An inequality with an algebraic fraction

Inequalities

A-level Further Maths (9FM0) · Further Pure 1

ORIGINAL

Never multiply through by an expression of unknown sign — collect to one fraction, find critical values, test the intervals.

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Question

Use algebra to find the set of values of

x

for which

\dfrac{6}{x+2} \le x - 3

Every step worked, with the reasoning.

1
Critical value from the denominator: $x = -2$
The fraction is undefined at $x = -2$ , and $x + 2$ can be negative, so you cannot multiply through by it.
2
$x - 3 - \dfrac{6}{x+2} \ge 0 \Rightarrow \dfrac{(x - 3)(x + 2) - 6}{x + 2} \ge 0$
Collect everything on one side as a single fraction.
3
$\dfrac{x^2 - x - 12}{x + 2} \ge 0 \Rightarrow \dfrac{(x - 4)(x + 3)}{x + 2} \ge 0$
$(x - 3)(x + 2) - 6 = x^2 - x - 6 - 6 = x^2 - x - 12$ , which factorises.
4
Critical values: $x = -3,\ -2,\ 4$
Where the numerator is zero and where the denominator is zero.
5
Sign of $\dfrac{(x - 4)(x + 3)}{x + 2}$ : $x < -3$ : $-$ ; $\ -3 < x < -2$ : $+$ ; $\ -2 < x < 4$ : $-$ ; $\ x > 4$ : $+$
Test one value in each interval, e.g. $x = -2.5$ gives $\dfrac{(-6.5)(0.5)}{-0.5} = 6.5 > 0$ .
6
Include $x = -3$ and $x = 4$ (the expression equals $0$ ) but exclude $x = -2$ .
The inequality is non-strict, but the fraction is undefined at $x = -2$ .

Answer: $\{x : -3 \le x < -2\} \cup \{x : x \ge 4\}$