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Math Help - Help please?

  1. #1
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    Help please?

    Find the set of values of x for which

    \frac{x}{x-1} > \frac{2}{3-x}

    thank you
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  2. #2
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    Quote Originally Posted by Silver View Post
    Find the set of values of x for which

    \frac{x}{x-1} > \frac{2}{3-x}

    thank you
    Wh
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  3. #3
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    Quote Originally Posted by janvdl View Post
    x(3-x) > 2(x-1)

    3x-x^2 > 2x-2

    -x^2 + x + 2 > 0

    -1(x^2 - x - 2) > 0

    (x^2 - x - 2) < 0

    (x - 2)(x + 1) < 0

    -1 < x < 2 BUT x \neq 1
    what about multiplying the both sides by the squared denominators?
    Ive answered a similar question like you did and my tutor said it was wrong and told me to use the squared method, which confused the hell outta me.

    (by squared denominator i mean this:

    (x-1)^2(3-x)^2 \frac{x}{x-1}>\frac{2}{3-x} (3-x)^2(x-1)^2)
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by janvdl View Post

    x(3-x) > 2(x-1)
    This has to be wrong. You can't multiply by x-1 or 3-x since we don't previously know what their signs are.

    We first gonna have to make some algebra:

    \begin{aligned}<br />
   \frac{x}{x-1}&>\frac{2}{3-x} \\ <br />
  \frac{x}{x-1}+\frac{2}{x-3}&>0 \\ <br />
  \frac{x^{2}-3x+2x-2}{(x-1)(x-3)}&>0 \\ <br />
  \frac{(x+1)(x-2)}{(x-1)(x-3)}&>0.<br />
\end{aligned}

    Now follow this and you'll get the answer.
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