College Algebra, Chapter 1, 1.6, Section 1.6, Problem 68

Solve the nonlinear inequality $\displaystyle \frac{x}{2} \geq \frac{5}{x+1} +4 $. Express the solution using interval notation and graph the solution set.

$
\begin{equation}
\begin{aligned}
\frac{x}{2} & \geq \frac{5}{x+1} +4\\
\\
0 & \geq \frac{5}{x+1} + 4 - \frac{x}{2} && \text{Subtract } \frac{x}{2}\\
\\
0 & \geq \frac{5(2) + 4(2)(x+1)-x(x+1)}{2(x+1)} && \text{Multiply the LCD } 2(x+1)\\
\\
0 & \geq \frac{10+8x+8-x^2-x}{2(x+1)} && \text{Simplify the numerator}\\
\\
0 & \geq \frac{-x^2 + 7x + 18}{2(x+1)} && \text{Factor out } -1\\
\\
0 & \geq \frac{-(x^2 - 7x - 18)}{2(x+1)} && \text{Divide } -1\\
\\
0 & \leq \frac{x^2 - 7x - 18}{2(x+1)} && \text{Factor out}\\
\\
0 & \leq \frac{(x-9)(x+2)}{2(x+1)} && \text{Multiply both sides by } 2\\
\\
0 & \leq \frac{(x-9)(x+2)}{(x+1)}
\end{aligned}
\end{equation}
$


The factors on the left hand side are $x-9$, $x+2$ and $x+1$. These factors are zero when $x$ is 9,-2 and -1 respectively. These numbers divide the real line into intervals
$(-\infty, -2], [-2,-1),(-1,9],[9,\infty)$




From the diagram, the solution of the inequality $\displaystyle 0 \leq \frac{(x-9)(x+2)}{(x+1)}$ are
$[-2,-1) \bigcup [9,\infty)$