AS & A Level · Practice questions · Simultaneous equations: one linear, one quadratic

Substitute, solve, pair up.

Six original Cambridge-style questions on simultaneous equations where one is linear and one is quadratic, finishing with a rectangle problem in context.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

One letter out, one quadratic in.

Rearrange the linear equation for the letter with the simplest coefficient.
Substitute everywhere that letter appears in the quadratic, including \(xy\) terms.
Expand squares fully, collect to \(ax^{2} + bx + c = 0\) and solve.
Back-substitute each root into the LINEAR equation to find its partner.
Pair the answers as coordinates; a repeated root means the line is a tangent.

Recall

[3 marks]

Solve the simultaneous equations \(y = 3x - 2\) and \(y = x^{2}\).

\(x^{2} = 3x - 2 \Rightarrow x^{2} - 3x + 2 = 0\). ✓
\((x - 1)(x - 2) = 0 \Rightarrow x = 1\) or \(x = 2\). ✓
\(x = 1,\ y = 1\) or \(x = 2,\ y = 4\). ✓

Calculation

[4 marks]

Solve the simultaneous equations \(x + 2y = 5\) and \(x^{2} + y^{2} = 10\).

\(x = 5 - 2y\), so \((5 - 2y)^{2} + y^{2} = 10\). ✓
\(25 - 20y + 4y^{2} + y^{2} = 10 \Rightarrow 5y^{2} - 20y + 15 = 0 \Rightarrow y^{2} - 4y + 3 = 0\). ✓
\((y - 1)(y - 3) = 0 \Rightarrow y = 1\) or \(y = 3\). ✓
\(y = 1\): \(x = 3\). \(y = 3\): \(x = -1\). Solutions \((3, 1)\) and \((-1, 3)\). ✓

Calculation

[4 marks]

Solve the simultaneous equations \(3x + y = 7\) and \(x^{2} + xy = 5\).

\(y = 7 - 3x\), so \(x^{2} + x(7 - 3x) = 5\). ✓
\(-2x^{2} + 7x - 5 = 0 \Rightarrow 2x^{2} - 7x + 5 = 0 \Rightarrow (2x - 5)(x - 1) = 0\). ✓
\(x = 1\): \(y = 4\). ✓
\(x = \dfrac{5}{2}\): \(y = 7 - \dfrac{15}{2} = -\dfrac{1}{2}\). ✓

Analysis

[5 marks]

The line \(y = x - 1\) meets the curve \(y^{2} = x + 5\) at the points A and B.

(a) Find the coordinates of A and B.

(b) Find the coordinates of the midpoint of AB and the exact length of AB.

(a) \((x - 1)^{2} = x + 5 \Rightarrow x^{2} - 2x + 1 = x + 5 \Rightarrow x^{2} - 3x - 4 = 0\). ✓
\((x - 4)(x + 1) = 0\): A\((4, 3)\) and B\((-1, -2)\). ✓
(b) Midpoint \(\left(\dfrac{4 + (-1)}{2},\ \dfrac{3 + (-2)}{2}\right) = \left(\dfrac{3}{2},\ \dfrac{1}{2}\right)\). ✓
AB \(= \sqrt{5^{2} + 5^{2}} = \sqrt{50} = 5\sqrt{2}\). ✓

Proof

[3 marks]

Show that the line \(y = 2x - 5\) and the curve \(y = x^{2} - 4x + 4\) meet at exactly one point, and find its coordinates.

\(x^{2} - 4x + 4 = 2x - 5 \Rightarrow x^{2} - 6x + 9 = 0\). ✓
Discriminant \(36 - 36 = 0\), so there is exactly one (repeated) root: the line is a tangent. ✓
\((x - 3)^{2} = 0 \Rightarrow x = 3,\ y = 2(3) - 5 = 1\). The point is \((3, 1)\). ✓

Analysis

[5 marks]

A rectangular sheet of metal has perimeter 34 cm. The length of a diagonal of the sheet is 13 cm.

(a) Taking the sides as \(x\) cm and \(y\) cm, write down two equations in \(x\) and \(y\).

(b) Solve your equations to find the dimensions of the sheet, and hence its area.

(a) \(2x + 2y = 34\), so \(x + y = 17\); by Pythagoras \(x^{2} + y^{2} = 169\). ✓
(b) \(y = 17 - x\): \(x^{2} + (17 - x)^{2} = 169 \Rightarrow 2x^{2} - 34x + 289 = 169\). ✓
\(2x^{2} - 34x + 120 = 0 \Rightarrow x^{2} - 17x + 60 = 0 \Rightarrow (x - 5)(x - 12) = 0\). ✓
\(x = 5,\ y = 12\) or \(x = 12,\ y = 5\): the sheet is 12 cm by 5 cm. ✓
Area \(= 12 \times 5 = 60\) cm². ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.