How do you solve the system of equations #3x + 3y = - 6# and #x + 6y = - 32#?

1 Answer
Jun 2, 2017

#(4,-6)#

Explanation:

First, we can simplify the first equation:

#3x+3y = -6#

#(3x)/3 + (3y)/3 = (-6)/3#

#x+y = -2#

Now, subtract the first equation from the second equation:

#color(white)"X"x+6y = -32#
#-(x+y = -2)#

#color(white)"XXX-" 5y = -30#

And solve for y:

#(5y)/5 = (-30)/5#

#y = -6#

Now that we know #y#, all we have to do is plug in #-6# for #y# and solve for #x#.

#x+y = -2#

#x-6=-2#

#color(white)"e"x = 4#

So our solution is:

#x=4,y=-6 color(white)"XX"# or #color(white)"XX" (4,-6)#

Final Answer