Question

Solve the following system of equations for x, y, and z using the substitution method. What is the value of y?

−6x−2y−6z=−16

−5x+3y+6z=−4

−12x−8y+2z=−40

(1 point)
Responses

2
2

This system has no solutions.
This system has no solutions.

0
0

This system has infinite solutions.

Answer 1

Add (1) and (2) to eliminate z:
-6x−2y−6z + (−5x+3y+6z) = −16 + (−4) → −11x + y = −20 → y = 11x − 20.

Substitute into (3):
−12x − 8(11x − 20) + 2z = −40 → −100x + 160 + 2z = −40 → −50x + z = −100 → z = 50x − 100.

Also substitute y into (1) (or (2)) to relate z and x:
−28x − 6z = −56 → 14x + 3z = 28 → z = (28 − 14x)/3.

Equate the two expressions for z:
50x − 100 = (28 − 14x)/3 → 164x = 328 → x = 2.

Then y = 11(2) − 20 = 2.

Answer: 2.