The system of equations is:
-28 = -40(q) - 14p (1)
10q+4 = 2p (2)
There are a number of ways to solve this problem. I personally, when it can be done easily, like to get one of the variables by itself and substitute it into the other equation. For example, let's get p by itself in equation (2):
2p = 10q + 4 Divide both sides by 2:
p = 5q + 2
Now we have an easy way to plug p into equation (1) and solve for q:
-28 = -40q - 14p Divide both sides by -2 to simplify:
14 = 20q + 7p Plug in (5q+2) for p
14 = 20q + 7(5q + 2) Distribute
14 = 20q + 35q + 14 Subtract 14 from both sides
0 = 55q
q = 0
We lucked out, and q= 0, so now we can easily plug q into either of our equations. Let's go with equation (1)
-28= -40q - 14p
-28 = 0 - 14p
-28 = -14 p Divide both sides by -14
p = 2
Our ordered pair (p,q), therefore, is (2,0)
Answer is D