What are #x# and #y# if #-x-3y= 15# and #2x+7y= -36#?

1 Answer
Mar 2, 2018

Answer:

#3# for #x# and #-6# for #y#

Explanation:

Let's solve for #x#:

#-x-3y = 15#

#-x = 15+3y#

#x = -15-3y#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *#

Now let's substitute that into the second equation

#2(-15-3y) + 7y = -36#

#-30 - 6y + 7y = -36#

#-6y + 7y = -6#

#y = -6#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *#

Now we need to solve for #x#:

#x = -15-3(-6)#

#x = -15 + 18#

#x = 3#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *#

We should still check our work:
Plug in #3# for #x# and #-6# for #y#

#-(3) - 3(-6) # should equal #15#

#-3 - (-18)#

#-3+18 = 15#

#15 = 15#, so we were right!!!