9.1 Solving linear equations — Learn

Prerequisites: inverse operations and simplifying arithmetic expressions.

Objectives: solve one-step and multi-step linear equations, including brackets and fractional coefficients.

This section covers solving linear equations in one unknown, including brackets, as set out in AQA reference A17.

An equation says that two expressions have the same value. The solution is the value of the unknown that makes the statement true. Keep the equation balanced: whatever operation you apply to one side, apply to the other.

Example 1. Solve $3x+5=20$.

Subtract $5$ from both sides: $3x=15$.

Divide both sides by $3$: $x=5$.

Check in the original equation: $3(5)+5=20$.

Example 2. Solve $2(x-4)=10$.

Divide both sides by $2$: $x-4=5$.

Add $4$ to both sides: $x=9$.

Check: $2(9-4)=10$.

Watch out. Dividing only one term breaks the balance. In $2x+6=14$, subtract $6$ first to get $2x=8$, then divide to get $x=4$.

A fractional coefficient works in the same way. For $x/3+2=7$, subtract $2$ to get $x/3=5$, then multiply both sides by $3$ to get $x=15$.

Progress check

The fixed fee example has a charge of £18 plus £6 per hour. Write its cost after $h$ hours: $18+6h$. You will compare it with another expression in Section 9.3.

You should now be able to

• solve one-step and two-step equations;
• solve equations with brackets or fractional coefficients;
• check a solution by substitution.