Now that we have one of the variables from Part A isolated, and written in terms of the other variable, we can now substitute this into the other of the two original equations.

2A + 3(3A-5) = -4

We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one of the original equations (or the expression derived in Part A) to solve for the other variable. When this is done with the system of two equations from Parts A and B, what is the solution?