# How do you solve (4x-5)/(x+3)>0?

May 10, 2015

$0 < \frac{4 x - 5}{x + 3} = \frac{\left(4 x + 12\right) - 17}{x + 3} = \frac{4 \left(x + 3\right) - 17}{x + 3} = 4 - \frac{17}{x + 3} .$

Add $\frac{17}{x + 3}$ to both sides to get:

$\frac{17}{x + 3} < 4$

There are now two permissible cases:

Case 1: When $x < - 3$, $x + 3 < 0$, so if we multiply both sides by $\left(x + 3\right)$ we have to reverse the inequality to get:

$17 > 4 \left(x + 3\right) = 4 x + 12$

Subtract 12 from both sides to get:

$5 > 4 x$.

Divide both sides by 4 to get:

$\frac{5}{4} > x$, i.e. $x < \frac{5}{4}$.

Since in this case we already know $x < - 3$, this condition is already fulfilled.

Case 2: When $x > - 3$, $x + 3 > 0$, so we can multiply both sides by $\left(x + 3\right)$ without reversing the inequality to get:

$17 < 4 \left(x + 3\right) = 4 x + 12$

Subtract 12 from both sides to get:

$5 < 4 x$.

Divide both sides by 4 to get:

$\frac{5}{4} < x$, i.e. $x > \frac{5}{4}$.

If $x > \frac{5}{4}$ then it satisfies $x > - 3$

So in Case 2, we just require $x > \frac{5}{4}$.

Combining the 2 cases, we find that $x < - 3$ or $x > \frac{5}{4}$.

Note that $x = - 3$ is not allowed due to the resulting division by 0, which is undefined.