How do you solve the system #w-2x+3y+z=3#, #2w-x-y+z=4#, #w+2x-3y-z=1#, #3w-x+y-2z=-4#?

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Answer 1 · 2017-11-26T21:38:52

#(w, x, y, z) = (2, 3, 1, 4)#

Given:

#{ (w-2x+3y+z=3),(2w-x-y+z=4),(w+2x-3y-z=1),(3w-x+y-2z=-4):}#

Adding the first and third equations together, we get:

#2w=4#

So:

#w=2#

Discarding the first equation, and substituting #w=2# into the others, we get:

#{ (-x-y+z=0),(2x-3y-z=-1),(-x+y-2z=-10):}#

From the first equation:

#z = x+y#

Substituting this into the second and third equations, we get:

#{ (x-4y=-1),(-3x-y=-10):}#

Multiplying the second equation by #-4# we get:

#12x+4y=40#

Adding this to the first equation, we get:

#13x = 39#

Hence:

#x = 3#

From the second equation:

#y = 10-3x = 10-9 = 1#

Then:

#z = x+y = 3+1 = 4#

So:

#(w, x, y, z) = (2, 3, 1, 4)#