Algebra

Rearranging an equation — whatever is done to one side must be done to the other. Work towards isolating the subject by undoing operations in reverse order.

Example: making a the subject of v² = u² + 2as. Subtract u² from both sides to get v² − u² = 2as, then divide both sides by 2s, giving a = (v² − u²) ÷ 2s. With v = 10, u = 4.0 and s = 6.0, a = (100 − 16) ÷ 12 = 7.0 m s⁻².

Rearrange before substituting — put the equation into the form you need while the symbols are still there, then put the numbers in once. Substituting first means carrying numbers through every step, which is slower and gives more opportunities for arithmetic slips.

Subjects inside a power or root — undo the power last, since it was applied last. To make v the subject of Eₖ = ½mv², multiply by 2, divide by m, then take the square root: v = √(2Eₖ/m).

Subjects in a denominator — multiply both sides by the denominator first. To make A the subject of R = ρL/A, multiply both sides by A to get RA = ρL, then divide by R, giving A = ρL/R.

Direct proportion — y ∝ x means y = kx for a constant k. Doubling x doubles y, and a graph of y against x is a straight line through the origin.

Inverse proportion — y ∝ 1/x means xy = k. Doubling x halves y, and a graph of y against 1/x is a straight line through the origin.

Proportion with powers — if y ∝ x², doubling x multiplies y by four; if y ∝ 1/x², doubling x divides y by four. Both appear constantly: kinetic energy with speed, and intensity with distance.

Example: braking distance is proportional to v². Going from 20 to 60 mph triples the speed, so the braking distance becomes nine times greater, not three. Recognising the power is what makes that answer immediate.

Solving with two unknowns — two independent equations are needed. Either substitute one into the other, or eliminate a variable by adding or subtracting the equations.