How do you solve $2x+2y-4z=8, 2x+y-z=6, -x-y+2z=2$ ?

Dec 15, 2017

This system of equations has no solution.

Given:

${ (2x+2y-4z=8), (2x+y-z=6), (-x-y+2z=2) :}$

Note that adding twice the third equation to the first we get:

$0 = 12$

which is false.

So this system is inconsistent and has no solutions.

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How do you solve 2x+2y-4z=8, 2x+y-z=6, -x-y+2z=22x+2y-4z=8, 2x+y-z=6, -x-y+2z=2?