SACE Board SACE 2023-style practice: Calculator-free. Solve |2x - 3| = x + 4.

The argument 2x - 3 changes sign at x = 32. Case 1 (x 32): 2x - 3 = x + 4, so x = 7. Since 7 32, valid. [1 mark] Case 2 (x < 32): -(2x - 3) = x + 4, so 3 - 2x = x + 4, giving -1 = 3x and x = -13. Since -13 < 32, valid. [1 mark] Both make the right side x + 4 non-negative, so the solutions are x = 7 and x = -13. [1 mark]

SACE Board SACE 2021-style practice: Calculator-free. Solve the inequality |3x + 2| < 8.

Use |A| < c -c < A < c: $-8 < 3x + 2 < 8.$ Subtract 2: -10 < 3x < 6. Divide by 3: -103 < x < 2. Marks: one for the double inequality, one for the solution -103 < x < 2.

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How does taking the absolute value reshape a graph, and how do we solve equations and inequalities involving it?

Sketch graphs involving the modulus (absolute value) function and solve equations and inequalities containing modulus expressions.

How the modulus reflects negative parts of a graph above the axis, the difference between |f(x)| and f(|x|), and solving modulus equations and inequalities by cases or squaring.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Definition and key properties
Sketching modulus graphs
Solving modulus equations
The triangle inequality
Solving modulus inequalities

What this dot point is asking

You need to sketch graphs involving the modulus function and to solve equations and inequalities that contain modulus expressions.

Definition and key properties

Sketching modulus graphs

Two distinct transformations get confused, so keep them separate:

$y = |f(x)|$ : sketch $y = f(x)$ , then reflect every part that is below the $x$ -axis up to above it. Points already at or above the axis stay put. The result never dips below the $x$ -axis and typically has sharp corners where $f$ crossed the axis.
$y = f(|x|)$ : keep the graph for $x \ge 0$ , then reflect it in the $y$ -axis to produce the $x < 0$ side. The result is symmetric about the $y$ -axis and ignores the original left half.

Solving modulus equations

The cleanest general method is to split at the points where each modulus argument changes sign, solve on each interval, and discard solutions that fall outside the interval assumed.

Solving a modulus equation by cases

Solve $|x - 3| = 2x - 1$

Step 1: Identify the sign-change point and set up cases

The expression $x - 3$ inside the modulus changes sign at $x = 3$ . This splits the number line into two regions where $|x-3|$ has a different algebraic form, so we treat each region separately.

The modulus argument $x - 3$ changes sign at $x = 3$ .

Step 2: Solve Case 1 and check

For $x \ge 3$ , the expression $x-3$ is non-negative, so $|x-3|=x-3$ . Solve the resulting linear equation and verify the solution lies in this region.

Case 1 ( $x \ge 3$ ): here $|x - 3| = x - 3$ , so

x - 3 = 2x - 1 \implies -2 = x.

But

x = -2

does not satisfy

x \ge 3

, so reject it.

Step 3: Solve Case 2 and check

For $x < 3$ , the expression $x-3$ is negative, so $|x-3|=-(x-3)=3-x$ . Solve the new equation, confirm the solution is in this region, and also verify that the right-hand side is non-negative (since a modulus cannot equal a negative number).

Case 2 ( $x < 3$ ): here $|x - 3| = -(x - 3) = 3 - x$ , so

3 - x = 2x - 1 \implies 4 = 3x \implies x = \tfrac{4}{3}.

Since

\tfrac{4}{3} < 3

, this is valid. Check the right side is non-negative:

2(\tfrac{4}{3}) - 1 = \tfrac{5}{3} > 0

. Good.

Final answer: $x = \dfrac{4}{3}$ .

For equations of the form $|A| = |B|$ , squaring is often faster, since $|A| = |B| \iff A^2 = B^2 \iff A = \pm B$ .

Solving an equation with two moduli

Solve $|x + 1| = |2x - 4|$

Step 1: Square both sides

When both sides of an equation are moduli (and hence non-negative), squaring is valid because $|A|=|B|$ is equivalent to $|A|^2=|B|^2$ , and squaring removes the modulus signs on both sides without introducing extraneous solutions from sign changes.

Both sides are moduli, so square: $(x + 1)^2 = (2x - 4)^2$ .

Step 2: Expand and simplify

Expand each side using the standard binomial identity and collect all terms on one side.

