Question 17

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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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