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Math Help - fraction equation inequality

  1. #1
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    fraction equation inequality

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

    <br />
\begin{array}{l}<br />
 \frac{{2x - 1}}{{x + 1}} < 3 \\ <br />
 2x - 1 < 3(x + 1) \\ <br />
 2x - 1 < 3x + 3 \\ <br />
  - 4 < x \\ <br />
 \end{array}<br />

    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:

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

    and

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

    Notice that 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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    \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 \frac{{2x - 1}}{{x + 1}} < 3 \

    <br />
\begin{array}{l}<br />
\frac{{2x - 1}}{{x + 1}} < 3 \\ <br />
2x - 1 < 3(x + 1) \\ <br />
2x - 1 < 3x + 3 \\ <br />
- 4 < x \\ <br />
\end{array}<br />

    I can't work out how to get the second solution of x > -1
    Follow Math Help Forum on Facebook and Google+

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