Evaluate $x$ in the equation $\frac{x^2-10}{x-3}+\frac{1}{x-3}=\frac{1}{5}$

Solution
First we need to find the Lowest Common Multiple (LCM) of the left hand side.

LCM=$x-3$

$\frac{x^2-10+1}{x-3}=\frac{1}{5}$
$\frac{x^2-9}{x-3}=\frac{1}{5}$
Recall that $(x-3)(x+3)=(x^2-9)$
$\frac{(x-3)(x+3)}{x-3}=\frac{1}{5}$
$x+3=\frac{1}{5}$
$x=-\frac{14}{5}$