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Chapter overview

Maths

Exam board

AQA

Number

Algebra

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Ratio proportion and rates of change

Shapes and area

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Statistics

Maths

Solving equations

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Solving equations

​​In a nutshell

​An equation can be solved by rearranging. You usually want to find the value of the unknown variable, e.g. xx. which makes the equation true. Think of an equation as a balancing scale, whatever you do to one side of the equation, you must do the same to the other.


Balancing scales

​x+5=13x+5=13

can be represented by

Maths; Algebra; KS4 Year 10; Solving equations

To solve the equation for xx, you need to rearrange the equation so that xx is by itself on one side of the equation, and all other numbers or terms should be on the other side. To get xx by itself, +5+5 needs to be moved to the other side. To do this, think of the inverse operation, or the opposite of +5+5. The opposite of +5+5 is −5-5 so subtract 55 from both sides of the equation. This gives​

​x=13−5x=8‾\begin {aligned} x &= 13 -5 \\ x &=\underline{8} \end {aligned}​​

This can be represented by

Maths; Algebra; KS4 Year 10; Solving equations



Solving algebraically

​When solving an equation, think about what to do to both sides of the equation to get the unknown variable by itself.


Example 1

x+3=7−3−3x=4‾\begin {aligned}\qquad x + 3 &= 7 \\-3 \qquad & \qquad -3\end {aligned}\\\quad \quad\underline {x=4}​​


When there is more than one number to move to the other side of the equation, it is important to move them in the right order. Use reverse BIDMAS to help.


Example 2

Solve the equation 

​3x+4=103x+4=10


Here, the 33 with the xx and +4+4 need to be moved to the other side. To decide which number to move across first, think about substituting a number in for xx. You would take the value of xx, multiply by 33 first, then +4+4 to the result. So when solving an equation, reverse the process. So move +4+4 to the other side first, then move 33 to the other side.

The opposite of +4+4 is −4-4, so subtract 44 from both sides. Then 33 is multiplying xx, so do the inverse and divide by 33 to both sides.

3x+4=10−4−43x=6÷3÷3x=2‾\begin {aligned}\qquad 3x+4 &=10 \\-4 \qquad & \qquad -4 \\3x &= 6 \\\div3 \qquad & \qquad \div 3 \end {aligned}\\\quad \quad\underline {x=2}​


Example 3

Solve 

​3x+32=11x−483x+32 = 11x-48​​


When there are multiple terms with xx in the equation, on both sides, first move the xx terms on one side, and the numbers onto the other side. Move the xx's onto the side where there are more of them.​

3x+32=11x−48−3x−3x32=8x−48+48+4880=8x÷8÷810=xx=10‾\begin {aligned}\quad 3x+32 &= 11x -48 \\-3x \qquad & \qquad -3x \\32 &= 8x - 48 \\+48 \qquad & \qquad +48 \\80 &= 8x \\\div 8 \qquad & \qquad \div 8 \\10 &= x \end {aligned}\\\quad \underline {x=10}