How do you solve #8x - 8y = 16# and #4x + 4y = -72 #?

1 Answer
MathFact-orials.blogspot.com
Jun 5, 2016

Work with one equation to get a term in the other, then substitute in, to get the system #x=-8, y=-10#

Explanation:

We have a system of 2 equations:

#8x-8y=16#
#4x+4y=-72#

Since this is in the substitution method category, we'll use that method.

If we take the second equation and multiply both sides by 2, we can get the term #8x#. We can take what #8x# will equal and substitute it into the first equation:

#(4x+4y)(2)=(-72)(2)#

#8x+8y=-144#

#8x=-144-8y#

So let's now substitute. Starting with the first equation:

#8x-8y=16#

#-144-8y-8y=16#

#-16y=160#

#y=-10#

We can now solve for x. I'm going to do it with both equations so that we can make sure we didn't make a mistake (if we get 2 different answers in checking the 2 equations, we did!)

#8x-8y=16#

#8x-8(-10)=16#

#8x+80=16#

#8x=-64#

#x=-8#

and now the other one:

#4x+4y=-72#

#4x+4(-10)=-72#

#4x-40=-72#

#4x=-32#

#x=-8#

#x=-8, y=-10#

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