Solve the system by finding x, y, and z. x-y+z=1 x+y+3z=-3 2x-y+2z=0

Question

asked Jul 12, 2020 203k views

1 Answer

Andreas Engedal · Answer 1 · 2020-07-18T06:13:03+0000

Answer:

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
$x=-1,\ y=-2,\ and\ z=0$