Question

How do you solve `-3x-y=0, y-4z=15, x-2y+z=-10`?

Answer 1

`x=-1"; "y=3"; "z=-3`

Explanation:

`((x,y,z,|,-),(-3,-1,0,|,0),(0,1,-4,|,15),(1,-2,1,|,-10))`
`color(brown)(" "R_3+2R_2)`
`color(brown)(" "darr)`

`((-3,-1,0,|,0),(0,1,-4,|,15),(1,0,-7,|,+20))`
`color(brown)(" "R_1+3R_3)`
`color(brown)(" "darr)`

`((0,-1,-21,|,60),(0,1,-4,|,15),(1,0,-7,|,+20))`
`color(brown)(" "R_1+R_2)`
`color(brown)(" "darr)`

`((0,0,-25,|,75),(0,1,-4,|,15),(1,0,-7,|,+20))`
`color(brown)(" "R_1-:(-25))`
`color(brown)(" "darr)`

`((0,0,1,|,-3),(0,1,-4,|,15),(1,0,-7,|,+20))`
`color(brown)(" "R_2+4R_1`
`color(brown)(" "darr)`

`((0,0,1,|,-3),(0,1,0,|,3),(1,0,-7,|,+20))`
`color(brown)(" "R_3+7R_1`
`color(brown)(" "darr)`

`((0,0,1,|,-3),(0,1,0,|,3),(1,0,0,|,-1))`
`color(brown)(" "R_3+7R_1`

`x=-1"; "y=3"; "z=-3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("check")`

`-3x-y=0" " ->" " -3(-1)-3= 0 " "color(red)(larr"True")`

`y-4z=15" " ->" "3-4(-3)=15" "color(red)(larr"True")`

`x-2y+z=-10" "->""-1-2(3)-3= -10" "color(red)(larr"True")`