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Math For College Readiness

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Simplify:
y - 3 / y - 2 - y + 1 / 2y - 5 + -4y + 7 / 2y^2 - 9y + 10

  • Math For College Readiness -

    I'm sure there are some missing parentheses. Making a guess, I'd say you meant

    (y-3)/(y-2) - (y+1)/(2y-5) + (-4y+7)/(2y^2-9y+10)

    putting all over a common denominator, we have

    [(y-3)(2y-5) - (y+1)(y-2) + (-4y+7)]/(2y^2-9y+10)

    ((2y^2-11y+15) - (y^2-y-2) + (-4y+7))/(2y^2-9y+10)

    = (y^2-14y+24)/(2y^2-9y+10)
    = (y-2)(y-12)/[(y-2)(2y-5)]
    = (y-12)/(2y-5)

  • Math For College Readiness -

    Thank you very much. Yeah, I should've added the parenthesis to make it look familiar.

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