0

# x+4y-z=63x+2y+3z=162x-y+z=3

system of equation with 3 equations using elimination

### 1 Answer by Expert Tutors

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (639 lesson ratings) (639)
1
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.