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.
