How do you solve the system of equations #3x + 3y = - 6# and #x + 6y = - 32#?
1 Answer
Jun 2, 2017
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
#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