Answer
$(-∞, -3)$ U $[2, 3)$
Work Step by Step
$(7x-14)/(x^2-9) \leq 0$
Numerator is zero when $7x-14=0$. Denominator is zero when $x^2-9=0$.
$7x-14=0$
$7x-14+14=0+14$
$7x=14$
$7x/7=14/7$
$x=2$
$x^2-9=0$
$x^2-9+9=0+9$
$x^2=9$
$\sqrt {x^2} = \sqrt 9$
$x= ±3$
Four regions to test: $(-∞, -3)$, $(-3, 2]$, $[2, 3)$, $(3, ∞)$
Let $x=-5$, $x=0$, $x=5/2$, $x=5$
$x=-5$
$(7x-14)/(x^2-9) \leq 0$
$(7*-5-14)/((-5)^2-9) \leq 0$
$(-35-14)/(25-9) \leq 0$
$-49/16 \leq 0$ (true)
$x=0$
$(7x-14)/(x^2-9) \leq 0$
$(7*0-14)/(0^2-9) \leq 0$
$(0-14)/(0-9) \leq 0$
$-14/-9 \leq 0$
$14/9 \leq 0$ (false)
$x=5/2$
$(7x-14)/(x^2-9) \leq 0$
$(7*5/2-14)/((5/2)^2-9) \leq 0$
$(35/2-14)/(25/4-9) \leq 0$
$(7/2)/(-11/4) \leq 0$
$3.5/-2.75 \leq 0$
$14/-11 \leq 0$
$-14/11 \leq 0$ (true)
$x=5$
$(7x-14)/(x^2-9) \leq 0$
$(7*5-14)/(5^2-9) \leq 0$
$(35-14)(25-9) \leq 0$
$21/16 \leq 0$ (false)
