college algebra

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Solving for a variable

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

  • college algebra -

    If 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 -

    the 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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