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

Answer:

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

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 #4# equations in #3# unknowns. So either there is some redundancy, or there is no solution.

Note that: (1) + (1) + (2) is #4x+y+z=8#, identical to (4). So there's our expected redundancy. We can ignore (4) and work with the first three equations.

Subtract #2 xx# (1) from (2) to get:

(5): #color(white)(X)-7y-3z=4#

Subtract #3 xx# (1) from (3) to get:

(6): #color(white)(X)-y+z=2#

Add #3xx# (6) to (5) to get:

#-10y = 10#

So #y = -1#

Hence from (6) we have #z = 2+y = 2-1=1#

Then from (1) we have #x = 1-2y-z = 1+2-1 = 2#

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