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

1 Answer
Alan P.
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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