Question 907729
What your professor probably means is that you tried to solve the equation
{{{ a/(a^2+3a-4)+4a/(a^2+7a+12) =0}}} ,
which has {{{a=1/5}}} as one of its solutions ( {{{a=0}}} is another solution).
while there was no equation to be solved.
Maybe the instructions for the problem just asked for the common denominator.
Maybe the instructions for the problem just asked to simplify the expression
{{{ a/(a^2+3a-4)+4a/(a^2+7a+12) }}} .
If simplifying was the task, and all work needed to be shown,
maybe it would have been enough to write:
{{{ a/(a^2+3a-4)+4a/(a^2+7a+12) =  a/((a+4)(a-1))+4a/((a+4)(a+3)) =  a(a+3)/((a+4)(a-1)(a+3))+4a(a-1)/((a+4)(a-1)(a+3)) = (a(a+3)+4a(a-1))/((a+4)(a-1)(a+3)) = (a^2+3a+4a^2-4a)/((a+4)(a-1)(a+3)) = (5a^2-a)/((a+4)(a-1)(a+3)) }}} .
Maybe you were asked to show a fully factored result, and you would have needed to factor the numerator and end with
{{{a(5a-1)/((a+4)(a-1)(a+3)) }}} .
You certainly knew how to do all that.
If you only had to enter the final result as an answer,
both {{{a(5a-1)/((a+4)(a-1)(a+3)) }}} and {{{(5a^2-a)/((a+4)(a-1)(a+3)) }}} may have been accepted answers.