Expanding gives $x^2 + 2x + 1 = 4x^2 - 16x + 16$ , so $3x^2 - 18x + 15 = 0$ , that is $x^2 - 6x + 5 = 0$ .

Step 3: Factorise, solve and verify

Factorise the quadratic to read off the solutions, then confirm each by substituting back into the original equation.

The equation factorises to $(x - 1)(x - 5) = 0$ . The solutions are $x = 1$ and $x = 5$ , both of which can be confirmed by substitution into the original equation.

Final answer: $x = 1$ and $x = 5$ .

The triangle inequality

A property worth knowing is the triangle inequality, $|a + b| \le |a| + |b|$ , with equality only when $a$ and $b$ have the same sign. It captures the idea that the distance of a sum is never more than the sum of the distances, and it underlies many bounds in later mathematics. While SACE rarely asks you to prove it, recognising the modulus as a measure of distance - $|x - a|$ being the gap between $x$ and $a$ on the number line - is the intuition that makes both the triangle inequality and the geometric reading of modulus equations natural.

Solving modulus inequalities

Two standard equivalences handle most cases, for $c > 0$ :

|x| < c \iff -c < x < c, \qquad |x| > c \iff x < -c \ \text{or} \ x > c.

For

|f(x)| < c

, replace

x

f(x)

and solve the resulting double inequality. A reliable alternative is to square (valid because both sides are non-negative) and solve the polynomial inequality, then sketch to read off the solution set.

A modulus inequality

Solve $|2x + 1| \le 5$

Step 1: Convert the modulus inequality to a double inequality

The rule $|A| \le c$ (for $c > 0$ ) is equivalent to $-c \le A \le c$ . This captures the idea that the expression $A$ must lie within distance $c$ of zero on the number line. Applying the rule to $|2x+1| \le 5$ gives a compound inequality with no modulus.

-5 \le 2x + 1 \le 5.

Step 2: Solve for $x$

Perform the same arithmetic operation across all three parts of the compound inequality to isolate $x$ .

Subtract $1$ : $-6 \le 2x \le 4$ . Divide by $2$ : $-3 \le x \le 2$ .

Final answer: $-3 \le x \le 2$ .

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20222 marksCalculator-assumed. For

f(x) = \dfrac{5}{(x-2)(x+3)}

, describe how to obtain the graph of

y = |f(x)|

from the graph of

y = f(x)

, and state the asymptotes.

Show worked answer →

Keep every part of $y = f(x)$ that is already at or above the $x$ -axis, and reflect every part below the $x$ -axis up across it, since the modulus makes all outputs non-negative. [1 mark]

For $f(x) = \dfrac{5}{(x-2)(x+3)}$ the curve has vertical asymptotes at $x = 2$ and $x = -3$ , and is negative on the middle branch $-3 < x < 2$ . After the modulus, that middle branch flips above the axis, so $y = |f(x)|$ lies at or above $y = 0$ , still with vertical asymptotes at $x = -3$ and $x = 2$ and horizontal asymptote $y = 0$ . [1 mark]

SACE 20233 marksCalculator-free. Solve

|2x - 3| = x + 4

Show worked answer →

The argument $2x - 3$ changes sign at $x = \tfrac{3}{2}$ .

Case 1 ( $x \ge \tfrac{3}{2}$ ): $2x - 3 = x + 4$ , so $x = 7$ . Since $7 \ge \tfrac{3}{2}$ , valid. [1 mark]

Case 2 ( $x < \tfrac{3}{2}$ ): $-(2x - 3) = x + 4$ , so $3 - 2x = x + 4$ , giving $-1 = 3x$ and $x = -\tfrac{1}{3}$ . Since $-\tfrac{1}{3} < \tfrac{3}{2}$ , valid. [1 mark]

Both make the right side $x + 4$ non-negative, so the solutions are $x = 7$ and $x = -\tfrac{1}{3}$ . [1 mark]

SACE 20212 marksCalculator-free. Solve the inequality

|3x + 2| < 8

Show worked answer →

Use $|A| < c \iff -c < A < c$ :

-8 < 3x + 2 < 8.

Subtract $2$ : $-10 < 3x < 6$ . Divide by $3$ : $-\tfrac{10}{3} < x < 2$ .

Marks: one for the double inequality, one for the solution $-\tfrac{10}{3} < x < 2$ .