Simultaneous equations linear and quadratic GCSE questions — a Higher-tier staple — are solved by substitution: rearrange the linear equation for one letter, substitute it into the quadratic equation, then solve the resulting quadratic, which usually gives two pairs of solutions, using factorising or the quadratic formula.
Why does substitution work for a linear and quadratic pair?
Elimination — adding or subtracting equations — relies on matching coefficients of the same power of x. A linear equation contains only x and y, while a quadratic equation contains an x² (or y²) term, so there is no matching power to cancel by subtraction. Substitution sidesteps this: it replaces one letter entirely, turning two equations into a single equation in one unknown, which you already know how to solve.
How do you solve one linear and one quadratic equation step by step?
- Rearrange the linear equation to make x or y the subject.
- Substitute that expression into the quadratic equation, replacing the matching letter.
- Expand any brackets and simplify into the form ax² + bx + c = 0.
- Solve the resulting quadratic by factorising or the quadratic formula.
- Substitute each x-value back into the linear equation to find its matching y-value.
- Write both solution pairs and check them in both original equations.
Worked example: solve the simultaneous equations y = x + 1 and y = x² − 5.
- Both equations equal y, so set the right-hand sides equal: x + 1 = x² − 5.
- Rearrange into standard form: x² − x − 6 = 0.
- Factorise: (x − 3)(x + 2) = 0.
- So x = 3 or x = −2.
- Substitute into y = x + 1: x = 3 gives y = 4; x = −2 gives y = −1.
- Check in y = x² − 5: x = 3 gives 9 − 5 = 4 ✓; x = −2 gives 4 − 5 = −1 ✓.
Answer: (3, 4) and (−2, −1).
How do you solve simultaneous equations with a circle equation?
Higher-tier papers sometimes pair a linear equation with a circle equation such as x² + y² = r². The method is identical — rearrange the linear equation and substitute — but you must expand a squared bracket carefully.
Worked example: solve x + y = 5 and x² + y² = 17.
- Rearrange the linear equation: y = 5 − x.
- Substitute into the circle equation: x² + (5 − x)² = 17.
- Expand: x² + 25 − 10x + x² = 17, giving 2x² − 10x + 8 = 0.
- Divide every term by 2: x² − 5x + 4 = 0.
- Factorise: (x − 1)(x − 4) = 0, so x = 1 or x = 4.
- Substitute into y = 5 − x: x = 1 gives y = 4; x = 4 gives y = 1.
Check: 1² + 4² = 1 + 16 = 17 ✓; 4² + 1² = 16 + 1 = 17 ✓.
Answer: (1, 4) and (4, 1).
What does the number of solutions tell you geometrically?
A linear and quadratic pair represents a straight line and a curve on the same grid. The number of solutions matches the number of times the line and curve meet, and it is controlled by the discriminant (b² − 4ac) of the final quadratic.
| Number of real solutions | Discriminant | Geometric meaning |
|---|---|---|
| 2 | b² − 4ac > 0 | Line crosses the curve at two distinct points |
| 1 | b² − 4ac = 0 | Line touches the curve at exactly one point (a tangent) |
| 0 | b² − 4ac < 0 | Line and curve never meet |
Recognising which case applies before solving can save time: if a question asks you to "show that a line is a tangent to a curve," you are being asked to prove the discriminant equals zero, not to find coordinates.
What if the resulting quadratic does not factorise neatly?
Not every quadratic from a substitution step has whole-number roots. When factorising does not work, use the quadratic formula:
x = (−b ± √(b² − 4ac)) ÷ 2a
Identify a, b and c from your rearranged equation ax² + bx + c = 0, substitute carefully — particularly the sign of b — and simplify the surd if it does not reduce to a whole number. Once you have both x-values, substitute each back into the linear equation exactly as before to find the paired y-values.
Frequently asked questions
Why can't you use elimination for linear and quadratic simultaneous equations?
Elimination cancels a variable by matching its coefficient in both equations, but a quadratic equation contains a squared term with no matching power in the linear equation. There is nothing to subtract away, so elimination cannot reduce the pair to one unknown. Substitution avoids this problem because it replaces a letter directly rather than relying on matching powers.
How many solutions can a linear and quadratic simultaneous equation system have?
Up to two, since substituting produces a quadratic equation, and a quadratic can have two, one, or zero real solutions. Geometrically this corresponds to a straight line crossing a curve twice, touching it once as a tangent, or missing it entirely. Always report every valid (x, y) pair the algebra produces, not just the first one you find.
What does it mean if the quadratic formula gives a negative number under the square root?
A negative value under the square root means the discriminant is negative, so there is no real solution — the straight line does not meet the curve anywhere on the graph. This is a valid and common GCSE answer; state clearly that there are no real solutions rather than trying to force a numerical result.
Which equation should you rearrange — the linear or the quadratic?
Always rearrange the linear equation, since it has only one power of each letter and is far easier to isolate for x or y. Substituting a rearranged linear expression into the quadratic keeps the algebra manageable; attempting to rearrange the quadratic equation first usually creates square roots and unnecessary complications.
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