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: $(2,1/4)$

How do you solve the system 4x-y=3y+74x-y=3y+7 and x+8y=4x+8y=4?