Question

Solve the inequality:

$\frac{3x+1}{x+4}\le 1$

Answer 1

$\frac{3x+1}{x+4}\le 1$

$\Rightarrow \frac{3x+1}{x+4}-1\le 0\Rightarrow \frac{3x+1}{x+4}-\frac{x+4}{x+4}\le 0$

$\Rightarrow \frac{3x-1-x-4}{x+4}\le 0\Rightarrow \frac{2x-3}{x+4}\le 0$

This means either Numerator is negative

and denominator is positive ar vice -versa

if $(2x-3)\le 0$ and $(x+4)>0$

$\Rightarrow x\le 3/2$ and $x>-4$

$\Rightarrow -4<x\le 3/2$

if $(2x-3)\ge (x+4)<0$

$\Rightarrow x\ge 3/2$ and $x<-4$ (not feasible)

Since a number can;t be simultaneously

greater then $\left(3/2\right)$ and less than $(-4)$

$-4<x\le \frac{3}{2}$