Solving single operation linear equations

  • EDEXCEL GCSE
  • AQA GCSE
  • OCR GCSE
  • EDUQAS GCSE

Video masterclass

Topic summary

To solve an equation, you are looking at finding the value of the letter. Your goal is to end up with \(x = \text{ something}\)

Solve with an addition

\[x+5=12\]

The inverse (or opposite) of \(+5\) is \(-5\). We must do this to both sides though.

\[x+5-5=12-5\]

\[x=7\]

Solve with a subtraction

\[x-7=10\]

The inverse (or opposite) of \(-7\) is \(+7\).

\[x-7+7=10+7\]

\[x=17\]

Solve with a multiplication

\[8x=40\]

\(8x\) means \(8 \times x\). The inverse (or opposite) of \(\times 8\) is \(\div 8\).

\[8x\div 8=40\div 8\]

\[x=5\]

Solve with a division

\[\frac{x}{3}=20\]

\(\dfrac{x}{3}\) means \(x \div 3\). The inverse (or opposite) of \(\div 3\) is \(\times 3\).

\[\frac{x}{3}\times 3=20\times 3\]

\[x=60\]

Extra questions (ultimate exclusive)

Ultimate members get access to four additional questions with full video explanations.