GCSE Maths Algebra Solving Equations

Equations with fractions

# 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.

E.g.

See also: Solving equations and linear equations

## How to solve equations with fractions

In order to solve equations with fraction:

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

## Equations with fractions examples

### Example 1: equations with one operation

Solve:

$\frac{x}{5}=4$

1. 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).

$\frac{x}{5}$

2Apply 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.

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

The final answer is:

$x=20$

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

$\frac{20}{5}=20\div5=4$

### Example 2: equations with one operation

Solve:

$\frac{x}{3}=8$

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).

$\frac{x}{3}$

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

The final answer is:

$x=24$

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

$\frac{24}{3}=24\div3=8$

### Example 3: equations with two operations

Solve:

$\frac{x+1}{2}=7$

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).

$\frac{x+1}{2}$

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.

The final answer is:

$x=13$

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

$\frac{13+1}{2}=\frac{14}{2}=14\div2=7$

### Example 4: equations with two operations

Solve:

$\frac{x}{4}-2=3$

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.

$\frac{x}{4}-2$

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.

The final answer is:

$x=20$

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

$\frac{20}{4}-2=20\div4 -2=5-2=3$

### Example 5: equations with three operations

Solve:

$\frac{3x}{5}+1=7$

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.

$\frac{3x}{5}+1$

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.

The final answer is:

$x=10$

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

$\frac{3\times10}{5}+1=\frac{30}{5}+1=6+1=7$

### Example 6: equations with three operations

Solve:

$\frac{2x-1}{7}=3$

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).

$\frac{2x-1}{7}$

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.

The final answer is:

$x=11$

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

$\frac{2\times11 -1}{7}=\frac{22-1}{7}=\frac{21}{7}=3$

### Example 7: equations with the unknown as the denominator

Solve:

$\frac{24}{x}=6$

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

$\frac{24}{x}$

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

The final answer is:

$x=4$

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

$\frac{24}{4}=24\div4=6$

### Example 8: equations with the unknown as the denominator

Solve:

$\frac{18}{x}-6=3$

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.

$\frac{18}{x}-6$

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.

The final answer is:

$x=2$

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

$\frac{18}{2}-6=9-6=3$

### 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

x=18

x=9

x=36

x=12

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

x=18

x=10

x=26

x=30

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

x=40

x=48

x=64

x=56

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

x=4

x=16

x=12

x=2

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

x=11

x=7

x=14

x=10

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

x=5

x=294

x=7

x=6

### Equations with fractions GCSE questions

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

(1 mark)

Show answer
x=15

for the correct answer

(1)

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

(2 marks)

Show answer
x-3=14

for the correct first step

(1)

x=17

for the correct answer

(1)

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

(3 marks)

Show answer

5a+6=46
for the correct first step

(1)

5a=40
for the correct second step

(1)

x=8
for the correct answer

(1)

## Learning checklist

You have now learned how to:

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

## Still stuck?

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#### GCSE Maths Papers - November 2022 Topics

Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

Designed to help your GCSE students revise some of the topics that are likely to come up in November exams.