algebra II

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I'm quite clueless as to how to solve this. I know the quadratic equation but am not sure how to get this equation into the format to plug in to the quadratic...

Solve the quadratic equation using the formula [1/(1+x)]-[1/(3-x)]=(6/35)
Show your work

  • algebra II -

    multiply both sides by 35(x+1)(x-3)

    then
    35(x-3)+35(x+1)=6(x-3)(x+1)

    and you should be able to assemble it from there.

  • algebra II -

    so should i just multiply that all out and then plug it in?

  • algebra II -

    Write the two terms on the left with a common denomimnator. Then convert to standard quadratic form.

    [(3-x) - (1+x)]/[(1+x)(3-x)] = 6/35
    (2-2x)/[(1+x)(3-x)] = 6/35
    (1-x) = (3/35)[(1+x)(3-x)]
    35 -35x = 3[-x^2 +2x +3)
    3x^2 -41x +26 = 0
    (3x -2)(x-13)
    Check my work. No guarantees

  • algebra II -

    We were lucky that one factored fairly easily. x = 2/3 or 13

  • algebra II -

    i got the same answer but used the quadratic :) thank you for all your help everyone!

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