## Intermediate Algebra (6th Edition)

$\{\frac{-7}{6},-3\}$
$\frac{10}{(2x+4)^2}-\frac{1}{(2x+4)} = 3$ This is equivalent to $10 \times \frac{1}{(2x+4)^2}-\frac{1}{(2x+4)} = 3$ $10 \times (\frac{1}{2x+4})^{2}-\frac{1}{(2x+4)} = 3$ Let $\frac{1}{2x+4} = y$, then the given equation becomes, $10y^{2} -y = 3$ $10y^{2} -y -3 =0$ By factoring, $10y^{2} +5y-6y -3 =0$ $5y(2y+1)-3(2y+1)=0$ $(5y-3)(2y+1)=0$ $y= \frac{3}{5}$ or $\frac{-1}{2}$ To find $x$, substitute $y$ values in $\frac{1}{2x+4} = y$ Let $y= \frac{3}{5}$ $\frac{1}{2x+4} = \frac{3}{5}$ $6x+12=5$ $6x+12-5=0$ $6x+7=0$ $6x=-7$ $x= \frac{-7}{6}$ Let $y= \frac{-1}{2}$ $\frac{1}{2x+4} = \frac{-1}{2}$ $-2x-4=2$ $-2x=2+4$ $-2x=6$ $x=-3$ $\{\frac{-7}{6},-3\}$ both values satisfy the given equation. So the solutions are $\{\frac{-7}{6},-3\}$