Revision for linear and quadratic inequalities, modulus equations and inequalities, equations that become quadratics after a substitution, and (4PM1) regions and linear programming. Both boards solve linear and quadratic inequalities. Modulus equations and inequalities, and the substitution method, are Cambridge 0606 only. Regions defined by inequalities and simple linear programming are Edexcel 4PM1 only.
Solve a linear inequality like an equation, but reverse the inequality sign when you multiply or divide by a negative number.
For a quadratic inequality, collect everything on one side, find the critical values, then sketch the parabola to see whether the answer is between them or outside them.
|ax + b| = c means ax + b = c or ax + b = −c. When the right-hand side contains x, check every answer in the original equation and reject any that fail.
|A| < |B| can be squared safely, because neither side is ever negative: A² < B² then gives a quadratic inequality.
Spot a hidden quadratic: in 2(e^x)² − 5e^x + 2 = 0 or x − 5√x + 6 = 0, let u = e^x or u = √x, solve for u, then go back to x.
Linear programming (4PM1): draw the boundary lines, shade the region that satisfies every inequality, then test the objective at each vertex. The best value is at a vertex, or at a nearby point with whole-number coordinates if the answer must be whole.
Watch out: Never divide both sides of an inequality by an expression such as (x + 3) that could be negative. Collect the terms on one side and factorise instead.
Worked example· 3 marks
Solve the inequality \(3x^2 - 2x - 1 \gt 2x^2 + x + 9\).
M1 collect terms: x² − 3x − 10 > 0
A1 critical values x = 5 and x = −2
A1 x < −2 or x > 5
Take 2x² + x + 9 from both sides: x² − 3x − 10 > 0.
Factorise: (x − 5)(x + 2) > 0, so the critical values are x = 5 and x = −2.
The parabola y = x² − 3x − 10 is U-shaped, so it is above the x-axis outside the roots: x < −2 or x > 5.
Answer: \(x \lt -2\) or \(x \gt 5\)
Equations and inequalities 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 · 3 marks0606no calculator
Solve the equation \(|2x - 5| = 7\).
Show the mark scheme and worked solution
M1 2x − 5 = 7 or 2x − 5 = −7
A1 x = 6
A1 x = −1
|2x − 5| = 7 means 2x − 5 = 7 or 2x − 5 = −7.
2x − 5 = 7 gives 2x = 12, so x = 6.
2x − 5 = −7 gives 2x = −2, so x = −1.
Answer: \(x = 6\) or \(x = -1\)
Question 2 · 5 marks4PM1no calculator
The region \(R\) is defined by the inequalities \(y \geq 1\), \(x + y \leq 6\) and \(y \leq 2x\).
(a) Find the coordinates of the vertices of \(R\).
(b) Find the area of \(R\).
Show the mark scheme and worked solution
M1 (a) solve two boundary equations at a time
A1 (a) (½, 1) and (5, 1)
A1 (a) (2, 4)
M1 (b) ½ × base × height with base 4½ along y = 1
A1 (b) 27/4
The boundaries are y = 1, x + y = 6 and y = 2x.
y = 1 and y = 2x meet where x = ½: the vertex (½, 1).
y = 1 and x + y = 6 meet where x = 5: the vertex (5, 1).
x + y = 6 and y = 2x: 3x = 6, so x = 2 and y = 4: the vertex (2, 4).
R is a triangle. Its base lies on y = 1 from x = ½ to x = 5, a length of 4½, and its height is 4 − 1 = 3.
IGCSE Math Revision