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Pearson Edexcel 4MA1 IGCSE  ·  Topic G3.14

G3.14 Solving Linear Equations

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01 — Explanation

How does the balancing method work?

Whatever you do to one side of an equation, you must do to the other — this keeps both sides equal while you isolate the unknown. Undo operations in the reverse order to how they were built (reverse BIDMAS): undo addition/subtraction before multiplication/division.

Example: 2x−3=72x-3=7. Add 3 to both sides: 2x=102x=10. Divide by 2: x=5x=5.

02 — Explanation

How do I solve an equation with brackets?

Expand the brackets first, then apply the balancing method.

Example: 2(x+3)=10  ⇒  2x+6=10  ⇒  2x=4  ⇒  x=22(x+3)=10 \;\Rightarrow\; 2x+6=10 \;\Rightarrow\; 2x=4 \;\Rightarrow\; x=2.

Common Misconception: Forgetting to multiply every term inside a bracket, especially with a negative outside it — −2(x−3)-2(x-3) expands to −2x+6-2x+6, not −2x−6-2x-6.

03 — Explanation

How do I solve an equation with the unknown on both sides?

Collect the xx-terms onto one side first (ideally eliminating the smaller one, to avoid a negative coefficient), then the constants, then divide.

Example: 3x+4=x+83x+4=x+8.

  1. Subtract xx from both sides: 2x+4=82x+4=8.
  2. Subtract 4: 2x=42x=4.
  3. Divide by 2: x=2x=2.

Exam Tip: Always move the smaller xx-term across, not the larger one — it keeps the remaining coefficient positive and avoids an easy sign slip in the final division.

04 — Explanation

How do I solve an equation with a fraction in it?

Multiply every term on both sides by the denominator to clear the fraction before doing anything else.

Example: x2+3=5\dfrac{x}{2}+3=5. Multiply every term by 2: x+6=10x+6=10. Subtract 6: x=4x=4.

05 — Explanation

How do I check my solution is actually correct?

Substitute your answer back into the original equation (not a simplified version) and confirm both sides match.

Example: Check x=2x=2 solves 3x+4=x+83x+4=x+8: LHS =3(2)+4=10=3(2)+4=10; RHS =2+8=10=2+8=10. Both sides agree, so x=2x=2 is correct.

Exam Tip: This check takes seconds and catches almost every slip — make it a habit on every equation question, especially ones with brackets or the unknown on both sides. If the two sides don't match, re-examine your working line by line rather than starting over from scratch — you'll usually spot exactly which step introduced the error.

06 — Explanation

How do I turn a word problem into an equation?

Define the unknown as xx, write an expression for each quantity described, and use the relationship given (total, perimeter, angle sum) to form the equation — then always check your final answer makes sense in context.

Example: A triangle has angles a°a°, (a+10)°(a+10)°, and (a+20)°(a+20)°. Angles in a triangle sum to 180°180°:

a+(a+10)+(a+20)=180  ⇒  3a+30=180  ⇒  3a=150  ⇒  a=50a+(a+10)+(a+20)=180 \;\Rightarrow\; 3a+30=180 \;\Rightarrow\; 3a=150 \;\Rightarrow\; a=50

Common Misconception: Stopping once you've found xx (or aa) without answering the actual question asked. If the question wants a side length or angle, substitute your solved value back into the original expression for that quantity — and always include the units.

07 — Worked example

Worked Example

A parallelogram has adjacent sides (x+4)(x+4) cm and (2x−1)(2x-1) cm, and a perimeter of 24 cm. Find the length of the longer side.

  1. Perimeter equation: 2(x+4)+2(2x−1)=242(x+4)+2(2x-1)=24.
  2. Expand: 2x+8+4x−2=242x+8+4x-2=24.
  3. Collect like terms: 6x+6=246x+6=24.
  4. Solve: 6x=186x=18, so x=3x=3.
  5. Substitute back: side lengths are x+4=7x+4=7 cm and 2x−1=52x-1=5 cm. The longer side is 7 cm.