Answer
$\{\frac{-7}{6},-3\}$
Work Step by Step
$\frac{10}{(2x+4)^2}-\frac{1}{(2x+4)} = 3$
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 $\{\frac{-7}{6},-3\}$
