Question
Solve the following system of equations for x, y, and z using the elimination method. What is the value of z?
3x+2y−z=−1
−2x−2y+3z=5
10x+4y−2z=6
(1 point)
z =
3x+2y−z=−1
−2x−2y+3z=5
10x+4y−2z=6
(1 point)
z =
Answers
Answered by
GPT-5 mini
z = 1
Work: 2*(3x+2y−z)= -2 gives 6x+4y−2z = -2. Subtracting from 10x+4y−2z = 6 gives 4x = 8 so x = 2.
Substitute into 3x+2y−z = −1: 6 + 2y − z = −1 ⇒ 2y − z = −7.
Substitute x into −2x−2y+3z = 5: −4 −2y +3z = 5 ⇒ −2y +3z = 9.
Add the last two: 2z = 2 ⇒ z = 1.
Work: 2*(3x+2y−z)= -2 gives 6x+4y−2z = -2. Subtracting from 10x+4y−2z = 6 gives 4x = 8 so x = 2.
Substitute into 3x+2y−z = −1: 6 + 2y − z = −1 ⇒ 2y − z = −7.
Substitute x into −2x−2y+3z = 5: −4 −2y +3z = 5 ⇒ −2y +3z = 9.
Add the last two: 2z = 2 ⇒ z = 1.
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