 Solving Equations | Questions and Revision | Maths Made Easy

# Solving Equations Worksheets, Questions and Revision

Level 1-3

## 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}

Level 1-3

## Type 2: Unknown appears more than once

This type is similar to Type 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}

Level 1-3

## 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}

Level 4-5

## 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}

Level 4-5

## 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}

Level 4-5

### Example Questions

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,

$x=\dfrac{1}{2}$

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,

$x=20$

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

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

\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

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