Answer
SOLUTIONS
$\frac{2x+3}{(x+1)(x+2)} \leq0$
INTERVAL
$(-\infty, -2) [\frac{-3}{2},-1)$
GRAPH
In the real number line put $-\infty$, -2, $\frac{-3}{2}$ and -1. Put a close circle in $\frac{-3}{2}$. In the -1 and -2 put an open circle. Connect $\frac{-3}{2}$ and -1. In -2 draw a line pointing towards $-\infty$
Work Step by Step
$\frac{1}{x+1}+\frac{1}{x+2} \leq0$
$\frac{1(x+1)+1(x+2)}{(x+1)(x+2)} \leq0$
$\frac{x+1+x+2}{(x+1)(x+2)} \leq0$
$\frac{2x+3}{(x+1)(x+2)} \leq0$
You have to find the solution from the numerator and denominator.
NUMERATOR x=$\frac{-3}{2}$
DENOMINATOR x=-1 x=-2
INTERVAL
$(-\infty, -2) (-2,\frac{-3}{2}] [\frac{-3}{2},-1) (-1, \infty)$
$(-\infty, -2)$ $y(-4)=\frac{2(-4)+3}{(-2+1)(-2+2)} \leq0)= y(-4)=-0.83$
Solution
$(-2,\frac{-3}{2}]$ $y(-1.8)=\frac{(2(-1.8)+3}{(-1.8+1)(-1.8+2)} \leq0= y(-1.8)=3.75$
No solution
$[\frac{-3}{2},-1)$ $y(-1.3)=\frac{2(-1.3)+3}{(-1.3+1)(-1.3+2)} \leq0= y(-1.3)=-1.90$
Solution
$(-1, \infty)$ $y(3)=\frac{2(3)+3}{(3+1)(3+2)} \leq0= y(3)=0.45$
No solution
CORRECT
$(-\infty, -2) [\frac{-3}{2},-1)$
GRAPH
In the real number line put $-\infty$, -2, $\frac{-3}{2}$ and -1. Put a close circle in $\frac{-3}{2}$. In the -1 and -2 put an open circle. Connect $\frac{-3}{2}$ and -1. In -2 draw a line pointing towards $-\infty$
