## Solving Linear Equations

**Linear equations** are a type of equation that appear all over the place in maths. They can look quite simple, like x+2=5 or they can look a little more complicated.

There are **5** key types of linear equation you will need to solve. Exam questions can contain multiple types to make it even harder.

Make sure you are happy with the following topics before continuing.

**Type 1:** Only 1 Unknown.

These are the simplest type of linear equation and can be solved easily.

Solve 6x-4 =26

**Step 1:** Rearrange so x's are alone on one side

\begin{aligned}(+4)\,\,\,\,\,\,\,\,\,6x-4 &=26 \\ 6x&= 30 \end{aligned}

**Step 2:** Divide both sides by the number before the unknown

\begin{aligned}(\div6) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 6x&= 30 \\ x&= 5\end{aligned}

**Type 2:** Unknown appears more than once

This type is similar to **T****ype 1 **but has one additional step.

Solve 12x +8 = 5x + 36

**Step 1:** Rearrange so that all x's are on one side

\begin{aligned}(-5x)\,\,\,\,\,\,\,\,\,12x +8 &= 5x+36 \\ 7x+8 &=36\end{aligned}

**Step 2:** Follow the steps for **Type 1**

\begin{aligned}(-8)\,\,\,\,\,\,\,\,\,7x+8 &=36 \\ (\div7)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x&=28 \\ x&=4\end{aligned}

## Type 3: Includes brackets

This example will be very similar to **Type 2**, but this time containing brackets. The brackets are a barrier to our solution so we need to expand any brackets first.

Solve 3(2x-6)=2(5x+3)

**Step 1:** Multiply out the brackets

\begin{aligned}3(2x-6) &= 2(5x+3)\\ 6x-18 &=10x+6\end{aligned}

**Step 2:** Follow the steps from **Type 2**

\begin{aligned}(-6)\,\,\,\,\,\,\,\,\, 6x-18 &=10x+6 \\ (-6x)\,\,\,\,\,\,\, 6x -24 &=10x \\(\div4) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-24 &= 4x \\ -6 &= x \end{aligned}

## Type 4: Includes a fraction(s)

Fractions make things a little more complicated. You always need to try and remove the fractions first before performing any other calculations.

Solve \dfrac{4x+5}{5} = \dfrac{x+17}{3}

**Step 1:** Multiply out the fractions

To do this we multiply the both sides of the equation by the denominator of the fractions, first (\times5) then (\times3).

\begin{aligned}(\times5) \,\,\,\,\,\,\,\,\, \dfrac{4x+5}{5} &= \dfrac{x+17}{3} \\\\ (\times3)\,\,\,\,\,\,\,\,\,\, 4x+5 &= \dfrac{5(x+17)}{3} \\\\ \,\,\,\,\,\,\,\,\, 3(4x+5) &= 5(x+17)\end{aligned}

**Step 2:** Follow the steps from **Type 3**

First multiply out the brackets,

\begin{aligned}3(4x+5) &= 5(x+17)\\ 12x+15 &= 5x+85\end{aligned}

Then solve the equation,

\begin{aligned}(-15)\,\,\,\,\,\,\,\,\,12x+15 &= 5x+85 \\(-5x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12x & = 5x +70 \\(\div7)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 7x &= 70 \\ x&=10\end{aligned}

## Type 5: Squares and Square roots

Removing a square or square root is often an extra **final step** to solving some equations.

**Example 1** – Including a square

Solve 5x^2 = 320

First we solve the equation the same as **Type 1**

\begin{aligned}(\div5)\,\,\,\,\,\,\,\,\,5x^2 &= 320 \\ \,\,\,\,\,\,\,\,\, x^2 &=64\end{aligned}

Finally we must perform the opposite operation to a square, a square root.

\begin{aligned}(\sqrt{})\,\,\,\,\,\,\,\,\,x^2 &= 64 \\ \,\,\,\,\,\,\,\,\, x &= \pm8\end{aligned}

**Example 2** – Including a square root

Solve 3\sqrt{x} = 15

First we solve the equation the same as **Type 1**

\begin{aligned}(\div3)\,\,\,\,\,\,\,\,\,3\sqrt{x} &= 15 \\ \,\,\,\,\,\,\,\,\, \sqrt{x} &=5\end{aligned}

Finally we must perform the opposite operation to a square root, a square.

\begin{aligned}(\,^2)\,\,\,\,\,\,\,\,\,\sqrt{x} &= 5 \\ \,\,\,\,\,\,\,\,\, x &=25\end{aligned}

## GCSE Maths Revision Cards

(222 Reviews) £8.99### Example Questions

**Question 1 **Solve the equation 2x+1=2

**[2 marks]**

To get the constant terms (numbers) only on one side of the equation, we will get rid of the +1 by subtracting 1 from both sides. Then, the equation becomes

2x=1

Then, all that remains to get x on its own by dividing both sides by 2 to get,

x=\dfrac{1}{2}

**Question 2:** Solve the equation \dfrac{1}{2}x-3=7

**[2 marks]**

To get the constant terms (numbers) only on one side of the equation, we will get rid of the -3 by adding 3 to both sides. Then, the equation becomes

\dfrac{1}{2}x=10

Then, all that remains to get x on its own by multiplying both sides by 2 to get,

x=20

**Question 3:** Solve the equation 12k-1=6k-25

**[2 marks]**

\begin{aligned} 12k-1&=6k-25 \\ 6k &= -24 \\ k &= -4 \end{aligned}

**Question 4:** Solve the equation 3(2m+6) =2(m-3)

**[2 marks]**

\begin{aligned} 3(2m+6) &=2(m-3) \\ 6m+18 &= 2m-6 \\ 4m &= -24 \\ m &= -6 \end{aligned}

**Question 5:** Solve the equation \dfrac{x^2}{5} = 31.25

**[2 marks]**

\begin{aligned} \frac{x^2}{5} &= 31.25 \\ x^2 &= 156.25 \\ x &= \sqrt{156.25} \\ x &= \pm 12.5 \end{aligned}

### Worksheets and Exam Questions

#### (NEW) Solving Equations (Foundation) Exam Style Questions - MME

Level 4-5 New Official MME### Drill Questions

#### Solving Linear Equations - Drill Questions

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