Question

solve:
\\(\frac{x + 9}{x^2 - 5x + 6} \leq 0\\)
\\(? < x < \square\\) or \\(x \leq \square\\)

Explanation:

Step1: Factor denominator

$x^2 -5x +6=(x-2)(x-3)$

Step2: Find critical points

Set numerator/denominator to 0:
$x+9=0 \implies x=-9$
$x-2=0 \implies x=2$
$x-3=0 \implies x=3$

Step3: Test intervals

Test $(-\infty,-9)$: $\frac{-}{(+)(+)}=-\leq0$ (valid)
Test $(-9,2)$: $\frac{+}{(-)(+)}=-\leq0$ (valid)
Test $(2,3)$: $\frac{+}{(+)(-)}=-\leq0$ (valid)
Test $(3,\infty)$: $\frac{+}{(+)(+)}=+
ot\leq0$ (invalid)

Step4: Check equality case

$x=-9$ makes numerator 0, so $x=-9$ is valid; $x=2,3$ make denominator 0 (undefined, excluded).

Answer:

$2 < x < 3$ or $x \leq -9$