IGCSE · Practice questions · Harder simultaneous equations

Pick the route, then pair the values.

Six original Cambridge-style questions on harder simultaneous equations: dividing one equation by the other, clearing fractions by multiplying by \(xy\), using \((x + y)^{2}\) and \((x - y)^{2}\), and a context where the pair has no solution.

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

The method

Spot the shape, choose the route, check the pairs.

Products of powers such as \(xy^{2}\) and \(xy\): divide (or multiply) the equations so a letter cancels.
Fractions: note \(x \ne 0,\ y \ne 0\), then multiply EVERY term by \(xy\); factorise if you can.
\(x^{2} + y^{2}\) and \(xy\): find \((x + y)^{2}\) and \((x - y)^{2}\), take \(\pm\) roots, then solve the linear pairs.
Pair each \(x\) with its own \(y\), using the simpler equation such as \(y = \dfrac{12}{x}\).
Check every pair in BOTH original equations, and reject lengths that are not positive.

Recall

[3 marks]

Solve the simultaneous equations \(xy^{2} = 45\) and \(xy = 15\).

Neither \(x\) nor \(y\) can be 0, so divide: \(\dfrac{xy^{2}}{xy} = \dfrac{45}{15}\), giving \(y = 3\). ✓
\(3x = 15\), so \(x = 5\). ✓
Check: \(xy^{2} = 5 \times 9 = 45\). Solution \(x = 5,\ y = 3\). ✓

Calculation

[4 marks]

Solve the simultaneous equations \(x^{3}y = 24\) and \(xy = 6\).

Divide: \(\dfrac{x^{3}y}{xy} = \dfrac{24}{6}\), so \(x^{2} = 4\) and \(x = 2\) or \(x = -2\). ✓
From \(xy = 6\): \(x = 2\) gives \(y = 3\); \(x = -2\) gives \(y = -3\). ✓
Check: \(8 \times 3 = 24\) and \((-8) \times (-3) = 24\). ✓
Solutions \(x = 2,\ y = 3\) and \(x = -2,\ y = -3\). ✓

Calculation

[5 marks]

Solve the simultaneous equations \(\dfrac{x}{y} + \dfrac{4y}{x} = 5\) and \(x + y = 10\).

\(x \ne 0,\ y \ne 0\). Multiply every term by \(xy\): \(x^{2} + 4y^{2} = 5xy\). ✓
\(x^{2} - 5xy + 4y^{2} = (x - y)(x - 4y) = 0\), so \(x = y\) or \(x = 4y\). ✓
\(x = y\): \(2y = 10\), so \(x = 5,\ y = 5\). ✓
\(x = 4y\): \(5y = 10\), so \(y = 2,\ x = 8\). ✓
Check: \(1 + 4 = 5\) and \(\dfrac{8}{2} + \dfrac{8}{8} = 4 + 1 = 5\). Solutions \((5, 5)\) and \((8, 2)\). ✓

Calculation

[5 marks]

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

\(x \ne 0,\ y \ne 0\). Multiply every term by \(xy\): \(3y + 2x = 2xy\). ✓
Substitute \(x = 5 - y\): \(3y + 10 - 2y = 2y(5 - y)\), so \(y + 10 = 10y - 2y^{2}\). ✓
\(2y^{2} - 9y + 10 = 0\), so \((2y - 5)(y - 2) = 0\), giving \(y = 2\) or \(y = \dfrac{5}{2}\). ✓
\(y = 2\) gives \(x = 3\); \(y = \dfrac{5}{2}\) gives \(x = \dfrac{5}{2}\). ✓
Check: \(1 + 1 = 2\) and \(\dfrac{6}{5} + \dfrac{4}{5} = 2\). Solutions \((3, 2)\) and \(\left(\dfrac{5}{2}, \dfrac{5}{2}\right)\). ✓

Analysis

[5 marks]

(a) Given that \(x^{2} + y^{2} = 34\) and \(xy = 15\), find the possible values of \(x + y\) and of \(x - y\).

(b) Hence solve the simultaneous equations, giving each solution as a pair.

(a) \((x + y)^{2} = 34 + 30 = 64\), so \(x + y = \pm 8\). ✓
\((x - y)^{2} = 34 - 30 = 4\), so \(x - y = \pm 2\). ✓
(b) \(x + y = 8\) with \(x - y = 2\) gives \((5, 3)\); with \(x - y = -2\) gives \((3, 5)\). ✓
\(x + y = -8\) with \(x - y = 2\) gives \((-3, -5)\); with \(x - y = -2\) gives \((-5, -3)\). ✓
Each pair has \(xy = 15\) and \(x^{2} + y^{2} = 34\): four solutions \((5, 3),\ (3, 5),\ (-3, -5),\ (-5, -3)\). ✓

Analysis

[6 marks]

A rectangular sheet of card has sides \(x\) cm and \(y\) cm, where \(x > y\). Its area is 120 cm² and its diagonal is 17 cm.

(a) Write down two equations in \(x\) and \(y\).

(b) Find \(x\) and \(y\), and the perimeter of the card.

(a) \(xy = 120\) and, by Pythagoras, \(x^{2} + y^{2} = 289\). ✓
(b) \((x + y)^{2} = 289 + 240 = 529\), so \(x + y = 23\) (lengths are positive). ✓
\((x - y)^{2} = 289 - 240 = 49\), so \(x - y = 7\) since \(x > y\). ✓
Adding: \(2x = 30\), so \(x = 15,\ y = 8\). Check: \(15 \times 8 = 120\), \(225 + 64 = 289\). Perimeter \(= 2 \times 23 = 46\) cm. ✓
(c) Here \(x^{2} + y^{2} = 100\) and \(xy = 60\), so \((x - y)^{2} = 100 - 120 = -20\). ✓
A square of a real number cannot be negative, so there are no real values of \(x\) and \(y\): no such rectangle exists. ✓

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.