Answer:
The solution to this system is (-2, 1, -3).
Step-by-step explanation:
Let's eliminate variable x first. Combine the first two equations, obtaining:
3x - 0y + 2z = -12.
Now subtract the third equation from this result:
3x - 0y + 2z = -12
-(3x − y − z = −4)
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y + 3z = -8
Similarly, combine the second and third original equations to eliminate x again. To do this, subtract 2(x + y + z = -4) from the first equation:
2x−y+z=−8
-2x - 2y - 2z = 8
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-3y - z = 0
Now we have eliminated x completely, and find from -3y - z = 0 that z = -3y. Substitute this -3y for z in the equation y + 3z = -8 found above:
y + 3(-3y) = -8. Then y - 9y = -8, and so y must = 1. From -3y - z = 0, substituting 1 for y, we find that z = -3(1), or z = -3.
Finally, subst. 1 for y and -3 for z in the second equation:
x + 1 - 3 = -4
So, x - 2 = -4, and thus x must be -2.
The solution to this system is (-2, 1, -3).
