Solving equations and inequalities

Solving a linear equation means using inverse operations to isolate the unknown, doing the same thing to both sides each time. With the unknown on both sides, collect the letters on one side first — usually the side with more of them, which avoids negatives — and the numbers on the other.

Always check by substituting back into the original equation. It takes seconds and catches sign errors, which are what go wrong here.

Every quadratic must be rearranged to equal zero before any method is used. x² = 5x + 6 becomes x² − 5x − 6 = 0, and only then can it be factorised to (x − 6)(x + 1) = 0, giving x = 6 or x = −1.

The reason factorising works is that if two things multiply to give zero, one of them must be zero — so each bracket is set to zero in turn.

When it will not factorise, use the quadratic formula:

x = ( −b ± √(b² − 4ac) ) ÷ 2a

The ± is what gives the two solutions. Completing the square is the third route, and is the one to use when the question asks for an exact answer in surd form or for the turning point.

Approximate solutions can also be read from a graph, as the points where the curve crosses the x-axis.

Two simultaneous equations are solved together, and the solution is the pair of values satisfying both — which on a graph is the point where the two lines cross.

For two linear equations, use elimination. Multiply one or both equations so that one letter has the same coefficient in each, then add or subtract the equations to eliminate it. Subtract when the signs match, add when they are opposite. Solve for the remaining letter, then substitute back to find the other.

For one linear and one quadratic, elimination does not work — use substitution instead. Rearrange the linear equation to make one letter the subject, substitute that into the quadratic, and solve the resulting quadratic. This usually gives two solutions, so there are two pairs of values, and each y must be paired with the x it came from.

Iteration finds an approximate solution to an equation that cannot be solved algebraically, by feeding an answer back into a formula until the values settle.

The method is mechanical. Substitute the given x₀ into the iterative formula to get x₁, substitute that to get x₂, and continue. On a calculator use the ANS key: enter the starting value, type the formula with ANS in place of x, and press equals repeatedly.

Stop when successive values agree to the accuracy asked for, writing each to more decimal places than the final answer needs so the rounding happens once.

A related question asks you to show a solution lies between two values. Substitute both into the equation rearranged to equal zero: if one gives a negative result and the other a positive one, the curve crosses zero between them. Saying that the sign changes is the reasoning the mark scheme wants.

Translating a situation into algebra is marked in its own right. Define the letter explicitly — "let x be the number of adult tickets" — write the relationships as equations, solve, then answer the question actually asked, in context and with units.

Inequalities are solved exactly like equations, with one exception: multiplying or dividing both sides by a negative reverses the sign. So −2x > 6 gives x < −3.

Show a solution set on a number line with an open circle for < or >, because that value is excluded, and a filled circle for ≤ or ≥. In set notation the same solution is written {x : x < −3}.

A quadratic inequality is solved by finding the roots, then sketching the parabola to see which region satisfies it. For x² − x − 6 < 0 the roots are −2 and 3, and the curve is below the axis between them, so the answer is −2 < x < 3. Above the axis it is two separate regions, written x < −2 or x > 3. Sketching settles it faster than reasoning, and explains why the answer is sometimes one region and sometimes two.