1. ## Help please?

Find the set of values of x for which

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

thank you

2. Originally Posted by Silver
Find the set of values of x for which

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

thank you
Wh

3. Originally Posted by janvdl
$\displaystyle x(3-x) > 2(x-1)$

$\displaystyle 3x-x^2 > 2x-2$

$\displaystyle -x^2 + x + 2 > 0$

$\displaystyle -1(x^2 - x - 2) > 0$

$\displaystyle (x^2 - x - 2) < 0$

$\displaystyle (x - 2)(x + 1) < 0$

$\displaystyle -1 < x < 2$ BUT $\displaystyle x \neq 1$
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:

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

4. Originally Posted by janvdl

$\displaystyle x(3-x) > 2(x-1)$
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.