Answer
See answer below
Work Step by Step
Let's find the solution for the equation:
$\frac{m}{m-3}+\frac{1}{4}=\frac{3}{m-3}$
$\frac{4m}{4(m-3)}+\frac{m-3}{4(m-3)}=\frac{12}{4(m-3)}$
$4m+m-3=12$
$5m=15$
$m=3$
Check:
$\frac{3}{3-3}+\frac{1}{4}=\frac{3}{3-3}$
$\frac{1}{4}\ne 0$
Therefore, he forgot to check the solution. $m=3$ does not satisfy the original equation, so this is an extraneous solution.
