How do you solve the system #4x-y=3y+7# and #x+8y=4#?

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Answer 1 · 2017-04-09T16:08:34

See below.

Let's first move the #y#'s to the left side in the first equation.

#4x-y=3y+7#
#4x-4y=7#

Dividing both sides by #4#,

#x-y=7/4#

Our other equation is:

#x+8y=4#

Subtracting the first equation from the second, we get:

#9y=9/4#

So #y=1/4#.

We plug this into one of the equations (I'm plugging into the second one).

#x+2=4#
#x=2#

So our ordered pair is: #(2,1/4)#