## Precalculus: Mathematics for Calculus, 7th Edition

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 $\infty$.
$\frac{x}{2} \geq \frac{5}{x+1}+4$ $\frac{x}{2}-\frac{5}{x+1}-4 \geq0$ $\frac{x(x+1)-5(2)-4(2x+2)}{2(x+1)} \geq0$ $\frac{x^{2}+x-10-8x-8}{2(x+1)} \geq0$ $\frac{x^{2}-7x-18}{2(x+1)} \geq0$ $\frac{(x-9)(x+2)}{2x+2} \geq0$ 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 $\infty$.