# How do you solve (x+5)/(x-2)>0?

Dec 13, 2016

$x \in \left(- \infty , - 5\right) \cup \left(2 , \infty\right)$

#### Explanation:

$\frac{x + 5}{x - 2} > 0$

This inequality will hold when both the numerator and denominator are positive or both negative.

The values of $x$ at which the numerator or denominator change sign are $x = - 5$ and $x = 2$.

So the intervals we need to consider are:

$\left(- \infty , - 5\right)$, $\left(- 5 , 2\right)$ and $\left(2 , \infty\right)$

When $x \in \left(- \infty , - 5\right)$ both the numerator and denominator are negative, so the quotient is positive as required.

When $x \in \left(- 5 , 2\right)$ the numerator is positive but the denominator is negative, so the quotient is negative.

When $x \in \left(2 , \infty\right)$ both the numerator and denominator are positive, so the quotient is positive.

So the solution space is:

$\left(- \infty , - 5\right) \cup \left(2 , \infty\right)$

graph{(y(x-2)-(x+5))(x+0.0001y-2)(y-1+0.001x) = 0 [-10.63, 9.37, -4.44, 5.56]}