Question

2x-3/x+1<=1 solve the inequalities​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{2x - 3}{x + 1} \leqslant 1 \: \: and \: x \: \ne \: \: - 1$

$\rm :\longmapsto\:\dfrac{2x - 3}{x + 1} - 1 \leqslant0$

$\rm :\longmapsto\:\dfrac{2x - 3 - (x + 1)}{x + 1}\leqslant0$

$\rm :\longmapsto\:\dfrac{2x - 3 - x - 1}{x + 1}\leqslant0$

$\rm :\longmapsto\:\dfrac{x - 4}{x + 1}\leqslant0$

$\rm :\longmapsto\:\dfrac{(x - 4)(x + 1)}{ {(x + 1)}^{2} }\leqslant0$

$\bf\implies \:x \: \in \: - 1 < x \leqslant 4$

$\bf\implies \:x \: \in \: ( - 1, \: 4]$

Additional information :-

$\rm :\longmapsto\:A < B\bf\implies \: - A > - B$

$\rm :\longmapsto\:A > B\bf\implies \: - A < - B$

$\rm :\longmapsto\:A \leqslant B\bf\implies \: - A \geqslant - B$

$\rm :\longmapsto\:A \geqslant B\bf\implies \: - A \leqslant - B$

$\rm :\longmapsto\:A < - B\bf\implies \: - \: A > B$

$\rm :\longmapsto\:A > - B\bf\implies \: - \: A < B$