How do you solve #x+y-2z=5#, #-x+2y+z=2#, #2x+3y-z=9# ?

Question

1322 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-09T21:06:18

#x=1#, #y=2# and #z=-1#

Given the system:

#{ (x+y-2z=5), (-x+2y+z=2), (2x+3y-z=9) :}#

We can eliminate #z# from the first equation by adding twice the second equation to it to get:

#-x+5y=9#

We can eliminate #z# from the second equation by adding the third equation to it to get:

#x+5y=11#

We can eliminate #y# from this last equation by subtracting the previous equation to get:

#2x = 2#

Then we can divide both sides by #2# to find:

#color(blue)(x = 1)#

We can substitute this value of #x# into the previous equation to find:

#1+5y = 11#

Subtracting #1# from both sides, we get:

#5y = 10#

Then dividing both sides by #5# we find:

#color(blue)(y = 2)#

Substituting these values for #x# and #y# into the second equation we were given, we find:

#-(1)+2(2)+z = 2#

That is:

#3+z = 2#

Subtracting #3# from both sides, we find:

#color(blue)(z = -1)#