How do you solve #9x-9y=-27# and #5x+27=y# using substitution?

1 Answer
Feb 9, 2017

Answer:

See the entire solution process below:

Explanation:

Step 1) Because the second equation is already solved for #y#, substitute #5x + 27# for #y# in the first equation and solve for #x#:

#9x - 9y = -27# becomes:

#9x - 9(5x + 27) = -27#

#9x - 45x - 243 = -27#

#-36x - 243 = -27#

#-36x - 243 + color(red)(243) = -27 + color(red)(243)#

#-36x - 0 = 216#

#-36x = 216#

#(-36x)/color(red)(-36) = 216/color(red)(-36)#

#(color(red)(cancel(color(black)(-36)))x)/cancel(color(red)(-36)) = -6#

#x = -6#

Step 2) Substitute #-6# for #x# in the second equation and calculate #y#:

#5x + 27 = y# becomes:

#(5 xx -6) + 27 = y#

#-30 + 27 = y#

#-3 = y#

#y = -3#

The solution is #x = -6# and #y = -3# or #(-6, -3)#