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Math Help - Solve the equation

  1. #1
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    Solve the equation

    3\cdot\left[\frac {4x+1}5\right]=7x-1 where [a] is the integer part of a.
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  2. #2
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    Hello, james_bond!

    I haven't solved it yet, but here's an approach.
    . . Maybe it will inspire someone . . .



    3\cdot\left[\frac {4x+1}5\right]=7x-1 .where [a] is the integer part of a

    We have: . \left[\frac{4x+1}{5}\right] \;=\;\frac{7x-1}{3} \quad\Rightarrow\quad \left[\frac{4x+1}{5}\right]\;=\;2x + \frac{x-1}{3}\;\;{\color{blue}[1]}


    Since the right side must be an integer, x-1 \:=\:3k\:\text{, for some integer }k
    . . That is: . x \:=\:3k+1


    Substitute into [1]: . \left[\frac{4(3k+1)+1}{5}\right] \;=\;2(3k+1) + \frac{(3k+1)-1}{3}

    . . which simplifies to: . \left[\frac{12k+5}{5}\right] \;=\;7k+2 \quad\Rightarrow\quad \left[2k+1+\frac{2k}{5}\right] \;=\;7k+2


    And that's it for now . . .

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  3. #3
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    Hello,

    Soroban made a mistake since x is not necessarily an integer. Instead, as he pointed out, it is true that (7x-1)/3 is an integer. Thus, x=(3k+1)/7 for some integer k.

    You have an inequality t-1<[t]\leq t.
    In this case, you get (4x+1)/5 - 1 < (7x-1)/3 <= (4x+1)/5 which means -7/23<x<=8/23.
    Now, substituting x=(3k+1)/7, you get k=-1, 0.

    Bye.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Exclamation Remarks on Soroban's post

    Quote Originally Posted by Soroban View Post
    Hello, james_bond!

    I haven't solved it yet, but here's an approach.
    . . Maybe it will inspire someone . . .




    We have: . \left[\frac{4x+1}{5}\right] \;=\;\frac{7x-1}{3} \quad\Rightarrow\quad \left[\frac{4x+1}{5}\right]\;=\;2x + \frac{x-1}{3}\;\;{\color{blue}[1]}


    Since the right side must be an integer, x-1 \:=\:3k\:\text{, for some integer }k
    . . That is: . x \:=\:3k+1


    Substitute into [1]: . \left[\frac{4(3k+1)+1}{5}\right] \;=\;2(3k+1) + \frac{(3k+1)-1}{3}

    . . which simplifies to: . \left[\frac{12k+5}{5}\right] \;=\;7k+2 \quad\Rightarrow\quad \left[2k+1+\frac{2k}{5}\right] \;=\;7k+2


    And that's it for now . . .

    In addition to the comments by wisterville, I have the followings.

    At the right-hand side of [1], we do not know that \frac{x-1}{3} is an integer, what we know is 2x+\frac{x-1}{3} is an integer exactly.
    Letting x=\frac{13}{7}, we indeed see that \frac{x-1}{3}=\frac{2}{7}\not\in\mathbb{Z} but we have 2x+\frac{x-1}{3}=6\in\mathbb{Z}.

    We clearly have [t]\leq t<[t]+1 for all t\in\mathbb{R}.
    Then, using the equation you gave with t=\frac{4x+1}{5}, we get
    \frac{7x-1}{3}\leq\frac{4x+1}{5}<\frac{7x-1}{3}+1.
    You can solve these inequalities by seperating into two parts.
    Last edited by bkarpuz; September 24th 2008 at 09:54 AM.
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