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

Nov 26, 2017

$\left(w , x , y , z\right) = \left(2 , 3 , 1 , 4\right)$

#### Explanation:

Given:

$\left\{\begin{matrix}w - 2 x + 3 y + z = 3 \\ 2 w - x - y + z = 4 \\ w + 2 x - 3 y - z = 1 \\ 3 w - x + y - 2 z = - 4\end{matrix}\right.$

Adding the first and third equations together, we get:

$2 w = 4$

So:

$w = 2$

Discarding the first equation, and substituting $w = 2$ into the others, we get:

$\left\{\begin{matrix}- x - y + z = 0 \\ 2 x - 3 y - z = - 1 \\ - x + y - 2 z = - 10\end{matrix}\right.$

From the first equation:

$z = x + y$

Substituting this into the second and third equations, we get:

$\left\{\begin{matrix}x - 4 y = - 1 \\ - 3 x - y = - 10\end{matrix}\right.$

Multiplying the second equation by $- 4$ we get:

$12 x + 4 y = 40$

Adding this to the first equation, we get:

$13 x = 39$

Hence:

$x = 3$

From the second equation:

$y = 10 - 3 x = 10 - 9 = 1$

Then:

$z = x + y = 3 + 1 = 4$

So:

$\left(w , x , y , z\right) = \left(2 , 3 , 1 , 4\right)$