Question

Solve the system by finding x, y, and z.

x-y+z=1
x+y+3z=-3
2x-y+2z=0

Answer 1

The solution is
`x=-1,\ y=-2,\ and\ z=0`

Explanation:

Given:

The system of equations given are:

`x-y+z=1\\x+y+3z=-3\\2x-y+2z=0`

Add first and second equations. This gives,

`x-y+z+x+y+3z=1-3\\2x+4z=-2\\\textrm{Divide both sides by 2}\\x+2z=-1-----\ 4`

Adding equations 2 and 3, we get

`x+y+3z+2x-y+2z=-3+0\\3x+5z=-3 --------------\ 5`

We need solve equations 4 and 5.

Multiplying equation (4) by -3, we get

`(x+2z=-1)* -3=-3x-6z=3`

Adding the above equation to equation (5), we get

`-3x-6z+3x+5z=3-3\\-z=0\\z=0`

Now, we plug in
`z=0` in equation (4). This gives,

`x+2(0)=-1\\x+0=-1\\x=-1`

Now, we plug in
`x=-1,z=0` in equation (1). This gives,

`-1-y+0=1\\-1-y=1\\-y=1+1\\-y=2\\y=-2`

Therefore, the solution is
`x=-1,\ y=-2,\ and\ z=0`