Answer
The solutions are $\frac{1}{7}$ and $1$.
Work Step by Step
The given equation can be written as:
$(y^{-1})^2-8y^{-1}+7=0$
Let $u=y^{-1}$.
Rewrite the equation above using $u$ to obtain:
$u^2-8u+7=0$
Factor the trinomial to obtain:
$(u-7)(u-1)=0$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
\\&u-7=0 &\text{or} &u-1=0
\\&u=7 &\text{or} &u=1
\end{array}
Since $u=y^{-1}$, then:
\begin{array}{ccc}
\\&u=7 &\text{or} &u=1
\\&y^{-1}=7 &\text{or} &y^{-1}=1
\\&(y^{-1})^{-1}=7^{-1} &\text{or} &(y^{-1})^{-1}=1^{-1}
\\&y=\frac{1}{7} &\text{or} &y=1
\end{array}
Therefore, the solutions are $\frac{1}{7}$ and $1$.
