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How do you solve the following linear system: $8x+3y-6z=4, x-2y-z=2, 4x+y-2z=-4$ ?

1 Answer

Dec 8, 2015

Combine $2$ pairs of equations to eliminate 1 of the variables from 2 equations; then solve the 2 equations for each of the remaining variables.
$color(white)("XXX")(x,y,z)=(-4,0,-6)$

Explanation:

Given:
[1] $color(white)("XXX")8x+3y-6z=4$
[2] $color(white)("XXX")x-2y-z=2$
[3] $color(white)("XXX")4x+y-2z=-4$

To eliminate $x$ from 2 equations:

Multiply [2] by $8$
[4] $color(white)("XXX")8x-16y-8z=16$
Subtract [4] from [1]
[5] $color(white)("XXX")19y+2z=-12$

Multiply [3] by $2$
[6] $color(white)("XXX")8x+2y-4z=-8$
Subtract [6] from [1]
[7] $color(white)("XXX")y-2z=12$

Now we have two equations in two unknowns:
[5] $color(white)("XXX")19y+2z=-12$
[7] $color(white)("XXX")y-2z=12$

Adding [5] and [7]
[8] $color(white)("XXX")20y = 0$
[9] $color(white)("XXX")y=0$

Substituting $0$ for $y$ in [7] (could have used [5])
[10] $color(white)("XXX")(0)-2z=12$
[11] $color(white)("XXX")z=-6$

Substituting $(-6)$ for $z$ and $0$ for $y$ in [2] (could have used [1] or [3])
[12] $color(white)("XXX")x-2*(0)-(-6) = 2$
[13] $color(white)("XXX")x=-4$

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