Answer
$\left\{\dfrac{4}{3},\dfrac{9}{4}\right\}$
Work Step by Step
Let $z=
(2w-3)^{-1}
$. Then the given equation, $
6=7(2w-3)^{-1}+3(2w-3)^{-2}
,$ is equivalent to
\begin{align*}
6&=7z+3z^2
\\
0&=7z+3z^2-6
\\
3z^2+7z-6&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+3)(3z-2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z+3=0 & 3z-2=0
\\
z=-3 & 3z=2
\\\\
& z=\dfrac{2}{3}
.\end{array}
Since $z=
(2w-3)^{-1}=\dfrac{1}{2w-3}
$, by back-substitution,
\begin{array}{l|r}
\dfrac{1}{2w-3}=-3 & \dfrac{1}{2w-3}=\dfrac{2}{3}
.\end{array}
By cross-multiplication, the equations above are equivalent to
\begin{array}{l|r}
-3(2w-3)=1(1) & 1(3)=2(2w-3)
\\
-6w+9=1 & 3=4w-6
\\
-6w=1-9 & 3+6=4w
\\
w=\dfrac{-8}{-6} & 9=4w
\\
w=\dfrac{4}{3} & \dfrac{9}{4}=w
\\
& w=\dfrac{9}{4}
.\end{array}
Hence, the solution set of the equation $
6=7(2w-3)^{-1}+3(2w-3)^{-2}
$ is $\left\{\dfrac{4}{3},\dfrac{9}{4}\right\}$.
