# How do you solve the following linear system:  x-2y+z=-14, y-2z=7, 2x+3y-z=-1 ?

Dec 31, 2017

$\left(x , y , z\right) = \left(- 6 , 3 , - 2\right)$

#### Explanation:

There are a bunch of ways in which we can solve the system of linear equations. Elimination method is illustrated here.

Let us number the equations and proceed

$x - 2 y + z = - 14 \to \left(1\right)$
$y - 2 z = 7 \to \left(2\right)$
$2 x + 3 y - z = - 1 \to \left(3\right)$

Because we can eliminate $z$ from equations (1) and (3) easily by summing them we eliminate $z$ from all the equations. You can arbitrarily choose $x$ or $y$ and you will get the same answer.

$\left(1\right) \setminus \times 2 + \left(2\right) \setminus \implies$

$2 x - 4 y + \setminus \cancel{2 z} = - 28$
$0 x + y - \setminus \cancel{2 z} = 7$

$2 x - 3 y = - 21 \to \left(4\right)$

$\left(1\right) + \left(3\right) \implies$

$x - 2 y + \cancel{z} = - 14$
$2 x + 3 y - \cancel{z} = - 1$

$3 x + y = - 15 \to \left(5\right)$

$\left(4\right) + 3 \setminus \times \left(5\right) \implies$

$2 x - \cancel{3 y} = - 21$
$9 x + \cancel{3 y} = - 45$
$11 x = - 66$
$x = - 6$

Substitute in $\left(5\right)$

$3 x + y = - 15$
$y = - 15 - 3 x$
$y = - 15 - 3 \left(- 6\right)$
$y = - 15 + 18$
$y = 3$

Substitute $y$ in $\left(2\right)$

$y - 2 z = 7$
$3 - 2 z = 7$
$- 2 z = 4$
$z = - 2$

Substitute in any of the equation to verify the results. Equation $\left(3\right)$ gives

$2 x + 3 y - z = - 1$

L.H.S $\implies$

$2 \left(- 6\right) + 3 \left(3\right) - \left(- 2\right) = - 12 + 9 + 2$
$= - 12 + 11$
$= - 1$

Hence our solution is $\left(x , y , z\right) = \left(- 6 , 3 , - 2\right)$