This time, we wanted to find the solution to this system of equations. The simplest way to do this is by using the first equation to find a in terms of b, and then substitute in the second equation. We will find that a=-b-3, and so…
-3b-b^2+3 = -b-3-74.We will find that this solves to the quadratic b^2+2b-80=0, which is equivalent to (b+10)(b-8)=0, and so has solutions b=-10 and b=8. Plug this in to our equation for a in terms of b and our solutions are as follows:
solutions: a=7,b=-10;a=-11, b=8