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In order to access this I need to be confident with:

Multiplication and division

Addition and subtraction

Simplifying algebraic expressions

Collecting like termsThis topic is relevant for:

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.

**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.

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.**

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

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

DOWNLOAD FREESolve:

\[\frac{x}{5}=4\]

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

The unknown variable is

Looking at the left hand side of the equation, the

\[\frac{x}{5}\]

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

The inverse of “dividing by

We need to multiply both sides of the equation by

3**Write 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.

\[\frac{20}{5}=20\div5=4\]

Solve:

\[\frac{x}{3}=8\]

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

The unknown variable is

Looking at the left hand side of the equation, the

\[\frac{x}{3}\]

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

The inverse of “dividing by

We need to multiply both sides of the equation by

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

The final answer is:

\[x=24\]

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

\[\frac{24}{3}=24\div3=8 \]

Solve:

\[\frac{x+1}{2}=7\]

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

The unknown variable is

Looking at the left hand side of the equation,

\[\frac{x+1}{2}\]

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

We need to do the inverse operations in the reverse order.

First we need to multiply both sides of the equation by

Then we need to subtract

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

The final answer is:

\[x=13\]

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

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

Solve:

\[\frac{x}{4}-2=3\]

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

The unknown variable is

Looking at the left hand side of the equation,

\[\frac{x}{4}-2\]

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

We need to do the inverse operations in the reverse order.

First we need to add

Then we need to multiply both sides of the equation by

**Write 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.

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

Solve:

\[\frac{3x}{5}+1=7\]

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

The unknown variable is

Looking at the left hand side of the equation,

\[\frac{3x}{5}+1\]

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

We need to do the inverse operations in the reverse order.

First we need to subtract

Then we need to multiply both sides of the equation by

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

The final answer is:

\[x=10\]

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

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

Solve:

\[\frac{2x-1}{7}+1=3\]

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

The unknown variable is

Looking at the left hand side of the equation,

\[\frac{2x-1}{7}+1\]

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

We need to do the inverse operations in the reverse order.

First we need to multiply both sides of the equation by

Then we need to add

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

The final answer is:

\[x=11\]

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

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

Solve:

\[\frac{24}{x}=6\]

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

The unknown variable is

Looking at the left hand side of the equation,

\[\frac{24}{x}\]

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

We need to multiply both sides of the equation by

Then we can divide both sides by

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

The final answer is:

\[x=4\]

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

\[\frac{24}{4}=24\div4=6\]

Solve:

\[\frac{18}{x}-6=3\]

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

The unknown variable is

Looking at the left hand side of the equation,

\[\frac{18}{x}-6\]

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

First we add

Then we need to multiply both sides of the equation by

Then we can divide both sides by

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

The final answer is:

\[x=2\]

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

\[\frac{18}{2}-6=9-6=3\]

**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

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.

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)**

You have now learned how to:

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

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