Equations and Inequalities

Inequalities appear throughout the H2 Maths 9758 syllabus — as constraints in differentiation problems, domains for inverse functions, and standalone examination questions. Mastering the sign-diagram technique and modulus inequalities early saves significant time across topics.

What you need to know

• Never cross-multiply by an unknown expression: if $f(x)$ can be negative, multiplying both sides by $f(x)$ reverses the inequality sign unpredictably. Always move everything to one side and use a sign diagram instead.
• Sign diagram (rational/polynomial inequalities): factorise the numerator and denominator completely, mark all critical values on a number line, then test the sign of the whole expression in each interval. The sign alternates at every simple root or pole.
• Modulus inequalities — two cases: $|A| < b$ (with $b > 0$) is equivalent to $-b < A < b$; $|A| > b$ is equivalent to $A < -b$ or $A > b$. For $|A| < |B|$, square both sides (both are non-negative) to get $A^2 < B^2$.
• Graphical method: sketch $y = f(x)$ and $y = g(x)$ on the same axes; the solution to $f(x) > g(x)$ is the set of $x$-values where the graph of $f$ lies above the graph of $g$.
• Systems of linear equations: for two equations in two unknowns, elimination or substitution gives an exact algebraic answer. For three equations in three unknowns the Graphing Calculator (GC) is expected in the examination — use Simultaneous Eqn Solver — but be ready to show one elimination step if asked.

Worked example 1 — rational inequality via sign diagram

Problem. Solve $\dfrac{x-1}{x+2} \ge 0$.

Solution. The expression equals zero at $x = 1$ and is undefined at $x = -2$. Critical values: $-2$ and $1$.

The quotient is $\ge 0$ on the positive intervals, and equals zero at $x = 1$ (included). $x = -2$ is excluded (undefined).

Worked example 2 — modulus inequality

Problem. Solve $|2x - 3| < 5$.

Solution. Apply the rule $|A| < b \Leftrightarrow -b < A < b$:

Add 3 throughout:

Divide by 2:

Worked example 3 — system of linear equations

Problem. Solve $2x + y = 7$ and $3x - 2y = 0$.

Solution. From the second equation, $y = \dfrac{3x}{2}$. Substituting into the first:

Then $y = \dfrac{3(2)}{2} = 3$.

Common mistakes

• Cross-multiplying by $(x + 2)$ in Example 1 without considering its sign — this gives the wrong solution set for the interval $x < -2$.
• Forgetting to exclude undefined points: at a pole such as $x = -2$, the inequality is undefined; that value must never be included even if the sign diagram shows a positive region up to it.
• Modulus: forgetting to flip the inequality for the negative branch: $|A| > b$ gives two separate inequalities; writing $-b > A > b$ (a single compound inequality) is a contradiction with no solution.
• Squaring modulus inequalities prematurely: $|A| < |B|$ may be squared safely (both sides non-negative), but $|A| < B$ may not — you must first verify or require $B > 0$.