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How do you solve $3x + 6z = 0$ , $-2x + y = 5$ , $y + 2z = 3$ ?

1 Answer

Dec 31, 2016

$(x,y,z)=(-2,1,1)$

Explanation:

Given
[1] $color(white)("XXX")color(white)(+)3xcolor(white)(+0y)+6z=0$
[2] $color(white)("XXX")-2x+color(white)(1)ycolor(white)(+0z)=5$
[3] $color(white)("XXX")color(white)(+0x+1)y+2z=3$

Statement [2] $rarr$
[4] $color(white)("XXX")y=2x+5$

Substituting from [4] into [3]
[5] $color(white)("XXX")2x+5+2z=3$
Simplifying
[6] $color(white)("XXX")2x+2z=-2$
or
[7] $color(white)("XXX")x+z=-1$
or
[8] $color(white)("XXX")z=-x-1$

Substituting from [8] into [1]
[9] $color(white)("XXX")3x+6(-x-1)=0$
Simplifying
[10] $color(white)("XXX")-3x=6$
or
[11] $color(white)("XXX")x=-2$

Substituting from [11] into [8]
[12] $color(white)("XXX")z=-(-2)-1=+2-1=1$

Substituting from [11] into [4]
[13] $color(white)("XXX")y=2(-2)+5=-4+5=1$

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