Papers 2 and 3. \(|x|\) is the size of \(x\) with its sign thrown away. Learn the V-shaped graph of \(y = |ax + b|\), then solve modulus equations and inequalities by squaring, by splitting into cases, or from a sketch.
\(|x|\) is the distance of \(x\) from 0. Two facts do most of the work: \(|a| = |b| \iff a^2 = b^2\), and \(|x - a| < b \iff a - b < x < a + b\). When one side is NOT a modulus, any answer found by splitting into cases must be checked in the original equation.
Fig. 1: y = |2x - 3| is y = 2x - 3 with the part below the x-axis (dashed) reflected up. It meets y = x at x = 1 and x = 3, so |2x - 3| < x for 1 < x < 3
Section 01
The graph of \(y = |ax + b|\) is a V.
Draw the straight line \(y = ax + b\) first, then reflect every part that lies below the \(x\)-axis in the \(x\)-axis. The result is a V whose vertex sits on the \(x\)-axis where \(ax + b = 0\).
Definition: \(|x| = x\) if \(x \ge 0\), and \(|x| = -x\) if \(x < 0\). So \(|5| = 5\) and \(|-5| = 5\).
Vertex: \(y = |ax + b|\) touches the \(x\)-axis at \(x = -\dfrac{b}{a}\).
\(y\)-intercept: put \(x = 0\) to get \((0, |b|)\).
Two arms: gradients \(a\) and \(-a\), so the V is symmetrical about the vertical line through the vertex.
Scope: on this syllabus you sketch \(|ax + b|\) only; \(|f(x)|\) for curves is not required.
Worked example Sketching a modulus graph
Sketch \(y = |2x - 3|\), stating the coordinates of the points where it meets the axes.
Vertex: \(2x - 3 = 0\) gives \(x = 1.5\), so the V touches the \(x\)-axis at \((1.5, 0)\).
For \(x \ge 1.5\) the graph is \(y = 2x - 3\); for \(x < 1.5\) it is the reflection \(y = 3 - 2x\).
Answer: A V with vertex \((1.5, 0)\) and \(y\)-intercept \((0, 3)\), arms of gradient \(2\) and \(-2\) (Fig. 1).
Worked example Reading an inequality from a sketch
Use Fig. 1 to solve \(|2x - 3| < x\).
Right arm meets \(y = x\): \(2x - 3 = x\), so \(x = 3\).
Left arm meets \(y = x\): \(3 - 2x = x\), so \(x = 1\).
The V is below the line \(y = x\) between the two crossing points.
Answer: \(1 < x < 3\)
Examiner trap
The vertex of \(y = |2x - 3|\) is at \(x = 1.5\), not \(x = 3\) or \(x = -1.5\). Solve \(ax + b = 0\) properly; the most common slip is forgetting to divide by \(a\).
Section 02
Equations: square, or split into cases.
When BOTH sides are moduli, use \(|a| = |b| \iff a^2 = b^2\), or equivalently \(a = b\) or \(a = -b\). When only ONE side is a modulus, split into the two cases and check each answer, because squaring or splitting can create false roots.
In Case 2, \(-(2x - 1)\) is \(-2x + 1\), not \(-2x - 1\). Put brackets round the whole side before you change its sign.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What is the value of \(|-7|\)?
Quick check+10 XP
Where does the graph of \(y = |2x + 6|\) touch the \(x\)-axis?
Quick check+10 XP
Which is equivalent to \(|x - 5| < 2\)?
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For real \(a\) and \(b\), \(|a| = |b|\) is equivalent to
Section 03
Inequalities: squeeze, square or sketch.
Pick the method that fits the shape of the question. A modulus less than a number becomes a double inequality. Two moduli can be squared safely because both sides are non-negative. A modulus against a straight line is best done with a sketch and the crossing points.
Form
Method
Result
\(|x - a| < b\)
double inequality
\(a - b < x < a + b\)
\(|x - a| > b\)
two separate pieces
\(x < a - b\) or \(x > a + b\)
\(|f| < |g|\)
square both sides
\(f^2 < g^2\), a quadratic inequality
\(|f| < \) a line
sketch, find crossings
read off where the V is below
Worked example Modulus less than a number
Solve \(|2x - 5| < 7\).
Write as a double inequality: \(-7 < 2x - 5 < 7\).
Add 5 throughout: \(-2 < 2x < 12\).
Divide by 2: \(-1 < x < 6\).
Answer: \(-1 < x < 6\)
Worked example Two moduli: square
Solve \(|x - 3| > |2x + 1|\).
Both sides are non-negative, so squaring keeps the inequality: \((x - 3)^2 > (2x + 1)^2\).
The line is steeper than the V, so it lies below the V only to the left of \(x = -2\). Check \(x = -3\): \(-1 < 2\). True.
Answer: \(x < -2\)
Examiner trap
Never square when one side might be negative: \(2x + 5 < |x + 1|\) squared gives a wrong answer, because squaring a negative left side flips which side is bigger. Square only when both sides are moduli (or are known to be non-negative).
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Solve the equation \(|2x + 1| = |x - 4|\).
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Solve the inequality \(|3x - 2| < 4\).
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Solve the inequality \(|x - 1| > |x + 3|\).
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Solve the inequality \(|3x - 6| < x + 2\).
Skill unlocked
Modulus, mastered.
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