List the critical values of the related function. Then solve the inequality.
$\frac{x - 5}{x + 3} - \frac{x + 4}{x - 2} \leq 0$
The critical value(s) is/are 0 .
(Simplify your answer. Type an integer or a simplified fraction. Type an exact ansu

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Answer

Therefore, the solution to the inequality $\frac{x^{2} - 3 x - 2}{(x + 3) (x - 2)} \geq 0$ is $x \in (- \infty, - 3] \cup [1, 2)$

Steps

Step 1 ：Given the inequality $\frac{x - 5}{x + 3} - \frac{x + 4}{x - 2} \leq 0$

Step 2 ：Simplify the inequality to get $\frac{- x^{2} + 3 x + 2}{(x + 3) (x - 2)} \leq 0$

Step 3 ：Rewrite the inequality as $\frac{x^{2} - 3 x - 2}{(x + 3) (x - 2)} \geq 0$

Step 4 ：Solve the equation $x^{2} - 3 x - 2 = 0$ to get the solutions $x = 1, 2$

Step 5 ：Fill in a sign chart for the intervals $x < - 3$ , $- 3 < x < 1$ , $1 < x < 2$ , and $2 < x$

Step 6 ：From the sign chart, we find that the expression is positive when $x \in (- \infty, - 3] \cup [1, 2)$

Step 7 ：Therefore, the solution to the inequality $\frac{x^{2} - 3 x - 2}{(x + 3) (x - 2)} \geq 0$ is $x \in (- \infty, - 3] \cup [1, 2)$