SOLUTION: What are the x, y, and z values for the problems:3x+2y+z=7, 5x+5y+4z=3, 3x+2y+3z=1, and how to solve to get those values?

Algebra.Com
Question 658034: What are the x, y, and z values for the problems:3x+2y+z=7, 5x+5y+4z=3, 3x+2y+3z=1, and how to solve to get those values?
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
3x+2y+z=7, 5x+5y+4z=3, 3x+2y+3z=1
3x+2y+z=7
3x+2y+3z=1 || Subtract 2nd from 1st
-2z = 6
z = -3
3x+ 2y = 10 ||multiply 1st EQ by 5 and subtract from 3 times the 2nd EQ
5x + 5y = 15
5y = -5
y = -1 and x = 4
and...Using 3x+2y+3z=1



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