# How do you solve the system x-y+z=3, 2x+y+z=8, and 3x+y-z=1?

Jan 27, 2017

Solution is $x = 1$, $y = 2$ and $z = 4$

#### Explanation:

We are given three equations in three variables, which are

$x - y + z = 3$ ..........................................(1)

$2 x + y + z = 8$ ..........................................(2)

and $3 x + y - z = 1$ ..........................................(3)

To solve them, what we need is to eliminate some vatiables to get the value of others.

It is apparent from (1) nd (3) that as signs of $y$ and $z$ are opposite, they will cancel out on adding. Hence adding them, we get

$x - y + z + 3 x + y - z = 3 + 1$ or $4 x = 4$ i.e. $x = 1$

Now subtracting (2) from (1), we get

$x - y + z - 2 x - y - z = 3 - 8$ or $- x - 2 y = - 5$.

But as $x = 1$, we have

$- 1 - 2 y = - 5$ i.e. $- 2 y = - 4$ or $y = \frac{- 4}{-} 2 = 2$

Putting values of $x$ and $y$ in (1), we get

$1 - 2 + z = 3$ or $z = 3 - 1 + 2 = 4$

Hence, solution is $x = 1$, $y = 2$ and $z = 4$