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Topic 5

Simultaneous equations

Syllabus 5.1 · Papers 1 and 2

🎯What you need to be able to do

  • Solve a linear and a non-linear equation simultaneously by substitution.
  • Pair each \(x\) with its own \(y\), and present the solutions as coordinate pairs.
  • Solve systems where elimination is quicker, for example by dividing one equation by the other.
  • Interpret the solutions as the points where a line meets a curve, and use them (lengths, midpoints).

📚The mathematics

Substitution from the linear equation

Make \(x\) or \(y\) the subject of the linear equation, choosing whichever avoids fractions, and substitute into the other one. You get a quadratic in one variable. Solve it, then find the other variable by substituting back into the linear equation.

Why the linear one? Each \(x\) value gives exactly one \(y\) from a straight line. Substituting back into a quadratic can give two \(y\) values for each \(x\), and pairs that are not solutions at all.

The closed curve x squared plus xy plus y squared = 7, a tilted oval centred on the origin, and the line y = x + 1 crossing it at A(1, 2) and B(minus 2, minus 1), with the midpoint M(minus a half, a half) of the chord AB marked.
The worked example: the two solutions are the two points where the line cuts the curve.

Other shapes you will meet

  • Products. \( xy = 6 \) and \( 2x + y = 7 \): substitute \( y = 7 - 2x \) into \( xy = 6 \).
  • Dividing equations. \( xy^{2} = 12 \) and \( xy = 3 \): neither is linear, but dividing the first by the second gives \( y = 4 \) at once (allowed because \( xy \ne 0 \)).
  • Indices. \( 2^{x} \times 4^{y} = 32 \) becomes \( 2^{x + 2y} = 2^{5} \), so \( x + 2y = 5 \): write every term as a power of the same base, then equate the powers.

The number of solutions is the number of intersection points, which the discriminant of the final quadratic predicts (Topic 2): no real roots means the line misses the curve.

✏️Worked example

The line \( y = x + 1 \) meets the curve \( x^{2} + xy + y^{2} = 7 \) at the points \(A\) and \(B\). (a) Find the coordinates of \(A\) and \(B\). [5] (b) Find the exact length of \(AB\) and the coordinates of its midpoint. [3]

(a) Substitute \( y = x + 1 \):

\[ x^{2} + x(x + 1) + (x + 1)^{2} = 7 \;\Longrightarrow\; 3x^{2} + 3x + 1 = 7 \;\Longrightarrow\; x^{2} + x - 2 = 0 \]

so \( (x + 2)(x - 1) = 0 \) and \( x = 1 \) or \( x = -2 \). From the line, \( x = 1 \Rightarrow y = 2 \) and \( x = -2 \Rightarrow y = -1 \). So \( A(1, 2) \) and \( B(-2, -1) \).

(b) \( AB = \sqrt{(1 - (-2))^{2} + (2 - (-1))^{2}} = \sqrt{9 + 9} = 3\sqrt{2} \). Midpoint: \( \left(\dfrac{1 + (-2)}{2}, \dfrac{2 + (-1)}{2}\right) = \left(-\tfrac{1}{2}, \tfrac{1}{2}\right) \).

Check it. Put both points into the curve: \( 1 + 2 + 4 = 7 \) ✓ and \( 4 + 2 + 1 = 7 \) ✓. The midpoint also lies on the line: \( \tfrac{1}{2} = -\tfrac{1}{2} + 1 \) ✓.
Crossed pairs. Listing \( x = 1, -2 \) and \( y = 2, -1 \) separately does not say which goes with which, and \( (1, -1) \) is not a solution. Write the answers as coordinate pairs.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Solve the simultaneous equations \( x + 2y = 5 \) and \( x^{2} + y^{2} = 10 \). [5]
\( x = 5 - 2y \): \( 25 - 20y + 4y^{2} + y^{2} = 10 \Rightarrow 5y^{2} - 20y + 15 = 0 \Rightarrow y^{2} - 4y + 3 = 0 \), so \( y = 1 \) or \( y = 3 \). Then \( x = 3 \) or \( x = -1 \). Solutions \( (3, 1) \) and \( (-1, 3) \).
2. (No calculator.) Solve \( xy = 6 \) and \( 2x + y = 7 \). [4]
\( y = 7 - 2x \Rightarrow x(7 - 2x) = 6 \Rightarrow 2x^{2} - 7x + 6 = 0 \Rightarrow (2x - 3)(x - 2) = 0 \). So \( (2, 3) \) and \( \left(\tfrac{3}{2}, 4\right) \).
3. (No calculator.) Solve \( xy^{2} = 12 \) and \( xy = 3 \). [2]
Neither \(x\) nor \(y\) can be zero, so divide: \( \dfrac{xy^{2}}{xy} = y = 4 \). Then \( x = \tfrac{3}{4} \). Solution \( \left(\tfrac{3}{4}, 4\right) \).
4. (No calculator.) Show that the line \( y = 3 - x \) does not meet the curve \( x^{2} - 3xy + y^{2} + 11 = 0 \). [4]
Substitute: \( x^{2} - 3x(3 - x) + (3 - x)^{2} + 11 = 5x^{2} - 15x + 20 = 0 \), i.e. \( x^{2} - 3x + 4 = 0 \). Discriminant \( 9 - 16 = -7 < 0 \): no real roots, so the line and curve do not meet.
5. (Calculator.) The line \( y = 2x + 3 \) meets the curve \( y = x^{2} - x - 1 \) at \(P\) and \(Q\). Find the length of \(PQ\). [5]
\( x^{2} - x - 1 = 2x + 3 \Rightarrow x^{2} - 3x - 4 = 0 \Rightarrow x = 4 \) or \( x = -1 \). Points \( (4, 11) \) and \( (-1, 1) \). \( PQ = \sqrt{5^{2} + 10^{2}} = \sqrt{125} = 11.2 \) (3 s.f.).
6. (No calculator.) Solve \( 2^{x} \times 4^{y} = 32 \) and \( \dfrac{3^{x}}{9^{y}} = \dfrac{1}{3} \). [4]
\( 2^{x} \times 2^{2y} = 2^{5} \Rightarrow x + 2y = 5 \). \( 3^{x - 2y} = 3^{-1} \Rightarrow x - 2y = -1 \). Adding: \( 2x = 4 \), \( x = 2 \), and \( y = \tfrac{3}{2} \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — type both equations to see the intersection points you should be finding
  • Khan Academy — systems of non-linear equations