2024 H2 Mathematics Paper 1 Question 4

Equations and Inequalities

Answers

$- 2 < x \leq - 1$ or $0 < x \leq 6.$

$- 6 \leq x < 0$ or $0 < x \leq 6.$

(a)

\begin{matrix} \frac{4}{x + 2} \geq \frac{x - 3}{x} \\ \frac{4}{x + 2} - \frac{x - 3}{x} \geq 0 \\ \frac{4 x - ((x + 2) (x - 3))}{(x + 2) x} \geq 0 \\ \frac{5 x - x^{2} + 6}{(x + 2) x} \geq 0 \\ \frac{x^{2} - 5 x - 6}{(x + 2) x} \leq 0 \\ \frac{(x + 1) (x - 6)}{(x + 2) x} \leq 0 \end{matrix}

number-line

- 2 < x \leq - 1 ∎ or 0 < x \leq 6 ∎

(b)

Replacing $x$ from part (a) with $| x |,$

- 2 < | x | \leq - 1 or 0 < | x | \leq 6

Not that $- 2 < | x | \leq - 1$ is not applicable as $| x | \geq 0$ for all real values of $x$

abs-graph

From the graph, solution to the inequality is

- 6 \leq x < 0 ∎ or 0 < x \leq 6 ∎