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

Factorising Expanding brackets Simplifying expressionsThis topic is relevant for:

Here we will learn how to solve linear and simple equations, including equations with one unknown, equations with an unknown on both sides, equations with brackets and equations with fractions.

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

A **linear **algebraic equation contains **variables **that have an **exponent **(are **raised to a power) **that is no higher than **one. **

All equations have an equals** **sign, which means that everything on the left-hand side of the = is exactly the same as everything on the right-hand side.

A **linear equation **will make a **straight line graph** and have a general form of ax + by + c = 0

This is often written as

E.g.

\[\begin{aligned}
2x-3y+4&=0\\
\\
y&=5x+2\\
\\
3y-7x&=8\\
\\
4x&=12
\end{aligned}\]

There are five main types of linear and simple equations:

**a Solve linear equations with one unknown**

**b Solve linear equations with an unknown on both sides**

**c Solve linear equations with brackets**

**d Solve linear equations with fractions**

**e Solve simple equations with powers (exponents) and roots**

In order to solve a **linear equation **or a **simple equation **we need to work out the value of the unknown variable by doing the opposite of what the operation tells us to do.

**Top tip:**

Leave the variable alone for as long as possible and deal with everything else first.

- If the equation has an
**addition**, to ‘undo it’ we need to use**subtraction**

- If the equation has a
**subtraction**, to ‘undo it’ we need to use**addition**

- If the equation has a
**multiplication**, to ‘undo it’ we need to use**division**

- If the equation has a
**division**, to ‘undo it’ we need to use**multiplication**

We can check that our answer is correct by substituting it back into the original equation.

**See also:** Solving equations and equations with fractions

Get your free linear equations worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free linear equations worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEIn order to solve linear equations with one unknown:

**Rearrange the equation so the unknown variable is on its own on one side****Work out the unknown variable by doing the opposite of what it says**

Example with two steps

To solve

\[3x + 6 = 18\]

we need to:

**1 Rearrange the equation so the unknown variable (x) is on its own on one side**

Here the opposite of

**2 Work out what the unknown variable (x) is by doing the opposite of what it says.**

Here

\[3x=3\times x\]

so we divide by

We can check that our solution is correct by substituting it into the original equation

**Fully worked out answer:**

In order to solve linear equations with an unknown on both sides:

**Rearrange the equation so the unknown variables are on the same side****Rearrange the equation so the unknown variable is on its own on one side****Work out what the unknown variable**(x) is by doing the opposite of what it says

Example with an unknown on both sides

To solve

\[5x + 6 = 2x + 9\]

we need to:

**1 Rearrange the equation so the unknown variables (x) are on the same side. (Top tip: eliminate the smallest variable).**

Here

**2 Rearrange the equation so the unknown variable (x) is on its own on one side (here we are subtracting 6 from both sides of the equation). **

**3 Work out what the unknown variable (x) is by doing the opposite of what it says.**

Here

\[3x=3\times x\]

so we divide by

We can check that our solution is correct by substituting it into the original equation

**Fully worked out answer:**

In order to solve linear equations with brackets:

**Expand the brackets****Solve the equation by rearranging to get the unknown variable by itself on one side and then doing the opposite of what it says.**

Example with brackets

To solve

\[4(x – 2) = 12\]

we need to:

**1 Expand the brackets**

\begin{aligned}
4(x-2) &=12 \\\\
4 x-8 &=12
\end{aligned}

**2 Solve the equation by rearranging to get the unknown variable (x) by itself on one side and then doing the opposite of what it says.**

Here the opposite of

Then

\[4x=4\times x\]

so we divide by

We can check that our solution is correct by substituting it into the original equation.

**Top Tip:**

We could have first divided both sides of the equation by 4 for this question as 4 is a factor of 12.

**Fully worked out answer:**

In order to solve linear equations with fractions:

**Multiply each fraction by the denominator on the other side of the = to get rid of the fractions; don’t forget to include brackets.****Expand the brackets****Rearrange the equation so the unknown variables are on the same side.****Solve the equation by rearranging to get the unknown variable by itself on one side and then do the opposite of what it says.**

Example with fractions

To solve

\[\frac{5 x-2}{4}=\frac{2 x+2}{2}\]

we need to:

**1 Multiply each fraction by the denominator on the other side of the = to get rid of the fractions; don’t forget to include brackets.**

**2 Expand the bracket**

\[\begin{array}{l}
2(5 x-2)=4(2 x+2) \\
10 x-4=8 x+8
\end{array}\]

**3 Rearrange the equation so the unknown variables (x) are on the same side. (Top tip: eliminate the smallest variable).**

Here

**4 Solve the equation by rearranging to get the unknown variable (x) by itself on one side and then do the opposite of what it says.**

We can check that our solution is correct by substituting it into the original equation.

