# math

Solve the following system of equations:
4x+y–2z=10
2x–y+2z=–9
x+y+z=6

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1. I noticed that the coefficients of y are ± 1 , so ...
add the first and the second:
6x = 1
x = 1/6 , well that was lucky

3x + 3z = -3
x + z = -1
1/6 + z = -1
z = -7/6

sub into the last:
1/6 + y - 7/6 = 6
times 6
1 + 6y - 7 = 36
y = 42/6 = 7

1. 👍 0
2. 👎 0
2. 4x+y–2z=10
2x–y+2z=–9
6 x = 1
x = 1/6

1/6 + y + z = 6
is
6 y + 6 z = 35

2(1/6)-y+2z = -9 --> -3 y + 6 z = -28
so use those two
6 y + 6 z = 35
-3y + 6 z = -28
--------------------subtract
9 y = 63
y = 7

1/6 + 7 + z = 6
1 + 42 + 6 z = 36
6 z = 36-43 = -7
z - -7/6

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2. 👎 0

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