For the following system use the second equation to make a substitution for y in the first equation 2x+y=5 y+3=2x

1. Substitution: x = y + 10
3x + 2y - 5 = 0
3(y + 10) + 2y = 5
3y + 2y = 5 - 30
5y = -25
y = -5 (answer)

2. Replace x with 2y - 8 in the first equation.
2y + 5y = 18
7y = 18

Simplest form: 7y - 18 = 0, or (c) (answer)

3.
q = 1 - 2p
3p + q + 1 = 0
q = -1 -3p
1 - 2p = -1 - 3p
p = -2

Substitute:

-1 - 3(-2) = -1 + 6 = q = 5

Solution set:
{(-2,5)}, or (a) (answer)

4.
y = 5x - 4
3x + 5x - 4 = 1, or (a) (answer)

5.
y = x + 1
2x + x + 1 = 7
3x = 6
x = 2
y = 1 + 2 = 3

Now replace x with 2 and y with 3 in all of the answers and see which one doesn't work.
a: 7 - 4 - 2 = 1 this works
b: 4 + 2 + 1 = 7 this works
c: 2(4) + 3 = 7 8 + 3 = 11, not 7

(c) would not be the result of the system. (answer)

6.
x = y - 3
3(y - 3) + 2y = 7, or (b) (answer)

The last answer choice

Kis the midpoint of bc