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:

$- z = 5 implies z = -5$

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

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

How do you solve the system x+y+z=-3x+y+z=-3, 3x+y-z=133x+y-z=13, and 3x+y-2z=183x+y-2z=18?