GCSE Maths Linear equations & graphs revision

A linear equation has one unknown to the power 1. Solve 4x + 3 = 15 by undoing: subtract 3 (4x = 12), divide by 4 (x = 3). Check by substitution: 4 × 3 + 3 = 15 ✓ — a five-second habit that catches nearly every slip.

With the unknown on both sides, 5x − 2 = 3x + 10, collect x where its coefficient is bigger: subtract 3x (2x − 2 = 10), add 2 (2x = 12), divide (x = 6). Brackets come down first: 3(x − 4) = 2x + 5 expands to 3x − 12 = 2x + 5, so x = 17. Fractions clear by multiplying every term by the common denominator: (2x + 1)⁄3 = 5 becomes 2x + 1 = 15, giving x = 7.

Every straight line is y = mx + c: the gradient m says how much y grows when x grows by 1, and c is where the line cuts the y-axis. In y = 2x − 5 the gradient is 2 (up 2 for each 1 across) and the line passes through (0, −5). To plot a line, make a small table: for y = 2x − 5, x = 0, 1, 2 gives y = −5, −3, −1 — three points pin the line, and a ruler joins them.

The gradient is rise over run between any two points: from (1, 3) to (4, 12) the rise is 9 and the run is 3, so m = 3. A negative gradient leans downhill: y = −½x + 4 falls 1 for every 2 across. Gradients mean things in context — £2 per ticket, 3 miles per hour — which is why interpreting m earns marks on worded questions.

To find a line through two points, compute the gradient, then substitute one point into y = mx + c to find c. Through (2, 5) and (6, 11): m = (11 − 5)⁄(6 − 2) = 1.5; then 5 = 1.5 × 2 + c gives c = 2, so y = 1.5x + 2. With one point and a gradient, the same substitution finishes it: through (1, 3) with gradient 3 gives 3 = 3 + c, so c = 0 and y = 3x.

Keep fractions exact while you work: through (−1, 2) and (5, −1) the gradient is (−1 − 2)⁄(5 − (−1)) = −3⁄6 = −½; then 2 = −½ × (−1) + c gives c = 1½, so y = −½x + 1½. Converting to decimals early invites rounding drift — exact fractions protect the final mark.

Parallel lines never meet, so their gradients are equal: y = 2x + 1 and y = 2x − 5 are parallel; y = 2x + 1 and y = 3x + 1 are not. For a Higher perpendicular question, turn the gradient upside down and change the sign: perpendicular to y = 3x + 1 is gradient −⅓, because 3 × (−⅓) = −1. Fractions are easiest: the perpendicular to y = −½x + 4 has gradient 2.

A typical Higher question chains the skills: find the line perpendicular to y = 3x + 1 through (6, 1). Gradient −⅓; then 1 = −⅓ × 6 + c gives c = 3, so y = −⅓x + 3. State the gradient product check (3 × −⅓ = −1) in the working — it shows the rule was used deliberately, which some mark schemes credit.

Solving 2x + 1 = x + 4 graphically means plotting y = 2x + 1 and y = x + 4 and reading the x-coordinate where they cross: here x = 3. The graph gives an approximate solution — read as carefully as the grid allows and quote what you see (“x ≈ 3”). The same idea solves any equation: to solve x² = 2x + 3 you would plot y = x² and y = 2x + 3 (that method returns in Chapter 8).