Home / IGCSE Add Maths 0606 / Simultaneous equations / Simultaneous equations: a line and a curve
Cambridge IGCSE · Topic 5.1
IGCSE Add Maths 0606 / Topic 5 / IGCSE

Substitute the line, then pair the answers.

When one equation is linear and the other has squared or product terms, elimination by adding and subtracting no longer works. Instead make one letter the subject of the linear equation, substitute it into the curve and solve the quadratic that appears. Each root gives one point where the line meets the curve.

Topic 5All simultaneous equations lessons →
The key idea

Rearrange the linear equation to make \(x\) or \(y\) the subject (choose the one with coefficient 1 to avoid fractions), substitute into the non-linear equation, expand and collect into a quadratic \(= 0\), and solve it. Substitute each root back into the LINEAR equation to find its partner, and give the answers as pairs \((x, y)\). Two distinct roots: the line cuts the curve at two points. One repeated root: the line is a tangent. No real roots: they do not meet.

x y O −5 −5 5 5 A(−4, −3) B(3, 4) M(−½, ½) x² + y² = 25 y = x + 1 Substitute for y: x² + x − 12 = 0 x = −4 or x = 3 Two roots, so two intersection points AB = 7√2 M is the midpoint
Fig. 1: the line y = x + 1 cuts the circle x² + y² = 25 at A(-4, -3) and B(3, 4). Substituting gives the quadratic x² + x - 12 = 0, whose two roots are the x-coordinates of A and B. The chord AB has midpoint M(-1/2, 1/2) and length 7√2
Section 01

Substitute, expand, pair.

With one linear and one non-linear equation, use substitution. The linear equation tells you exactly what one letter is in terms of the other, so replace that letter everywhere in the non-linear equation. You are left with a quadratic in a single letter. Its roots are only half of each answer: put every root back into the linear equation to find the matching value of the other letter.

Worked example A line and a parabola
Solve the simultaneous equations \(y = x^{2} - 3x + 4\) and \(y = 2x - 2\).
  1. Both equations give \(y\), so set them equal: \(x^{2} - 3x + 4 = 2x - 2\).
  2. Collect terms: \(x^{2} - 5x + 6 = 0\), so \((x - 2)(x - 3) = 0\).
  3. \(x = 2\): \(y = 2(2) - 2 = 2\). \(x = 3\): \(y = 2(3) - 2 = 4\).
  4. Check in the curve: \(4 - 6 + 4 = 2\) and \(9 - 9 + 4 = 4\).
Answer: \(x = 2,\ y = 2\) or \(x = 3,\ y = 4\), i.e. the points \((2, 2)\) and \((3, 4)\)
Worked example Substituting into an xy term
Solve the simultaneous equations \(x - y + 1 = 0\) and \(x^{2} + xy + y^{2} = 7\).
  1. From the linear equation, \(y = x + 1\).
  2. Substitute: \(x^{2} + x(x + 1) + (x + 1)^{2} = 7\).
  3. Expand: \(x^{2} + x^{2} + x + x^{2} + 2x + 1 = 7\), so \(3x^{2} + 3x - 6 = 0\).
  4. Divide by 3: \(x^{2} + x - 2 = 0\), so \((x + 2)(x - 1) = 0\) and \(x = 1\) or \(x = -2\).
  5. \(x = 1\): \(y = 2\). \(x = -2\): \(y = -1\). Check: \(1 + 2 + 4 = 7\) and \(4 + 2 + 1 = 7\).
Answer: \(x = 1,\ y = 2\) or \(x = -2,\ y = -1\)
Your turn
Solve the simultaneous equations \(y = 2x + 1\) and \(y = x^{2} - 2\).
Examiner trap

The answer is PAIRS, not a list. Writing '\(x = 1\) or \(-2\), \(y = 2\) or \(-1\)' does not say which \(y\) goes with which \(x\). And find \(y\) from the linear equation: substituting \(x\) into a curve with \(y^{2}\) in it gives extra values of \(y\) that are not solutions.

Section 02

Pick the letter that avoids fractions.

You can make either letter the subject, but one choice is usually much easier. Look at the coefficients in the linear equation: if \(x\) has coefficient 1, make \(x\) the subject. In \(x + 2y = 5\), \(x = 5 - 2y\) is clean, but \(y = \dfrac{5 - x}{2}\) brings fractions into every line. Then expand with care: \((5 - 2y)^{2}\) has three terms, \(25 - 20y + 4y^{2}\). Geometrically, every pair you find is a point where the line meets the curve, whether the curve is a parabola or a circle (Fig. 1).

