Beginning Algebra With Applications, Chapter 3, 3.2, Section 3.2, Problem 172

Suppose $3n- 7 = 5 (2n + 7)$, evaluate $\displaystyle \frac{n^2}{2n - 6}$.

We solve for $n$ in the equation $3n- 7 = 5 (2n + 7)$


$
\begin{equation}
\begin{aligned}

3n - 7 =& 10n + 35
&& \text{Apply Distributive Property}
\\
3n - 10n =& 35 + 7
&& \text{Subtract $10n$ and add } 7
\\
-7n =& 42
&& \text{Simplify}
\\
\frac{\cancel{-7}n}{\cancel{-7}} =& \frac{42}{-7}
&& \text{Divide by } -7
\\
n =& -6
&&
\end{aligned}
\end{equation}
$


Now we substitute $n = -6$ into the equation $\displaystyle \frac{n^2}{2n - 6}$


$
\begin{equation}
\begin{aligned}

\frac{(-6)^2}{2(-6) - 6} =& \frac{36}{-18}
&& \text{Substitute } n = -6
\\
=& -2
&&

\end{aligned}
\end{equation}
$