How do you solve $\frac{x}{x - 2} > 9$ $x/(x-2)>9$ ?

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Answer 1 · 2017-02-23T15:39:43

The solution is $x \in] 2, \frac{9}{4} [$ $x in ]2,9/4[$

We cannot do crossing over.

We rearrange the equation

$\frac{x}{x - 2} > 9$ $x/(x-2)>9$

$\frac{x}{x - 2} - 9 > 0$ $x/(x-2)-9>0$

$\frac{x - 9 (x - 2)}{x - 2} > 0$ $(x-9(x-2))/(x-2)>0$

$\frac{x - 9 x + 18}{x - 2} > 0$ $(x-9x+18)/(x-2)>0$

$\frac{18 - 8 x}{x - 2} > 0$ $(18-8x)/(x-2)>0$

$\frac{2 (9 - 4 x)}{x - 2} > 0$ $(2(9-4x))/(x-2)>0$

Let $f (x) = \frac{2 (9 - 4 x)}{x - 2}$ $f(x)=(2(9-4x))/(x-2)$

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{2}$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaaaa)$ $2$ $color(white)(aaaaaaaa)$ $9/4$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-2$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $9-4x$ $color(white)(aaa)$ $+$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $||$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

Therefore,

$f(x)>0$ when $x in ]2,9/4[$

How do you solve x/(x-2)>9xx−2>9x/(x-2)>9?