Answer
a. Restriction values $: -4$
b. Solution set $ :\{-3\}$
Work Step by Step
$\frac{3}{x+4} - 7 = \frac{-4}{x+4}$
a. $x =-4 $ makes the denominator zero. So, $x \ne -4$
b. $\frac{3}{x+4} - 7 = \frac{-4}{x+4} ; x \ne -4;$
$\frac{3-7(x+4)}{x+4} = \frac{-4}{x+4} ; x \ne -4;$
$\frac{3-7x-28}{x+4} = \frac{-4}{x+4} ; x \ne -4;$
$\frac{-7x-25}{x+4} = \frac{-4}{x+4} ; x \ne -4;$
Multiply both sides by $(x+4)$ to clear fraction.
$(x+4)(\frac{-7x-25}{x+4} )=(x+4)( \frac{-4}{x+4}) ; x \ne -4;$
$-7x-25 = -4$
$-7x = -4+25$
$-7x =21$
Divide both sides by $-7$
$x=-3$
