24 3 4 x = 24 is 16 . 16 . Explain why Richard is wrong.

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

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

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Equations With Fractions

Here we will learn about equations with fractions, including solving equations with fractions where the unknown is the denominator of a fraction.

There are also equations with fractions worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are equations with fractions?

Equations with fractions involve solving equations where the unknown variable is part of the numerator and/or the denominator of the fraction.  

To solve equations with fractions we need to work out what the value of the unknown variable. We solve equations by using the “balancing method” by applying the inverse operation to both sides of the equation.

The inverse operation of addition is subtraction .

The inverse operation of subtraction is addition .

The inverse operation of multiplication is division .

The inverse operation of division is multiplication .

For example,

How to solve equations with fractions

In order to solve equations with fraction:

• Identify the operations that are being applied to the unknown variable.
• Apply the inverse operations, one at a time, to both sides of the equation.
• Write the final answer, checking that it is correct.

Equations with fractions worksheet

Get your free Equations with fractions worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on solving equations

Equations with fractions  is part of our series of lessons to support revision on  solving equations . You may find it helpful to start with the main solving equations lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Solving equations
• Quadratic equations  
• Linear equations
• Forming and solving equations
• Iteration maths

Equations with fractions examples

Example 1: equations with one operation.

Identify the operations that are being applied to the unknown variable .

The unknown variable is x .  Looking at the left hand side of the equation, the  x is divided by 5 (the denominator of the fraction).

2 Apply the inverse operations, one at a time, to both sides of the equation .

The inverse of “dividing by 5 ” is “multiplying by 5 ”.  We need to multiply both sides of the equation by 5 .

3 Write the final answer, checking that it is correct .

The final answer is:

We can check the answer by substituting the answer back into the original equation.

Example 2: equations with one operation

The unknown variable is x . Looking at the left hand side of the equation, the x is divided by 3 (the denominator of the fraction).

Apply the inverse operations, one at a time, to both sides of the equation .

The inverse of “dividing by 3 ” is “multiplying by 3 ”.  We need to multiply both sides of the equation by 3 .

Write the final answer, checking that it is correct .

Example 3: equations with two operations

Solve: 

The unknown variable is x . Looking at the left hand side of the equation, 1 is added to x and then divided by 2 (the denominator of the fraction).

We need to do the inverse operations in the reverse order. First we need to multiply both sides of the equation by 2 . Then we need to subtract 1 from both sides.

Example 4: equations with two operations

The unknown variable is x .  Looking at the left hand side of the equation, x is divided by 4 (the denominator of the fraction) and then 2 is subtracted.

We need to do the inverse operations in the reverse order. First we need to add 2 to both sides of the equation. Then we need to multiply both sides of the equation by 4 .

Example 5: equations with three operations

The unknown variable is x .  Looking at the left hand side of the equation, x is multiplied by 3 , then divided by 4 (the denominator of the fraction) and then 1 is added.

We need to do the inverse operations in the reverse order. First we need to subtract 1 to both sides of the equation. Then we need to multiply both sides of the equation by 5 and finally divide both sides by 3 .

Example 6: equations with three operations

The unknown variable is x . Looking at the left hand side of the equation, x is multiplied by 2 , then 1 is subtracted. Then we divide by 7 (the denominator).

We need to do the inverse operations in the reverse order. First we need to multiply both sides of the equation by 7 .  Then we need to add 1 to both sides and finally divide both sides by 2 .

Example 7: equations with the unknown as the denominator

The unknown variable is x . Looking at the left hand side of the equation, x is the denominator. 24 is divided by x .

We need to multiply both sides of the equation by x . Then we can divide both sides by 6 .

Example 8: equations with the unknown as the denominator

The unknown variable is x . Looking at the left hand side of the equation, x is the denominator. 18 is divided by x and then 6 is subtracted.

First we add 6 to both sides of the equation. Then we need to multiply both sides of the equation by x .  Then we can divide both sides by 9 .

Common misconceptions

• Types of number

The solution to an equation can be different types of number. The unknown does not have to be an integer (whole numbers), it can also be a fraction or a decimal and can be positive or negative.

• The side of the equation that th unknown is on 

The unknown variable, represented by a letter, is often on the left hand side of the equations however it doesn’t have to be. It could also be on the right hand side of an equation.

• Multiplying both sides of an equation

When multiplying each side of the equation of a number, it is a common mistake to forget to multiply every term. E.g. Solve: \frac{x}{2}+3=9

Here we have not multiplied the +3 by 2 resulting in the incorrect answer:

Here we have correctly multiplied each term by the denominator:

• Lowest common denominator (LCD)

It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting we need to work out the lowest/least common denominator (sometimes called the lowest common multiple or lcm) whereas when we solve equations involving fractions we need to multiply both sides of the equation by the denominator of the fraction.

Practice equations with fractions questions

1. Solve: \frac{x}{6}=3

2. Solve: \frac{x+4}{2}=7

3. Solve: \frac{x}{8}-5=1

4. Solve: \frac{3x+2}{4}=2

5. Solve: \frac{4x}{7}-2=6

6. Solve: \frac{42}{x}=7

Equations with fractions GCSE questions

1. Solve \frac{x}{5}=3

for the correct answer

2. Solve \frac{x-3}{7}=2

for the correct first step

3. Solve \frac{5a+6}{2}=23

5a+6=46 for the correct first step

5a=40 for the correct second step

x=8 for the correct answer

Learning checklist

You have now learned how to:

• Solve equations when there are fractions
• Solve fractions where the unknown is the denominator

The next lessons are

• Factorising
• Rearranging equations
• Simultaneous equations

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