How do you solve $\frac{8}{x - 1} < 1$ $8/(x-1)<1$ ?

Question

How do you solve $\frac{8}{x - 1} < 1$ $8/(x-1)<1$ ?

1 Answer

Impact of this question

2148 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-25T07:33:40

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

Explanation:

We cannot do crossing over

So,

$\frac{8}{x - 1} < 1$ $8/(x-1)<1$ , $\Rightarrow$ $=>$ , $\frac{8}{x - 1} - 1 < 0$ $8/(x-1)-1<0$

$\frac{8 - (x - 1)}{x - 1} < 0$ $(8-(x-1))/(x-1)<0$

$\frac{9 - x}{x - 1} < 0$ $(9-x)/(x-1)<0$

Let $f (x) = \frac{9 - x}{x - 1}$ $f(x)=(9-x)/(x-1)$

We can do a sign chart

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

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

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

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

Therefore,

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

Science

Math

Humanities

... and beyond

How do you solve $\frac{8}{x - 1} < 1$ $8/(x-1)<1$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 8/(x-1)<18x−1<18/(x-1)<1?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $\frac{8}{x - 1} < 1$ $8/(x-1)<1$ ?