# How do you solve the system x+y-z=0-1, 4x-3y+2z=16, 2x-2y-3z=5?

Dec 28, 2015

$x + y - z = 0 - 1$$\implies x + y - z = - 1. \ldots \ldots \ldots \ldots . . \left(i\right)$
$4 x - 3 y + 2 x = 16. \ldots \ldots \ldots . . \left(i i\right)$
$2 x - 2 y - 3 z = 5. \ldots \ldots \ldots \ldots \ldots . \left(i i i\right)$

Multiply $\left(i\right)$ by -4 and add in $\left(i i\right)$
$\implies - 4 x - 4 y + 4 z = 4$
$+ 4 x - 3 y + 2 z = 16$
$- 7 y + 6 z = 20$$\ldots \ldots \ldots \ldots \ldots . \left(i v\right)$

Again multiply $\left(i\right)$ by -2 and add in $\left(i i i\right)$
$\implies - 2 x - 2 y + 2 z = 2$
$+ 2 x - 2 y - 3 z = 5$

$- 4 y - z = 7. \ldots \ldots \ldots . . \left(v\right)$

Mutiply $\left(v\right)$ by $6$ and add in $\left(i v\right)$
$\implies - 24 y - 6 z = 42$
$- 7 y + 6 z = 20$

$- 31 y = 62$
$\implies y = - 2$

put $y = - 2$ in $\left(v\right)$
$\implies - 4 \left(- 2\right) - z = 7$
$\implies 8 - z = 7$
$\implies z = 1$

Now put $y = - 2$ and $z = 1$ in $\left(i\right)$
$\implies x - 2 - 1 = - 1$
$\implies x - 3 = - 1$
$\implies x = 2$