How do you solve the system x-y-z=6, -x+3y+2z=-11, and 3x+2y+z=1?

1 Answer
Dec 4, 2016

The solution is (2, -1, -3)

Explanation:

(A) x-y-z=6
(B) -x+3y+2z=-11
(C) 3x+2y+z=1

Add equations A and B together to eliminate x.

color(white)(a^2)x-color(white)ay-color(white)az=color(white)(aa^2)6
ul(-x+3y+2z=-11)
color(white)(aaaaa)2y+color(white)az=-5color(white)(aaa)Equation D

Multiply equation B by 3 and add it to equation C to also eliminate z.

3*(-x+3y+2z=-11)

-3x+9y+6z=-33
ul(color(white)(a^2)3x+2y+color(white)az==1)
color(white)(aaaaa)11y+color(white)a7z=-32color(white)(aaa)Equation E

Multiply equation D by -7 and add it to equation E to eliminate z.

-7*(2y+color(white)az=-5)

-14y-7z=color(white)(aa^2)35
color(white)(a^2)ul(11y+7z=-32)
color(white)l-3ycolor(white)(aaaa)=color(white)a3

(-3y)/-3=3/-3
y=-1

Substitute #y

=-1# into either equation D or E to solve for z. I choose equation D.

2(-1)+z=-5

-2+z=-5
+2color(white)(aaaaaa)+2
z=-3

Substitute y=-1 and z=-3 into equation A, B or C to find x.
I choose equation A.

x-(-1)-(-3)=6
x+1+3=6
x+4color(white)(aaa^2)=6
color(white)(a)-4color(white)(aaaa)-4
x=2

The solution is (2, -1, -3)