Equations

Video lesson5 worked examples

Theory

You must be able to solve linear equations, including those with rational/fractional coefficients and variables on both sides.

The primary method involves using inverse operations to add or subtract from both sides. The goal is to isolate the variable (the letter) on one side and the numerical constants on the other.

The Golden Rule: whatever you do to one side, do to the other. When brackets appear, expand them first; when fractions appear, multiply every term by the lowest common denominator to clear them before solving.

⚠️ Common Examiner Traps

Clearing fractions incompletely: when you multiply through by the common denominator, every term must be multiplied — including any whole numbers, not just the fractions.
Sign slips moving terms: a term crossing the equals sign changes sign. Collect the letters on the side that keeps their coefficient positive to avoid negatives.
Not expanding brackets first: $4(x-3)$ must become $4x - 12$ before collecting terms.
Fraction answers are fine: not every equation has a whole-number solution — leave the answer as a simplified fraction if that is what it is.

Worked examples

Example 1

Variables on Both Sides

Solve $5p - 4 = 2p + 11$ .

Step 1: Subtract 2p from both sides, and add 4 to both sides: $3p = 15$ .

Step 2: Divide by 3. Answer: $p = 5$ .

Example 2

With Brackets

Solve $4(x - 3) = 2x + 6$ .

Step 1: Expand the bracket first: $4x - 12 = 2x + 6$ .

Step 2: Subtract $2x$ from both sides and add 12 to both: $2x = 18$ .

Step 3: Divide by 2. Answer: $x = 9$ .

Example 3

Clearing a Single Fraction

Solve $\dfrac{3x - 1}{4} = 5$ .

Step 1: Multiply both sides by 4 to clear the denominator: $3x - 1 = 20$ .

Step 2: Add 1, then divide by 3: $3x = 21$ . Answer: $x = 7$ .

Example 4

Two Fractions

Solve $\dfrac{x + 3}{2} = \dfrac{2x - 1}{5}$ .

Step 1: With a single fraction equal to a single fraction, cross-multiply (or multiply both sides by 10, the LCM):

5(x + 3) = 2(2x - 1)

Step 2: Expand both sides: $5x + 15 = 4x - 2$ .

Step 3: Subtract $4x$ and 15 from both sides: $x = -17$ .

Answer: $x = -17$ .

Example 5

🎯 Exam-style

Solve $\dfrac{x}{4} + 1 = \dfrac{x + 5}{6}$ .

Step 1: The denominators are 4 and 6, so multiply every term by their LCM, 12 — including the $+1$ :

12 \times \dfrac{x}{4} + 12 \times 1 = 12 \times \dfrac{x + 5}{6}

Step 2: Simplify each term: $3x + 12 = 2(x + 5)$ .

Step 3: Expand and solve: $3x + 12 = 2x + 10 \implies x = -2$ .

Answer: $x = -2$ .

Ready to practise?

14 Linear Equations questions with answers and video solutions