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Math Help - part of a fractional equation w/ binomial denominator

  1. #1
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    part of a fractional equation w/ binomial denominator

    8(5-y)(6/5-y)

    I don't get how to solve this part.. the result should be 48.. would anyone explain, thanks
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  2. #2
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    Is the question 8(5-y)(\frac{6}{5}-y)=0 or something?

    (clearly it isn't this since y=48 is not a solution).
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  3. #3
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    no, the whole equation is

    3/8=6/5-y

    LCD=8(5-y)

    (3/8)8(5-y)=8(5-y)(6/5-y)

    3(5-y)=48

    y=-11


    to clarify, the fraction is 6/5-y, where 5-y is the binomial denominator and it's not as this shows it: 6/5-y
    I dont get how to show it in the proper way
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  4. #4
    Super Member Showcase_22's Avatar
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    Sorry but I can't really see the problem.

    You've just worked it out yourself to get y=-11, isn't that the question done?

    Anyway, if you have <br />
8(5-y)(6/5-y)<br />
it must be equal to something in order for it to have solutions.

    Are you trying to solve:

    \frac{3}{8}=\frac{6}{5}-y?

    You can solve this by:

    y=\frac{6}{5}-\frac{3}{8}

    y=\frac{48-15}{40}=\frac{33}{40}
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  5. #5
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    Talking

    Quote Originally Posted by strunz View Post
    no, the whole equation is

    3/8=6/5-y

    LCD=8(5-y)
    From your last line above, I will guess that the equation is actually "3/8 = 6/(5 - y)", rather than "3/8 = 6/5 - y".

    A good start for solving this proportion is to "cross-multiply" to get:

    . . . . .3(5 - y) = 6(8)

    Solve the resulting linear equation by first multiplying things out:

    . . . . .15 - 3y = 48

    Rearrange to get:

    . . . . .15 - 48 = 3y

    . . . . .-33 = 3y

    Then divide through. To confirm your solution, plug it back in to the original equation:

    . . . . .left-hand side: 3/8

    . . . . .right-hand side: 6/(5 - [-11]) = 6/(5 + 11) = 6/16 = 3/8

    And you're done!
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  6. #6
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    Yeah that's perfect, thanks.
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