## Precalculus (6th Edition) Blitzer

$a.\quad x\neq-4$ $b.\quad$ Solution set = $\{-3\}$
$a.$ The denominators in the equation must not be zero: $x+4\Rightarrow x\neq-4 \qquad(*)$ $b.$ Multiply both sides with the LCD,$\quad(x+4)$ $(x+4) \displaystyle \left[ \frac{3}{x+4}-7 \right] = (x+4) \left[ \dfrac{-4}{x+4} \right]\quad$ ... distribute and simplify $3-7(x+4)=-4\quad$ ... distribute and simplify $3-7x-28=-4$ $-7x-25=-4\quad$ ... add $25$ to both sides $-7x=21$ $x=-3 \quad$ ... satisfies (*), a valid solution. Solution set = $\{-3\}$