How do you solve the system 3x-y z=5, x 3y 3z=-6, and x 4y-2x=12?

1 answer:

8 0

3x - y + z = 5 . . . (1)
x + 3y + 3z = -6 . . . (2)
x + 4y - 2z = 12 . . . (3)

From (2), x = -6 - 3y - 3z . . . (4)
Substituting for x in (1) and (3) gives
3(-6 - 3y - 3z) - y + z = 5 => -18 - 9y - 9z - y + z = 5 => -10y - 8z = 23 . . (5)
-6 - 3y - 3z + 4y - 2z = 12 => y - 5z = 18 . . . (6)

(6) x 10 => 10y - 50z = 180 . . . (7)
(5) + (7) => -58z = 203
z = 203/-58 = -3.5

From (6), y - 5(-3.5) = 18 => y = 18 - 17.5 = 0.5
From (4), x = -6 - 3(0.5) - 3(-3.5) = -6 - 1.5 + 10.5 = 3

x = 3, y = 0.5, z = -3.5

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