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system of equation with 3 equations using elimination
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I will restate your equations with spacing that is a bit easier to see
x +  4y    -z = 6           first equation
3x +2y + 3z = 16      second equation
2x   -y    +z  =  3    third equation
The first step is to add the third equation to first one this gives
3x  +3y           = 9       This eliminates z, so we look to do this again. 
  So triple the first equation and add to it the second  equation.   This gives
6x +14 y           = 34        Now we have two new equations involving only x and y
    Double the  first new equation and subtract it from the second new equation.  This gives
8y  = 16        So  y = 2.      Substitute y = 2 into the first new equation to get x =  1
  The substitute   x = 1 and y = 2 into any of the original equations to get z = 3.