## Precalculus: Mathematics for Calculus, 7th Edition

(-$\infty$,-1) $\cup$ (-$\frac{1}{2}$,$\infty$)
$\frac{x-4}{2x+1}$ $\lt$ 5 $\frac{x-4}{2x+1}$ -$\frac{5(2x+1)}{2x+1}$ $\lt$ 0 $\frac{-9x-9}{2x+1}$ $\lt$ 0 Factors are zero when x=-1 or x=-$\frac{1}{2}$, we can't include -$\frac{1}{2}$ in the solution because the denominator can't be zero (endpoints don't satisfy the inequality anyway) , these two points divide the number line into three intervals, plug in the value from each interval to decide which intervals are the solution.