JAMB 2014 · UTME · Q17

What is the solution of $\dfrac{x - 5}{x + 3} < -1$ ?

Worked solution (try it first)

Add 1 to both sides to get 0 on the right: $\dfrac{x - 5}{x + 3} + 1 < 0$ .
Use the common denominator: $\dfrac{x - 5 + x + 3}{x + 3} = \dfrac{2x - 2}{x + 3} < 0$ .
The critical values are $x = 1$ (top zero) and $x = -3$ (bottom zero).
The fraction is negative when the top and bottom have opposite signs, which is between them.
So $-3 < x < 1$ , option B.