Question 658034
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Hi,
3x+2y+z=7, 5x+5y+4z=3, 3x+2y+3z=1

3x+2y+z=7
<u>3x+2y+3z=1</u>  || Subtract 2nd from 1st
  -2z = 6
    z = -3

3x+ 2y = 10    ||multiply 1st EQ by 5 and subtract from 3 times the 2nd EQ
<u>5x + 5y = 15</u>
       5y = -5
        y = -1  and x = 4
and...Using 3x+2y+3z=1
 {{{12 - 2 - 9 = 1}}}