Revision for solving one linear and one non-linear equation by substitution, pairing the solutions correctly, and problems that lead to simultaneous equations. Both boards solve one linear and one quadratic equation together. Cambridge 0606 also sets pairs where neither equation is linear, such as xy = 4 with x² + y² = 17.
Rearrange the linear equation to make x or y the subject, then substitute into the other equation to get a quadratic in one letter.
Choose the letter that avoids fractions: from 2x + y = 5, use y = 5 − 2x.
Each root of the quadratic gives one solution pair. Find the other letter from the linear equation, not the quadratic, so you do not pick up extra pairs.
Write the answers as pairs, for example x = 4, y = −3 and x = −1.5, y = 8. Mixing up the pairs loses the final mark.
If the quadratic has a repeated root, the line is a tangent to the curve; if it has no real roots, they do not meet.
When neither equation is linear (0606), look for a way to eliminate: divide one equation by the other, or use (x + y)² = x² + 2xy + y².
Watch out: Check each pair in both original equations. A pair that satisfies only one of them comes from a slip in the substitution.
Worked example· 5 marks
Solve the simultaneous equations
\(2x + y = 5\)
\(xy = -12\)
M1 substitute y = 5 − 2x
A1 2x² − 5x − 12 = 0
M1 solve the quadratic
A1 x = 4, y = −3
A1 x = −3/2, y = 8
From the first equation, y = 5 − 2x.
Substitute: x(5 − 2x) = −12, so 5x − 2x² = −12 and 2x² − 5x − 12 = 0.
Factorise: (2x + 3)(x − 4) = 0, so x = 4 or x = −3/2.
y = 5 − 2x: x = 4 gives y = −3; x = −3/2 gives y = 8.
Answer: \(x = 4, y = -3\) and \(x = -\frac{3}{2}, y = 8\)
Simultaneous equations questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 4 marks06064PM1no calculator
The line \(y = 2x - 2\) meets the curve \(y = x^2 - 3x + 4\) at the points \(A\) and \(B\).
Find the coordinates of \(A\) and \(B\), and the exact length of \(AB\).
Show the mark scheme and worked solution
M1 x² − 3x + 4 = 2x − 2
A1 x = 2 and x = 3
A1 (2, 2) and (3, 4)
B1 AB = √5
Set the equations equal: x² − 3x + 4 = 2x − 2, so x² − 5x + 6 = 0.
(x − 2)(x − 3) = 0, so x = 2 or x = 3.
On the line, x = 2 gives y = 2 and x = 3 gives y = 4: the points (2, 2) and (3, 4).
AB = √((3 − 2)² + (4 − 2)²), which is √5.
Answer: \((2, 2)\) and \((3, 4)\); \(AB = \sqrt{5}\)
Question 2 · 4 marks06064PM1no calculator
A rectangular garden has an area of 84 m² and a perimeter of 38 m.
Find its length and width.
Show the mark scheme and worked solution
B1 l + w = 19 and lw = 84
M1 substitute: w(19 − w) = 84
A1 w² − 19w + 84 = 0
A1 12 m by 7 m
Let the length be l m and the width w m. Perimeter: 2l + 2w = 38, so l + w = 19. Area: lw = 84.
l = 19 − w, so (19 − w)w = 84 and w² − 19w + 84 = 0.
(w − 7)(w − 12) = 0, so w = 7 or w = 12.
Either way the sides are 12 m and 7 m: length 12 m, width 7 m.
Answer: length 12 m, width 7 m
Question 3 · 3 marks06064PM1no calculator
Show that the line \(y = 4x - 1\) is a tangent to the curve \(y = x^2 + 2x\), and find the coordinates of the point of contact.
Show the mark scheme and worked solution
M1 x² + 2x = 4x − 1
A1 (x − 1)² = 0: one repeated root
A1 (1, 3)
Set the equations equal: x² + 2x = 4x − 1, so x² − 2x + 1 = 0.
This is (x − 1)² = 0: one repeated root, so the line meets the curve at only one point and is a tangent.
x = 1 gives y = 4(1) − 1, which is 3: the point of contact is (1, 3).
Answer: \((1, 3)\)
Question 4 · 5 marks06064PM1no calculator
Solve the simultaneous equations
\(x + y = 4\)
\(x^2 - xy + 2y^2 = 14\)
giving your answers in the form \(a \pm b\sqrt{7}\).
IGCSE Math Revision