Asked by neal
Solve the system by substitution.
12x+6y+7z=−35
7x−5y−6z=200
x+y=−10
12x+6y+7z=−35
7x−5y−6z=200
x+y=−10
Answers
Answered by
mathhelper
1st times 6: 72x + 36y + 42z = -210
2nd times 7: 49x - 35y - 42z = 1400
add them,
121x + y = 1190
subtract the 3rd from that,
120x = 1200
x = 10
in the 3rd: x+y = -10
10 + y = -10
y = -20
back into the 1st:
120 - 120 + 7z = -35
z = -5
x = 10, y = -20, z = -5
2nd times 7: 49x - 35y - 42z = 1400
add them,
121x + y = 1190
subtract the 3rd from that,
120x = 1200
x = 10
in the 3rd: x+y = -10
10 + y = -10
y = -20
back into the 1st:
120 - 120 + 7z = -35
z = -5
x = 10, y = -20, z = -5
Answered by
oobleck
since y = -x-10, use that to get
12x+6(-x-10)+7z = -35
7x-5(-x-10)-6z = 200
rearrange that to be
6x+7z = 25
2x-z = 25
now, since z = 2x+25, use that to get
6x+7(2x-25) = 25
x = 10
so y = -20 and z = -5
12x+6(-x-10)+7z = -35
7x-5(-x-10)-6z = 200
rearrange that to be
6x+7z = 25
2x-z = 25
now, since z = 2x+25, use that to get
6x+7(2x-25) = 25
x = 10
so y = -20 and z = -5
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