How do you solve the system of equations #2x-y+3z=-1#, #x+2y-4z=-1# and #y-2z=0# by elimination?

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Answer 1 · 2017-09-07T17:23:21

From third equation #y=2z#

After plugging y into second one,

#x+2*2z-4z=-1# or #x=-1#.

After plugging x and y into first one,

#2*(-1)-2z+3z=-1#

#-2+z=-1#

#z=1#. Consequently, #y=2*1=2#

1) I found y in terms of z from third equation.

2) I plugged y into second one for finding x.

3) I plugged x and y into first one for finding z.

4) I found y.