Worked example Make x the subject
Solve the simultaneous equations \(x + 2y = 5\) and \(x^{2} + 3y^{2} = 13\).
  1. \(x\) has coefficient 1, so \(x = 5 - 2y\).
  2. Substitute: \((5 - 2y)^{2} + 3y^{2} = 13\), so \(25 - 20y + 4y^{2} + 3y^{2} = 13\).
  3. Collect: \(7y^{2} - 20y + 12 = 0\), which factorises as \((7y - 6)(y - 2) = 0\).
  4. \(y = 2\): \(x = 5 - 4 = 1\). \(y = \dfrac{6}{7}\): \(x = 5 - \dfrac{12}{7} = \dfrac{23}{7}\).
  5. Check the second pair: \(\dfrac{529}{49} + 3 \times \dfrac{36}{49} = \dfrac{637}{49} = 13\).
Answer: \(x = 1,\ y = 2\) or \(x = \frac{23}{7},\ y = \frac{6}{7}\)
Worked example A line and a circle
Find the coordinates of the points where the line \(y = x - 1\) meets the circle \((x - 2)^{2} + (y + 1)^{2} = 10\).
  1. Substitute \(y = x - 1\): \(y + 1 = x\), so \((x - 2)^{2} + x^{2} = 10\).
  2. Expand: \(x^{2} - 4x + 4 + x^{2} = 10\), so \(2x^{2} - 4x - 6 = 0\).
  3. Divide by 2: \(x^{2} - 2x - 3 = 0\), so \((x - 3)(x + 1) = 0\).
  4. \(x = 3\): \(y = 2\). \(x = -1\): \(y = -2\).
  5. Check: \(1^{2} + 3^{2} = 10\) and \((-3)^{2} + (-1)^{2} = 10\).
Answer: \((3, 2)\) and \((-1, -2)\)
Your turn
Solve the simultaneous equations \(2x + y = 3\) and \(x^{2} + y^{2} = 5\).
Examiner trap

\((5 - 2y)^{2}\) is NOT \(25 + 4y^{2}\) or \(25 - 4y^{2}\): write it as \((5 - 2y)(5 - 2y)\) and you get the middle term \(-20y\). A lost middle term or a sign slip here changes both answers, and every later mark goes with it.

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

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To solve \(x + 3y = 10\) and \(x^{2} + y^{2} = 10\), which substitution avoids fractions?

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Expand \((3 - 2x)^{2}\).

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After substituting a line into a curve, the quadratic has one repeated root. What does this mean?

Section 03

The chord AB and the tangent case.

A common follow-up: the line meets the curve at \(A\) and \(B\); find the midpoint or the length of \(AB\). Once you have both points, use the coordinate formulae: the midpoint is \(\left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right)\) and the length is \(\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\), left as an exact surd. The number of roots of the quadratic also tells you how the line and curve meet. As in topic 2.3 on the discriminant: two distinct roots, two points; a repeated root, the line is a tangent touching the curve at ONE point; no real roots, the line misses the curve.

Worked example Midpoint and length of a chord
The line \(y = x + 1\) meets the circle \(x^{2} + y^{2} = 25\) at \(A\) and \(B\). Find the midpoint of \(AB\) and the exact length of \(AB\).
  1. Substitute: \(x^{2} + (x + 1)^{2} = 25\), so \(2x^{2} + 2x - 24 = 0\) and \(x^{2} + x - 12 = 0\).
  2. \((x + 4)(x - 3) = 0\): \(x = -4\) gives \(y = -3\), \(x = 3\) gives \(y = 4\). So \(A(-4, -3)\), \(B(3, 4)\) (Fig. 1).
  3. Midpoint: \(\left(\dfrac{-4 + 3}{2}, \dfrac{-3 + 4}{2}\right) = \left(-\dfrac{1}{2}, \dfrac{1}{2}\right)\).
  4. Length: \(\sqrt{(3 - (-4))^{2} + (4 - (-3))^{2}} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}\).
Answer: Midpoint \(\left(-\frac{1}{2}, \frac{1}{2}\right)\), \(AB = 7\sqrt{2}\)
Worked example One repeated root: a tangent
Show that the line \(y = 2x + 5\) is a tangent to the circle \(x^{2} + y^{2} = 5\), and find the point of contact.
  1. Substitute: \(x^{2} + (2x + 5)^{2} = 5\), so \(x^{2} + 4x^{2} + 20x + 25 = 5\).
  2. Collect: \(5x^{2} + 20x + 20 = 0\), and divide by 5: \(x^{2} + 4x + 4 = 0\).
  3. \((x + 2)^{2} = 0\) has the single repeated root \(x = -2\), so the line meets the circle at only one point: it is a tangent.
  4. \(y = 2(-2) + 5 = 1\). Check: \((-2)^{2} + 1^{2} = 5\).
Answer: Repeated root, so the line touches the circle at \((-2, 1)\)
Your turn
The line \(y = x - 1\) meets the curve \(y = x^{2} - 4x + 3\) at \(A\) and \(B\). Find the midpoint of \(AB\) and the exact length of \(AB\).
Examiner trap

The length of \(AB\) uses BOTH the \(x\)-difference and the \(y\)-difference: the gap between the roots of the quadratic (7 in the worked example) is only the horizontal distance. Subtract with brackets, \(3 - (-4) = 7\), not \(3 - 4\). On Paper 1 leave the answer as a simplified surd, \(7\sqrt{2}\), not \(9.90\).

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Solve the simultaneous equations \(y = x - 2\) and \(x^{2} + 2xy - y^{2} = 14\).

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The line \(y = 3 - x\) meets the circle \(x^{2} + y^{2} = 29\) at \(A\) and \(B\). Find the midpoint of \(AB\).

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The line \(y = 2x + 1\) meets the curve \(y = x^{2} + 1\) at \(P\) and \(Q\). Find the exact length of \(PQ\).

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The line \(y = 4x - 4\) and the curve \(y = x^{2}\) are drawn on the same axes. Which statement is true?

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Six original Cambridge-style questions
A line with a parabola, an \(xy\) term, a choice of subject that avoids fractions, the midpoint and length of a chord of a circle, a tangent, and a constant chosen to make a line touch a curve.