The solution to the system $4y=3x+7$ and $9x+4y-139=0$ is shown below.
I solved for the solution and found that the answer is correct, and is $(11, 10)$. But, what is the mistake that is made here? I'm assuming the strategy is incorrect because it does not look like it would work with different situations involving systems.
As Oussama mentioned, there is an easier way to look at this. Write the first equation as: $$3x-4y=-7,$$ and the second equation as: $$9x+4y=139.$$ Then, we have: \begin{align} 9x+4y &= 139\\ 3x-4y &= -7. \end{align} Add the two equations together to get: \begin{align} 9x+4y &= 139\\ 3x-4y &= -7\\ \hline \\ 12x&= 132. \end{align} Notice that this method didn't require as many algebraic manipulations. Now, solve for $y$ just as before and you are done.