Solve a linear and a quadratic equation together
Substitute one relation into the other, recover every coordinate pair and check both original equations.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Paper 1 is part of every route. Written practice is self-checked; it does not award official examination marks.
Before you begin
- Rearrange a linear equation for one variable.
- Expand a squared bracket and solve a quadratic.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Substitute y = x + 1 into x² + y² = 13. Which equation is correct?
What you will learn
- Use substitution when one simultaneous equation is linear and the other quadratic.
- Recover and check paired values, including tangent intersections.
Replace one variable everywhere
A linear relation such as y = x + 1 lets you replace every y in x² + y² = 13 with x + 1. Keep brackets around the substituted expression: x² + (x + 1)² = 13 becomes 2x² + 2x − 12 = 0. Its roots x = 2 and x = −3 must each be taken back to y = x + 1.
These roots create the pairs (2, 3) and (−3, −2). Pairing x = 2 with y = −2 incorrectly mixes two different solutions. Check each pair in both original equations; finding only the x-values does not finish a simultaneous-equations problem.
Intersections and repeated solutions
The same method equates a line y = mx + c to a quadratic curve y = f(x). The resulting quadratic can have two, one repeated or no real roots, corresponding to two intersections, tangency or no intersection. For y = 2x − 1 and y = x², x² − 2x + 1 = 0 gives the single tangent point (1, 1).
Substitution also works when the quadratic relation includes xy rather than a graph already isolated for y. If x + y = 5 and xy = 6, use y = 5 − x to obtain x(5 − x) = 6. Avoid dividing by x unless you have checked that x = 0 is impossible; dividing can discard a legitimate solution in other systems.
Pause and explain
If x = −3 and y = x + 1, which paired solution follows?
Worked example
Solve y = x + 1 and x² + y² = 13 simultaneously.
Show the worked solution
- Substitute y = x + 1: x² + (x + 1)² = 13.
- Simplify to x² + x − 6 = 0 and factor (x + 3)(x − 2) = 0.
- For x = 2, y = 3; for x = −3, y = −2. Both pairs give x² + y² = 13.
Answer (x, y) = (2, 3) or (−3, −2).
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the intersection of y = 2x − 1 and y = x². Explain why there is one point.
Give me a hint
Equate the two expressions for y and look for a repeated root.
Compare my reasoning
- Write x² = 2x − 1, or x² − 2x + 1 = 0.
- Factor to (x − 1)² = 0, giving one distinct x-value, x = 1.
- The line gives y = 1; both expressions agree and the repeated root shows tangency.
One tangent point: (1, 1).
Look for these in your work
- Recognise the repeated root as one distinct point.
- Recover y and check both expressions.
Solve x + y = 5 and xy = 6 simultaneously.
Give me a hint
Use the linear equation to write y = 5 − x.
Compare my reasoning
- Substitute to x(5 − x) = 6.
- Rearrange to x² − 5x + 6 = 0, giving x = 2 or 3.
- Recover y = 3 or 2 respectively; both pairs have sum 5 and product 6.
(x, y) = (2, 3) or (3, 2).
Look for these in your work
- Reduce to one quadratic by substitution.
- Keep both correct ordered pairs.
A rectangle has perimeter 20 cm and diagonal √58 cm. Find its positive side lengths using simultaneous equations.
Give me a hint
Let the sides be x and y; use the perimeter and Pythagoras.
Compare my reasoning
- The perimeter gives x + y = 10; the diagonal gives x² + y² = 58.
- Substitute y = 10 − x: 2x² − 20x + 42 = 0, or (x − 3)(x − 7) = 0.
- The pairs are (3, 7) and (7, 3), describing the same rectangle with swapped side names. Both give perimeter 20 and diagonal squared 58.
The side lengths are 3 cm and 7 cm.
Look for these in your work
- Use half the perimeter for the sum of adjacent sides.
- Check positivity and both geometric constraints.
Common mistakes
- Square the whole substituted expression, including its cross term.
- Keep each x-value paired with its own y-value.
- Check both original equations rather than only the rearranged quadratic.
What can you explain now?
Solve y = 2x + 1 and y = x² − 2 simultaneously.
Compare with the explanation
(x, y) = (3, 7) or (−1, −1).
Equating gives x² − 2x − 3 = 0, with roots 3 and −1. Substituting into the line gives 7 and −1.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain why two roots of the substituted equation must be converted into two pairs. Show how tangency appears in the algebra.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.