One of the variables, either #x# or # y# , must be isolated . In this case, I will rearrange the second equation, # x+3y=-28# and isolate the #x# variable.

#x+3y=-28# Now subtract #3y# from each side of the equation

#x+3y-3y=-28-3y#

#x=-3y-28#

Now that I have isolated the #x# variable in your second equation, I will substitute it in your first equation.

#3x-y=-4#

#3(-3y-28)-y=-4#

#-9y-84-y=-4# collect like terms

#-10y-84 = -4# Add 84 to each side

#-10y-84+84=-4+84#

#-10y=80# Divide by #-10#

#y=-8#

Now simply substitute #-8# in for #y# in either equation.

#x+3y=-28#

#x+3(-8)=-28#

#x-24= -28# Now add #24# to each side of the equation

#x-24+24=-28+24#

#x=-4#

#(-4,-8)#

Now there is one small problem, and that is, how do I know this is the correct solution? Since I used your second equation, #x+3y=-28# to calculate the value of #y#, I can't use this same equation to check my answer. It would always appear to be correct. I must go to the other equation, in your case the first one, to verify that I have the correct answer. So:

# 3x-y=-4#

#3(-4)-(-8)=-4#

#-12+8=-4#

#-4=-4#

This confirms that I have the solution correct.