How do you solve #-8x - 8y = 0# and #-8x + 8y = -128#?

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Answer 1 · 2017-09-03T14:10:57

See a solution process below:

Step 1) Solve the first equation for #x#:

#-8x - 8y = 0#

#-8x - 8y + + color(red)(8y) = 0 + color(red)(8y)#

#-8x - 0 = 8y#

#-8x = 8y#

#(-8x)/color(red)(-8) = (8y)/color(red)(-8)#

#1x = -1y#

#x = -y#

Step 2) Substitute #-y# for #x# in the second equation and solve for #y#:

#-8x + 8y = -128# becomes:

#(-8 * -y) + 8y = -128#

#8y + 8y = -128#

#16y = -128#

#(16y)/color(red)(16) = -128/color(red)(16)#

#y = -8#

Step 3) Substitute #-8# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = -y# becomes:

#x = - -8#

#x = 8#

The Solution Is: #x = 8# and #y = -8# or #(8, -8)#