How do you solve the system #x+y+z=-3#, #3x+y-z=13#, and #3x+y-2z=18#?

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Answer 1 · 2017-04-25T08:00:41

The augmented matrix looks something like this:

#((1,1,1),(3,1,-1),(3,1,-2)) ((-3),(13),(18))#

Then: #R2 to R2 - 3 R1, R3 to R3 - 3 R1#

#((1,1,1),(0,-2,-4),(0,-2,-5)) ((-3),(22),(27))#

Then: #R3 to R3 - R2#

#((1,1,1),(0,-2,-4),(0,0,-1)) ((-3),(22),(5))#

So now we back substitute:

#- z = 5 implies z = -5#

#- 2 y - 4z = 22 implies y= -1#

#x + y + z = -3 implies x = 3#