Algebra one help!!!

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okay it's due tomorrow.
so here's a few problems that I need clarified.
please&thankyou.

Get the y by itself.
-3x+2y=2

x-4y=8

2x+y=-3
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Hey, Silver, let me see if I can help you out. I'm going to assume that your task is not only to get y on its own side of equality sign, but to solve for y as well.

-3x+2y=2

Okay so our first task, here, is to get the x value away from the y and across the equality sign. To do this, we add 3x to both sides of the equation (remember that, on the y side, we're trying to bring x to zero).

-3x+3x+2y=2+3x
0x+2y=2+3x
2y=2+3x

We still need to rid of the lefthand 2, in order to achieve a lonely y, and so we divide both sides of the equation by that value: 2.

2y/2=2/2+3x/2

The 2's on the lefthand side cancel out to leave y, and we simplify the righthand fractions.

y=1+3x/2


x-4y=8

Same thing here, but because x is positive, we are going to subtract it. Recall that a lone variable is considered to have a numerical value of 1.

x-x-4y=8-x
0x-4y=8-x
-4y=8-x
-4y/-4=8/-4-(x/-4)
y=-2+x/4

(Two negatives equal a positive, so -x/-4 becomes x/4.)


2x+y=-3

2x-2x+y=-3-2x
0x+y=-3-2x
y=-3-2x

. . .

Easy peasy, right? If you have any questions, or if I have left anything unclear, feel free to PM. ^_^




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yea pretty much what was said above



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