Jambmaths question:
What is the solution of $\frac{x-5}{x+3}<-1$ ?
Option A:
$x<-3\text{ or }3>1$
Option B:
$-3<x<5$
Option C:
$x<-3\text{ or }x>5$
Option D:
$-3<x<1$
Jamb Maths Solution:
$\begin{align} & \frac{x-5}{x+3}<-1 \\ & \text{To keep the sign unchanged, multiply both sides by }{{(x+3)}^{2}} \\ & \frac{(x-5){{(x+3)}^{2}}}{(x+3)}<-1{{(x+3)}^{2}} \\ & (x-5)(x+3)<-1({{x}^{2}}+6x+9) \\ & {{x}^{2}}-2x-15<-{{x}^{2}}-6x-9 \\ & 2{{x}^{2}}+4x-6<0 \\ & {{x}^{2}}+2x-3<0 \\ & (x+1)(x-3)<0 \\ & \text{Using table to determine the range or interval of the inequality} \\ & \text{From the range table},\text{ the interval for the inequality is}~\text{1}<x<\text{3} \\\end{align}$
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