How do you solve the system of equations algebraically #x-y-z=7, x+2y+3z=-12, 3x-2y+7z=30#?

1 Answer
Mar 29, 2017

Answer:

Multiply the first equation by -3 first. Then go ahead, x=2/26; y=-226/26 and z=46/26

Explanation:

When you multiply the first equation by -3, you will get:
#-3x+3y+3z=-21#
Then combine this equation with the second. Yielding,
#y=9-10z#.

When you see y (in the first (original equation)), write this and for the second original equation do the same.

You will get #x+9z=16# and #x-17z=-30#. You can multiply either the first equation by -1 or the second by -1. You will get
#z=46/26#

Since #y=9-10z#, you can find y now. #y=9-(460/26)# or #y=-226/26#.

Finally, you can easily solve using any original equation. #x=2/26#.