IGCSE Maths Equations & inequalities notes

An equation is a balance. To solve 3x + 5 = 17, subtract 5 from both sides (3x = 12), then divide both sides by 3 (x = 4). The order is reverse: undo the +/− before the ×/÷. With brackets, expand first: 4(x − 2) = 12 becomes 4x − 8 = 12, so 4x = 20 and x = 5. With the unknown on both sides, collect the letters on the side that keeps the coefficient positive: 4x + 3 = 2x + 11 → 2x + 3 = 11 → 2x = 8 → x = 4.

Watch negative answers: 5 − 2x = 11 gives −2x = 6, so x = −3. Dividing by −2 flips nothing here (division by a negative only flips *inequalities*), but candidates who “tidy” −2x = 6 into x = 3 throw away the sign. Fractions clear the same way: 2x⁄3 = 8 → multiply both sides by 3 → 2x = 24 → x = 12.

Two unknowns demand two independent equations. For elimination, scale the equations until one letter has matching coefficients, then add or subtract. For x + 2y = 11 and 3x − y = 12: multiply the second by 2 to get 6x − 2y = 24, add to the first to kill y: 7x = 35, so x = 5, and back-substitute: 5 + 2y = 11 gives y = 3. For substitution, use an equation already solved for one letter, like y = x + 2: drop it into the other equation, 2x + (x + 2) = 11 → 3x = 9 → x = 3, y = 5.

Word problems hide the pair: “two tickets and three popcorns cost $23; one of each costs $9” is 2t + 3p = 23 and t + p = 9. Define your letters in a sentence — that line often carries a mark — then solve as above: t = 9 − p, so 18 − 2p + 3p = 23 → p = 5, t = 4.

Extended. A quadratic has two roots — write both. First try factorising: x² − 5x + 6 = 0 needs two numbers multiplying to 6 and adding to −5: −2 and −3, so (x − 2)(x − 3) = 0 and x = 2 or 3. When the numbers are unfriendly, the quadratic formula (printed on the paper) takes over: x = (−b ± √(b² − 4ac)) ⁄ 2a. For x² + 4x + 1 = 0: x = (−4 ± √(16 − 4))⁄2 = (−4 ± √12)⁄2 = −2 ± √3, which is −0.27 or −3.73 to 2 d.p.

Quote roots exactly (surd form) when the question says “exact”, and to the stated accuracy otherwise — usually 2 d.p. Keep the calculator’s full value between the two roots, computing each root separately from x = (−b + √…)⁄2a and x = (−b − √…)⁄2a. Negative coefficients need care: for x² − 4x − 7 = 0, b = −4, so −b = 4, and x = (4 ± √(16 + 28))⁄2 = 2 ± √11, i.e. 5.32 or −1.32 to 2 d.p.

Extended. Rearranging a formula is solving an equation where the “answer” is a letter. To make x the subject of y = 3x − 7: add 7 to both sides, then divide by 3: x = (y + 7)⁄3. Undo operations in the same reverse order as equations, and apply each move to the whole side. For v = u + at, making t the subject: subtract u, then divide by a: t = (v − u)⁄a.

When the letter sits under a square, undo the square last by taking the root of the whole side — and remember the ± if the context allows both signs: A = πr² becomes r = √(A⁄π). When the subject appears twice, collect its terms first. Harder cases can take two rearrangements — do not rush; write one operation per line.

Solve inequalities exactly like equations, with one exception: multiplying or dividing by a negative number reverses the inequality sign. For 5 − x ⩾ 2: subtract 5 to get −x ⩾ −3, then divide by −1 and flip: x ⩽ 3. Alternatively move the x across first: 5 − 2 ⩾ x, so 3 ⩾ x — same conclusion with no flip needed.

On a number line, a filled circle marks an endpoint that is included (⩽, ⩾); an open circle marks one that is not (<, >), with an arrow showing the direction. The compound statement −1 < x ⩽ 4 means “between −1 and 4, excluding −1, including 4” — open at −1, filled at 4. Extended questions also describe regions: a broken line means its boundary inequality is strict (y < 2x + 1), a solid line means it is included (y ⩾ …), and “the unwanted region is shaded” tells you the answer lives in the unshaded part.