Quadratic simultaneous equations

The idea

Substitute the linear equation into the quadratic one. The quadratic you get has up to two solutions, and each gives one point where the line meets the curve.

Make $y$ (or $x$ ) the subject of the linear equation.
Substitute into the quadratic, rearrange to $= 0$ and solve.
Put each $x$ back into the linear equation to get its $y$ .

Watch out An answer is a pair: give both coordinates for each solution.

See it worked

Worked example

Solve $y = x + 1$ and $x^2 + y^2 = 13$ .

Substitute: $x^2 + (x + 1)^2 = 13$ .
$2x^2 + 2x - 12 = 0$ , so $x^2 + x - 6 = 0$ and $(x + 3)(x - 2) = 0$ .
$x = 2$ gives $y = 3$ ; $x = -3$ gives $y = -2$ .

Answer $(2, 3)$ and $(-3, -2)$

Your turn

Question 1Non-calculator3 marks

The line $y = 2x - 1$ meets the curve $y = x^2 - 4x + 7$ at two points. Find their $x$ -coordinates.

Question 2Non-calculator3 marks

Solve the simultaneous equations $x + y = 3$ and $xy = 2$ . Give the $x$ -values.

Question 3Non-calculator3 marks

The line $y = x + k$ is a tangent to the curve $y = x^2 + 3x$ . Find $k$ .

In past papers

23 official questions in 21 papers since 2018. The latest:

All A-level Maths past papers

A-level Maths Pro, Pearson Edexcel 9MA0

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