How do you solve the following system?: #-x -2y =1, x -y = -1#

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Answer 1 · 2017-04-12T12:25:56

See the entire solution process below:

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

#x - y = -1#

#x - y + color(red)(y) = -1 + color(red)(y)#

#x - 0 = -1 + y#

#x = -1 + y#

Step 2) Substitute #-1 + y# for #x# in the first equation and solve for #y#:

#-x - 2y = 1# becomes:

#-(-1 + y) - 2y = 1#

#1 - y - 2y = 1#

#1 - 1y - 2y = 1#

#1 - 3y = 1#

#-color(red)(1) + 1 - 3y = -color(red)(1) + 1#

#0 - 3y = 0#

#-3y = 0#

#(-3y)/color(red)(-3) = 0/color(red)(-3)#

#(color(red)(cancel(color(black)(-3)))y)/cancel(color(red)(-3)) = 0#

#y 0#

Step 3) Substitute #0# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = -1 + y# becomes:

#x = -1 + 0#

#x = -1#

The solution is: #x = -1# and #y = 0# or #(-1, 0)#