**Fully worked out answer:**

In order to solve simple equations with powers and roots:

**Rearrange the equation so the unknown variable is on its own on one side****Work out the unknown variable by doing the opposite of what it says**

Example with a power

To solve

\[4x^{2}=64\]

we need to:

**1 Rearrange to get the unknown variable (x**

Here

\[4x^{2}=4\times x^{2}\]

so we divide by

**2 Work out the unknown variable by doing the opposite of what it says.**

The opposite of squaring is to take the square root, so do this to both sides.

(Remember square roots have a positive and a negative answer)

**Fully worked out answer:**

Example with a square root

To solve

\[5\sqrt{x}=30\]

we need to:

**1 Rearrange to get the unknown variable (√x) by itself on one side.**

Here

\[5\sqrt{x}=5\times \sqrt{x}\]

so we divide by

**2 Work out the unknown variable by doing the opposite of what it says. **

The opposite of taking the square root is squaring, so do this to both sides.

We can check that our solution is correct by substituting it into the original equation.

**Fully worked out answer:**

**We must do the opposite of what the operation tells us to do**

E.g.

means

The opposite of

E.g.

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

means

The opposite of

**We must do the same thing to both sides of the equal sign**

Equals means that both sides of the

E.g.

If we

**When cross multiplying we must use brackets to multiply every term in the numerator**

We only multiply the numerator, not the denominator.

E.g.

**When expanding brackets we need to multiply out each term**

E.g.

\[\quad 2(6x-3)=12x-6\]
\[2\times 6x=12 \quad \quad 2\times -3=-6\]

Be careful when multiplying the coefficients

**Remember a polynomial is an algebraic expression that consists of two or more algebraic terms**

1. Solve:

2x+8=20

x=14

x=8

x=6

x=10

2x+8=20

Subtract 8 from both sides

2x=12

Divide both sides by 2

x=6

2. Solve:

5x-3=12

x=5

x=-8

x=3

x=-12

5x-3=12

Add 3 to both sides

5x=15

Divide both sides by 5

x=3

3. Solve:

2x-7=x+5

x=7

x=12

x=-7

x=4

2x-7=x+5

Subtract x from both sides

x-7=5

Add 7 to both sides

x=12

4. Solve:

7x-3=4x+15

x=4

x=7

x=6

x=3

7x-3=4x+15

Subtract 4x from both sides

3x-3=15

Add 3 to both sides

3x=18

Divide both sides by 3

x=6

5. Solve:

5(x-3)=10

x=2

x=3

x=5

x=-3

5(x-3)=10

Divide both sides by 5

x-3=2

Add 3 to both sides

x=5

6. Solve:

3(x-3)=6(x+5)

x=3

x=-5

x=-13

x=13

3(x-3)=6(x+5)

Expand the brackets

3x-9=6x+30

Subtract 3x from both sides

-9=3x+30

Subtract 30 from both sides

-39=3x

Divide both sides by 3

x=-13

7. Solve:

\frac{x+3}{2}=\frac{x+4}{3}

x=-3

x=-4

x=-2

x=-1

\frac{x+3}{2}=\frac{x+4}{3}

Multiply both sides by 6 and simplify

3x+9=2x+8

From both sides, subtract 2x and subtract 9

x=-1

8. Solve:

\frac{3x+1}{2}+\frac{2x-2}{4}

x=-2

x=2

x=-1

x=1

\frac{3x+1}{2}+\frac{2x-2}{4}

Multiply both sides by 4 and simplify

6x+2=2x-2

From both sides, subtract 2x and subtract 2

x=-1

9. Solve:

16{x}^2=64

x=\pm4

x=\pm2

x=\pm3

x=\pm16

16{x}^2=64

Divide both sides by 16

x^{2}=4

Square root both sides

x=\pm2

10. Solve:

5\sqrt{3x}=30

x=36

x=3

x=12

x=9

5\sqrt{3x}=30

Divide both sides by 5

\sqrt{3x}=6

Square both sides

3x=36

Divide both sides by 3

x=12

1. Solve:

3 x^{2} = 27

** (1 mark)**

Show answer

y=\pm 3

** (1)**

2. Solve:

4(3 – x) = 32

** **

** (2 marks)**

Show answer

12 – 4x = 32 or 3 – x = 8

** (1)**

x = -5

** (1)**

3. Solve:

\frac{2x + 3}{5}=\frac{x – 5}{2}

** **

** (3 marks)**

Show answer

2(2x + 3) = 5(x – 5)

** (1)**

4x + 6 = 5x – 25

** (1)**

x=31

** (1)**

You have now learned how to:

- Use algebraic methods to solve linear equations.

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