How do you solve $8/(x-1)<1$ ?

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Answer 1 · 2016-12-25T07:33:40

The answer is $x in ] -oo,1 [ uu ] 9, +oo[$

We cannot do crossing over

So,

$8/(x-1)<1$ , $=>$ , $8/(x-1)-1<0$

$(8-(x-1))/(x-1)<0$

$(9-x)/(x-1)<0$

Let $f(x)=(9-x)/(x-1)$

We can do a sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $1$ $color(white)(aaaa)$ $9$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-1$ $color(white)(aaaaa)$ $-$ $color(white)(aaa)$ $+$ $color(white)(aaa)$ $+$

$color(white)(aaaa)$ $9-x$ $color(white)(aaaaa)$ $+$ $color(white)(aaa)$ $+$ $color(white)(aaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaa)$ $-$ $color(white)(aaa)$ $+$ $color(white)(aaa)$ $-$

Therefore,

$f(x)<0$ , when $x in ] -oo,1 [ uu ] 9, +oo[$

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