How do you solve the system of equations #-8x - 8y = 8# and #- 7x - 8y = 16# using elimination?

Question

3002 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-29T13:33:30

#x=8# and #y=-9#

Both have -8y so we can eliminate it by adding equation:

#quad\ quad-8x-8y=8#
#-quad-7x-8y=16#

#quadquad\ \ -x quadquadquadquadquad\ = -8#

Solving for x would get you #x = 8#.

Now plug this value into first equation:
#-8(8)-8y = 8#

#-64-8y = 8#

#-8y = 8+64#

#-8y = 72#

#y = -9#

So we have #x=8# and #y=-9#

Answer 2 · 2018-03-29T13:40:33

#x=8, y=-9#

#[[-8x-8y,=,8],[-7x-8y,=,16]]#

multiply first row by #-1#

#[[8x+8y,=,-8],[-7x-8y,=,16]]#

Now add the rows:

#(8-7)x+(8-8)y=-8+16#

#1x+0y=8#

#x=8#

Now, we can use any row to obtain #y#:

#8x+8y=-8#

Using #x=8#:

#8color(red)((8))+8y=-8#

#64+8y=-8#

#8y=-8-64#

#8y=-72#

#y=-72/8#

#y=-9#