# Solve the following inequalities: (2x+3)/(x-5)>0

Solve the following inequalities:
$\frac{2x+3}{x-5}>0$
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talhekh
$\frac{2x+3}{x-5}>0$
$2x=3=0$; $x-5=0$
$x=-\frac{3}{2}$; $x=5$
$x=-10:\frac{2\left(-10\right)+3}{-10-5}>0$
$+\mathrm{#}>0$
$x=0$
$x=10:\frac{2\left(10\right)+3}{10-5}>0$
Solution: $\left(-\mathrm{\infty };-\frac{3}{2}\right)\cup \left(5;\mathrm{\infty }\right)$

Jeffrey Jordon

Answer is given below (on video)