I need help with systems of equations

Question

What is the solution for z in the following system of equations? (Record only the numerical value of your answer)

3x+4y-z=-32

3x-y+2z=-13

5y+2z=-4

Answer 1

Subtracts Equation #1 MINUS Equation #2:

5y - 3z = -19

The 2x2 system is:

5y - 3z = -19

5y + 2z = -4

Subtracts the BOTTOM MINUS THE TOP, 5y CANCELS:

2z - (-3z) = -4 - (-19)

2z + 3z = -4 + 19

5z = 15

z = 3

Plugging into the 2x2:

5y - 3(3) = -19 AND 5y + 2(3) = -4

5y - 9 = -19 AND 5y + 6 = -4

5y = -10 5y = -10

y = -2 y = -2

Plugging into the original system:

3x + 4(-2) - 3 = -32

3x + -8 - 3 = -32

3x + -11 = -32

3x = -21

x = -7

check, per original second equation:

3(-7) - (-2) + 2(3) =

-21 + 2 + 6 = -13 <-- YES it checks

the solution set is {x=-7, y=-2, z=3}

Answer 2

3x + 4y - z = -23

3x - y + 2z = -13

5y + 2z = -4

5y - 3z = -19 Eq 1 - Eq 2

5z = 15 Eq 3 - EQ 4