## Precalculus (6th Edition) Blitzer

$a.\quad \left\{\begin{array}{l} x\neq 4\\ x\neq-2 \end{array}\right.$ $b.\quad$ Solution set = $\emptyset$
Factor the denominators: For $a^{2}+bx+c$, we search for two factors ob whose sum is c. Here, we find $-4$ and $+2$. $\displaystyle \frac{1}{x-4}-\frac{5}{x+2}=\frac{6}{(x-4)(x+2)}$ $a.$ The denominators in the equation must not be zero: $\left\{\begin{array}{l} x-4\neq 0\\ x+2\neq 0\\ (x-4)(x+2)\neq 0 \end{array}\right\}\Rightarrow\left\{\begin{array}{l} x\neq 4\\ x\neq-2 \end{array}\right. \qquad(*)$ $b.$ Multiply both sides with the LCD,$\quad (x-4)(x+2)$ $(x-4)(x+2) \left[ \displaystyle \frac{1}{x-4}-\frac{5}{x+2} \right]= (x-4)(x+2) \displaystyle \left[\frac{6}{(x-4)(x+2)} \right]\quad$ ... distribute and simplify $1(x+2)-5(x-4)=6\quad$ ... distribute and simplify $x+2-5x+20=6$ $-4x+22=6\quad$ ... add $-22$ to both sides $-4x=-16$ $x=4 \quad$ ... does not satisfy (*), not a valid solution. Solution set = $\emptyset$