Answer
SOLUTION
$\frac{(x-9)(x+2)}{2x+2} \geq0$
INTERVAL
$[-2,-1) [9,\infty)$
GRAPH
In the real number line put 3 numbers -2, -1 and 9. Draw a close circle in -2 and 9 and an open circle -1. Connect the numbers -2 and 1 by a line. Then draw a line starting in 9 pointing to $\infty$.
Work Step by Step
$\frac{x}{2} \geq \frac{5}{x+1}+4$
$\frac{x}{2}-\frac{5}{x+1}-4 \geq0$
$\frac{x(x+1)-5(2)-4(2x+2)}{2(x+1)} \geq0$
$\frac{x^{2}+x-10-8x-8}{2(x+1)} \geq0$
$\frac{x^{2}-7x-18}{2(x+1)} \geq0$
$\frac{(x-9)(x+2)}{2x+2} \geq0$
In order to solve you need to look at the solutions given by the inequality.
NUMERATOR x=9 x=-2
DENOMINATOR x=-1
Now with this number you can create intervals to find the solution.
INTERVAL
$(-\infty,-2] [-2,-1) (-1, 9] [9,\infty)$
$(-\infty,-2] y(-4)=\frac{(-4-9)(-4+2)}{2(-4)+2} \geq0= y(-4)=-4.33$
No solution
$[-2,-1)$ $y(-1.5)=\frac{(-1.5-9)(-1.5+2)}{2(-1.5)+2} \geq0= y(-1.5)=5.25$
Solution
$(-1, 9]$ $y(5)=\frac{(5-9)(5+2)}{2(5)+2} \geq0= y(5)=-2.33$
No solution
$[9,\infty)$ $y(10)=\frac{(10-9)(9+2)}{2(9)+2} \geq0= y(10)=0.54$
Solution
CORRECT
$[-2,-1) [9,\infty)$
GRAPH
In the real number line put 3 numbers -2, -1 and 9. Draw a close circle in -2 and 9 and an open circle -1. Connect the numbers -2 and 1 by a line. Then draw a line starting in 9 pointing to $\infty$.
