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

1 Answer
Dec 4, 2016

Answer:

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)#