Answer
SOLUTION
$\frac{-2x-1}{(x+3)(x-2)} \lt0$
INTERVAL
$(-3,\frac{-1}{2}] (2,\infty)$
GRAPH
In the number line put 3 numbers -3, $\frac{-1}{2}$ and 2. You shoould put a close circle $\frac{-1}{2}$ and an opne circle in -3 and 2. You should connect -3 and $\frac{-1}{2}$. Also draw a line on 2 and pointed to $\infty$.
Work Step by Step
$\frac{x+2}{x+3} \lt\frac{x-1}{x-2}$
$\frac{x+2}{x+3}-\frac{x-1}{x-2} \lt0$
$\frac{(x+2)(x-2)-(x-1)(x+3)}{(x+3)(x-2)} \lt0$
$\frac{x^{2}-4-(x^{2}+2x-3)}{(x+3)(x-2)} \lt0$
$\frac{x^{2}-4-x^{2}-2x+3}{(x+3)(x-2)} \lt0$
$\frac{-2x-1}{(x+3)(x-2)} \lt0$
Now you have to look for the solutions given by the inequality.
NUMERATOR $x=\frac{-1}{2}$
DENOMINATOR x=-3 x=2
INTERVALS
Options: $(\infty,-3) (-3,\frac{-1}{2}] [\frac{-1}{2},2) (2,\infty)$
$(\infty,-3)$ Y(-4)=\frac{-2(-4)-1}{(-4+3)(-4-2)} \lt0= y(-4)=1.16
No solution
$(-3,\frac{-1}{2}]$ $Y(-2)=\frac{-2(-2)-1}{(-2+3)(-2-2)} \lt0= y(-2)=-0.75$
Solution
$[\frac{-1}{2},2)$ $y(0)=\frac{-2(0)-1}{(0+3)(0-2)} \lt0= y(0)=0.16$
No solution
$(2,\infty)$ $Y(5)=\frac{-2(5)-1}{(5+3)(5-2)} \lt0= y(5)=-0.458$
Solution
CORRECT
$(-3,\frac{-1}{2}] (2,\infty)$
GRAPH
In the number line put 3 numbers -3, $\frac{-1}{2}$ and 2. You shoould put a close circle $\frac{-1}{2}$ and an opne circle in -3 and 2. You should connect -3 and $\frac{-1}{2}$. Also draw a line on 2 and pointed to $\infty$.
