JAMB 1993 · UME · Q21

If $x$ is negative, what is the range of values of $x$ within which $\frac{x + 1}{3} > \frac{1}{x + 3}$ ?

Worked solution (try it first)

Bring everything to one side and use a common denominator: $\frac{x + 1}{3} - \frac{1}{x + 3} = \dfrac{(x + 1)(x + 3) - 3}{3(x + 3)}$ .
Expand the top: $x^2 + 4x + 3 - 3 = x(x + 4)$ .
So you need $\dfrac{x(x + 4)}{3(x + 3)} > 0$ .
The critical values are $-4$ , $-3$ and $0$ .
Test a negative $x$ in each interval: $x < -4$ gives a negative value, $-4 < x < -3$ positive, and $-3 < x < 0$ negative.
So $-4 < x < -3$ , option B.