How do you solve the system of equations $x+y+z=1$ , $x+2y+z=2$ , $x-y+z=-5$ ?

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Answer 1 · 2017-03-18T00:11:32

This system has no solutions.

Given:

${(x+y+z=1), (x+2y+z=2), (x-y+z=-5):}$

Adding the first and third equations, we get:

$2x+2z = -4$

Hence:

$x+z = -2" "...$ (i)

Adding twice the third equation to the second equation, we get:

$3x+3z = -8$

Hence:

$x+z = -8/3" "...$ (ii)

Equations (i) and (ii) are incompatible, since $-2 != -8/3$

So there is no solution to this system of equations.

How do you solve the system of equations x+y+z=1x+y+z=1, x+2y+z=2x+2y+z=2, x-y+z=-5x-y+z=-5 ?