MATH SOLVE

6 months ago

Q:
# 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

-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