Find the set of values of x for which

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

thank you

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- May 31st 2008, 06:39 AM #1

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- May 31st 2008, 06:57 AM #2

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- May 31st 2008, 07:08 AM #3

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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:

$\displaystyle (x-1)^2(3-x)^2 \frac{x}{x-1}>\frac{2}{3-x} (3-x)^2(x-1)^2$)

- May 31st 2008, 07:14 AM #4
This has to be wrong. You can't multiply by $\displaystyle x-1$ or $\displaystyle 3-x$ since we don't previously know what their signs are.

We first gonna have to make some algebra:

$\displaystyle \begin{aligned}

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

\frac{x}{x-1}+\frac{2}{x-3}&>0 \\

\frac{x^{2}-3x+2x-2}{(x-1)(x-3)}&>0 \\

\frac{(x+1)(x-2)}{(x-1)(x-3)}&>0.

\end{aligned}$

Now follow this and you'll get the answer.