Home / A-Level Maths 9709 / Quadratics / Simultaneous equations: one linear, one quadratic
Cambridge International AS & A Level · Topic 1.1
A-Level Maths 9709 / Topic 1.1 / AS & A Level

Where a line meets a curve.

When one equation is linear and the other is quadratic, make one letter the subject of the linear equation and substitute. The result is a quadratic, and each of its roots gives one point where the line meets the curve.

Topic 1.1All quadratics lessons →
The key idea

Rearrange the LINEAR equation to make \(x\) or \(y\) the subject, substitute into the quadratic, solve the resulting quadratic, then substitute each root back into the linear equation. Solutions come in pairs \((x, y)\): one pair for each point of intersection.

x y (3, −4) (−4, 3) x² + y² = 25 x + y + 1 = 0 Substitute y = −x − 1: x² + (x + 1)² = 25 x² + x − 12 = 0 x = 3 or x = −4
Fig. 1: the line x + y + 1 = 0 cuts the circle x² + y² = 25 at (3, -4) and (-4, 3); the two roots of the substituted quadratic are the x-coordinates of these points
Section 01

Substitute from the linear one.

Elimination by adding or subtracting equations does not work when one equation has squared terms. Instead, substitute. Choose whichever letter is easiest to make the subject of the linear equation, ideally one with coefficient 1, to avoid fractions.

Worked example Line and parabola
Solve the simultaneous equations \(y = x + 2\) and \(y = x^{2} - 4\).
  1. Both expressions equal \(y\), so \(x^{2} - 4 = x + 2\).
  2. Rearrange: \(x^{2} - x - 6 = 0\), so \((x - 3)(x + 2) = 0\).
  3. \(x = 3\): \(y = 3 + 2 = 5\). \(x = -2\): \(y = -2 + 2 = 0\).
Answer: \(x = 3,\ y = 5\) or \(x = -2,\ y = 0\)
Worked example Line and circle
Solve \(x + y + 1 = 0\) and \(x^{2} + y^{2} = 25\) (Fig. 1).
  1. From the linear equation, \(y = -x - 1\).
  2. Substitute: \(x^{2} + (-x - 1)^{2} = 25 \Rightarrow x^{2} + x^{2} + 2x + 1 = 25\).
  3. \(2x^{2} + 2x - 24 = 0 \Rightarrow x^{2} + x - 12 = 0 \Rightarrow (x + 4)(x - 3) = 0\).
  4. \(x = 3\): \(y = -4\). \(x = -4\): \(y = 3\).
Answer: \((3,\ -4)\) and \((-4,\ 3)\)
Your turn
Solve \(y = 3x - 2\) and \(y = x^{2}\).
Examiner trap

\((-x - 1)^{2} = x^{2} + 2x + 1\), not \(x^{2} + 1\). Write the square as a product and expand it fully before collecting terms.

Section 02

Equations with an xy term.

Some quadratic equations contain a product \(xy\). The method is the same: substitute for one letter everywhere it appears, including inside the \(xy\) term, then expand carefully.

Worked example Fractional solution
Solve the simultaneous equations \(2x + y = 5\) and \(x^{2} + 2xy = 8\).
  1. Make \(y\) the subject: \(y = 5 - 2x\).
  2. Substitute into both places: \(x^{2} + 2x(5 - 2x) = 8 \Rightarrow x^{2} + 10x - 4x^{2} = 8\).
  3. \(-3x^{2} + 10x - 8 = 0 \Rightarrow 3x^{2} - 10x + 8 = 0 \Rightarrow (3x - 4)(x - 2) = 0\).
  4. \(x = 2\): \(y = 5 - 4 = 1\). \(x = \dfrac{4}{3}\): \(y = 5 - \dfrac{8}{3} = \dfrac{7}{3}\).
Answer: \(x = 2,\ y = 1\) or \(x = \dfrac{4}{3},\ y = \dfrac{7}{3}\)
Worked example Substituting for x
Solve \(x + 3y = 10\) and \(x^{2} + 3y^{2} = 28\).
  1. \(x\) has coefficient 1, so \(x = 10 - 3y\).
  2. \((10 - 3y)^{2} + 3y^{2} = 28 \Rightarrow 100 - 60y + 9y^{2} + 3y^{2} = 28\).
  3. \(12y^{2} - 60y + 72 = 0 \Rightarrow y^{2} - 5y + 6 = 0 \Rightarrow y = 2\) or \(y = 3\).
  4. \(y = 2\): \(x = 4\). \(y = 3\): \(x = 1\).
Answer: \((4,\ 2)\) and \((1,\ 3)\)
Examiner trap

Find each partner from the LINEAR equation, not the quadratic. Substituting \(y = 2\) into \(x^{2} + 3y^{2} = 28\) gives \(x = \pm 4\), and \((-4, 2)\) is not a solution of \(x + 3y = 10\).

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To solve \(y = x + 1\) and \(x^{2} + y^{2} = 13\), the best first step is to

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Solve \(y = 2x\) and \(y = x^{2}\).

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Substituting \(y = 3 - x\) into \(xy = 2\) gives

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After substituting, the quadratic has a repeated root. What does this tell you about the line and the curve?

Section 03

What the roots mean.

Each root of the substituted quadratic is the \(x\)-coordinate (or \(y\)-coordinate) of a point where the line meets the curve. So the number of roots tells you how the line and curve meet, and the discriminant can tell you before you solve.

Worked example A tangent
Show that the line \(y = 4x - 4\) is a tangent to the curve \(y = x^{2}\), and find the point of contact.
  1. Substitute: \(x^{2} = 4x - 4 \Rightarrow x^{2} - 4x + 4 = 0\).
  2. Discriminant: \(16 - 16 = 0\), a repeated root, so the line touches the curve.
  3. \((x - 2)^{2} = 0 \Rightarrow x = 2\), and \(y = 4(2) - 4 = 4\).
Answer: Tangent at \((2,\ 4)\)
Worked example In context
A rectangle has perimeter 26 cm and area 40 cm2. Find its length and width.
  1. Let the sides be \(x\) and \(y\) cm: \(2x + 2y = 26\) and \(xy = 40\).
  2. From the linear equation, \(y = 13 - x\), so \(x(13 - x) = 40\).
  3. \(x^{2} - 13x + 40 = 0 \Rightarrow (x - 5)(x - 8) = 0\), so \(x = 5,\ y = 8\) or \(x = 8,\ y = 5\).
Answer: 8 cm by 5 cm (both solutions describe the same rectangle)
Examiner trap

Give the answers as matched pairs. Writing '\(x = 3\) or \(-2\), \(y = 5\) or \(0\)' does not say which \(y\) goes with which \(x\), and examiners withhold the final mark.

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Solve the simultaneous equations \(x - y = 2\) and \(x^{2} + y^{2} = 34\).

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Find the coordinates of the points where the line \(y = 2x - 1\) meets the curve \(x^{2} + xy = 10\).

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The line \(y = x + 1\) cuts the circle \(x^{2} + y^{2} = 25\) at A and B. Find the length AB.

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

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Go deeper · practice
Six original Cambridge-style questions
Line and parabola, line and circle, an \(xy\) term, the midpoint and length of a chord, a tangent proof, and a rectangle from its perimeter and diagonal.