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Linear EquationsEdexcel IGCSE Maths: Revision notes

Section 1

What does it mean to "solve" an equation?

A linear equation says two expressions are equal, and your job is to find the single value of (x) that makes this true.

The golden rule: whatever you do to one side, you must do to the other. This keeps the equation balanced, like a set of scales.

Work through equations in this order:

  1. Expand any brackets
  2. Clear any fractions
  3. Collect the unknown (xx) on one side
  4. Collect the numbers on the other side
  5. Divide by the coefficient of xx
Key termslinear equationunknowncoefficient
Exam tip

Always do the same operation to BOTH sides in the same line of working — never just to one side.

Section 2

How do I solve equations with xx on both sides?

When xx appears on both sides, e.g. 5x+3=2x+185x + 3 = 2x + 18, move all the xx terms to one side and all the numbers to the other.

5x+3=2x+185x + 3 = 2x + 18 5x−2x=18−35x - 2x = 18 - 3 3x=153x = 15 x=5x = 5

Always move the smaller xx term across so you avoid negative coefficients where possible — it reduces the chance of sign errors.

Key termslike termsisolate
Example

Solve 7x−4=3x+207x - 4 = 3x + 20. Subtract 3x3x: 4x−4=204x - 4 = 20. Add 4: 4x=244x = 24. Divide by 4: x=6x = 6.

Common mistake

Forgetting to change the sign when moving a term across the equals sign — moving +2x+2x to the other side makes it −2x-2x, not +2x+2x.

Section 3

How do I deal with brackets?

Expand (multiply out) brackets before doing anything else. Multiply every term inside the bracket by the number or term directly outside it.

3(2x+5)=273(2x + 5) = 27 6x+15=276x + 15 = 27 6x=126x = 12 x=2x = 2

If there are brackets on both sides, or a minus sign in front of a bracket, expand both before collecting terms.

Key termsexpanddistributive law
Example

Solve 4(x−3)=2(x+5)4(x - 3) = 2(x + 5). Expand both: 4x−12=2x+104x - 12 = 2x + 10. So 2x=222x = 22, giving x=11x = 11.

Common mistake

With −2(x−4)-2(x - 4), both signs inside flip: it becomes −2x+8-2x + 8, not −2x−8-2x - 8.

Section 4

How do I solve equations with fractions?

The cleanest method is to multiply every term by the denominator (or the lowest common multiple of all denominators) to clear the fractions entirely, then solve as normal.

x3+2=5\frac{x}{3} + 2 = 5

Multiply every term by 3: x+6=15x + 6 = 15 x=9x = 9

For two different denominators, e.g. x+12=x−13\frac{x+1}{2} = \frac{x-1}{3}, multiply both sides by 6 (the LCM of 2 and 3): 3(x+1)=2(x−1)3(x+1) = 2(x-1) 3x+3=2x−23x + 3 = 2x - 2 x=−5x = -5

Key termsdenominatorlowest common multiple (LCM)
Exam tip

When you multiply a fraction like x+12\frac{x+1}{2} by 6, the whole numerator (x+1)(x+1) must be multiplied — keep it in brackets to avoid sign errors.

Think of it like this

Clearing a fraction is like scaling up a recipe — multiply every ingredient (term) by the same factor so the ratios stay identical.

Section 5

How do I check my answer?

Substitute your value of xx back into the ORIGINAL equation (before you expanded or cleared fractions). If both sides give the same number, your answer is correct.

For 5x+3=2x+185x + 3 = 2x + 18 with x=5x = 5: left side =5(5)+3=28= 5(5)+3 = 28; right side =2(5)+18=28= 2(5)+18 = 28. They match, so x=5x = 5 is correct.

Key termssubstituteverify
Exam tip

In an exam, always leave 30 seconds to check your solution by substitution — it catches almost all arithmetic slips.

Must Know

  • Always expand brackets first, before collecting terms or clearing fractions.
  • Clear fractions by multiplying every term by the denominator (or LCM of denominators).
  • Move terms across the equals sign by doing the inverse operation, and always flip the sign.
  • Collect xx terms on one side and numbers on the other, then divide by the coefficient of xx.
  • A negative sign in front of a bracket flips the sign of every term inside when expanded.
  • Check your answer by substituting it back into the original equation.

That's the notes covered.

Carry on to the next subtopic.