How do you solve (1) #x + 2y + z = 1#, (2) #2x - 3y - z = 6#, (3) #3x + 5y + 4z = 5#, (4) #4x + y + z = 8#?
1 Answer
Aug 6, 2016
Explanation:
Given:
(1):
#color(white)(X)x+2y+z=1# (2):
#color(white)(X)2x-3y-z=6# (3):
#color(white)(X)3x+5y+4z=5# (4):
#color(white)(X)4x+y+z=8#
First note that we have
Note that: (1) + (1) + (2) is
Subtract
(5):
#color(white)(X)-7y-3z=4#
Subtract
(6):
#color(white)(X)-y+z=2#
Add
#-10y = 10#
So
Hence from (6) we have
Then from (1) we have
So: