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Thread: fraction equation inequality

  1. #1
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    fraction equation inequality

    I'm trying to solve the equality $\displaystyle \frac{{2x - 1}}{{x + 1}} < 3 \$

    $\displaystyle
    \begin{array}{l}
    \frac{{2x - 1}}{{x + 1}} < 3 \\
    2x - 1 < 3(x + 1) \\
    2x - 1 < 3x + 3 \\
    - 4 < x \\
    \end{array}
    $

    I can't work out how to get the second solution of x > -1
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  2. #2
    MHF Contributor Quick's Avatar
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    You messed up in your second step.

    When you multiply both sides of an inequality by a negative number (which happens when x is less than -1) you have to flip the sign, meaning that the second step would actually yield two inequalities:

    $\displaystyle 2x-1<3(x+1)\quad\text{when}\quad x>-1$

    and

    $\displaystyle 2x-1>3(x+1)\quad\text{when}\quad x<-1$

    Notice that $\displaystyle x \neq -1$ because then you would have division by zero, which isn't allowed.
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  3. #3
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    $\displaystyle \begin{array}{rcl} \frac{2x-1}{x+1} - 3 & < & 0 \\ \frac{2x - 1 - 3(x+1)}{x+1} & < & 0 \\ \frac{2x - 1 - 3x - 3}{x+1} & < & 0 \\ \frac{-x - 4}{x+1} & < & 0 \\ \frac{x+4}{x+1}& > & 0\end{array}$

    (Divided -1 from both sides in that last step there which switches the inequality).

    So you have your critical points x = -4 and -1. Test below, in between, and above to see which regions are greater than 0 and you should have your answer.
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  4. #4
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    ineoualities with fraction

    Quote Originally Posted by Craka View Post
    I'm trying to solve the equality $\displaystyle \frac{{2x - 1}}{{x + 1}} < 3 \$

    $\displaystyle
    \begin{array}{l}
    \frac{{2x - 1}}{{x + 1}} < 3 \\
    2x - 1 < 3(x + 1) \\
    2x - 1 < 3x + 3 \\
    - 4 < x \\
    \end{array}
    $

    I can't work out how to get the second solution of x > -1
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