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Math Help - solving rational equation

  1. #1
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    solving rational equation

    hello, I cannot seem to understand the process for solving this.
    Can somone please show me the steps involved.

    Thanks for any help.

    (1/x)+5 > 1/3x

    I was thinking:
    1/x > (1/3x)-5
    (1/x) - 1/3x > -5
    1/x > -5
    x > 5

    but I know my attempt is incorrect.
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  2. #2
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    Re: solving rational equation

    Quote Originally Posted by fran1942 View Post
    hello, I cannot seem to understand the process for solving this.
    Can somone please show me the steps involved.

    Thanks for any help.

    (1/x)+5 > 1/3x

    I was thinking:
    1/x > (1/3x)-5
    (1/x) - 1/3x > -5
    1/x > -5
    x > 5

    but I know my attempt is incorrect.
    Is the inequality \displaystyle \frac{1}{x} + 5 > \frac{1}{3}x or \displaystyle \frac{1}{x} + 5 > \frac{1}{3x}?
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  3. #3
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    Re: solving rational equation

    It is the second equation you quoted.

    thanks.
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  4. #4
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    Re: solving rational equation

    First note that \displaystyle x \neq 0.

    \displaystyle \begin{align*} \frac{1}{x} + 5 &> \frac{1}{3x} \\ \frac{1}{x} - \frac{1}{3x} &> -5 \\ \frac{3}{3x} - \frac{1}{3x} &> -5 \\ \frac{2}{3x} &> -5 \end{align*}

    Now you need to consider two cases, \displaystyle x < 0 and \displaystyle x > 0.

    Case 1:

    \displaystyle \begin{align*} \frac{2}{3x} &> -5 \\ 2 &< -15x \\ -15x &> 2 \\ x &< -\frac{2}{15} \end{align*}

    And since we know \displaystyle x < 0, that means \displaystyle x < -\frac{2}{15} is acceptable.

    Case 2:

    \displaystyle \begin{align*} \frac{2}{3x} &> -5 \\ 2 &> -15x \\ -15x &< 2 \\ x &> -\frac{2}{15} \end{align*}

    And since we know \displaystyle x > 0, that means \displaystyle x > 0 is acceptable.

    So the solution is \displaystyle x \in \left(-\infty, -\frac{2}{15}\right) \cup (0, \infty)
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