Mr. Lee asks a student to solve the following system of linear equations. Which ordered pair (p, q) is a solution to the system?
Accepted Solution
A:
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