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July 24, 2014

July 24, 2014

Posted by **Anonymous** on Monday, February 18, 2013 at 8:37pm.

4x^2+3y=y+1/3 I'm stuck thanks

- college algebra -
**Anonymous**, Monday, February 18, 2013 at 9:35pmIf all you need to do is solve for x and y then this is what I would do:

first solving for y

4x^2+3y=y+(1/3) multiply everything by 3 to get rid of (1/3)

12x^2+9y=3y then isolate y

(12x^2)/(-6)=y

then solving for x

go back a few steps to get 12x^2=-6y (this is just rearranging the y= equation)

solve that for x and you should get x=sqrt(-y/2)

i hope this helps- it seems to simple for college algebra, but there you go.

- correction - college algebra -
**Reiny**, Monday, February 18, 2013 at 10:13pmthe 4th line should have been

12x^2+9y=3y + 1

you probably wanted to solve for y, thus expressing it as a function:

6y = -12x^2 + 1

y = -2x^2 + 1/6

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