1. ## [SOLVED] Inequalities

$\displaystyle 2/x<x-1$
Since Im not allowed to multiply by x I can´t work it out.
Never done this kind of stuff before.

$\displaystyle 2/x<x-1$ -> $\displaystyle 2/x -x +1< 0$
I can wright it as (2-x^2+x)/x <0

Multiplying $\displaystyle (-1)$ gives me $\displaystyle (x^2-x-2)/x>0$

Whatever I do I can´t get it right....
All help apprecieted.

2. ## I got it

Never mind solved

3. Just wondering, can you tell me what was your solution set?

4. ## Sure

Sure I did it this way ( if I understood it correctly.

$\displaystyle 2/x > x-1$ / add: -x+1
$\displaystyle 2/x-x+1 > 0$ /make denominator same
$\displaystyle (2-x^2+x)/x > 0$ /*(-1)
$\displaystyle (x^2-x-2)/x < 0$

Now I solved for x with:$\displaystyle X^2-x-2 -> x = 1/2 +- 1.5$
X1 = 2
X2 = (-1)

Now So I got the intervalls (-1);(0);(2)
So i picked a number in all intervalls + a number below (-1) and one above (2).

That gave me x>(-1)
or x>2
and not 0 cause that is undefined...

5. Originally Posted by Henryt999
$\displaystyle 2/x<x-1$
Note that you solved this inequality flipped.

6. ## I´m sorry what

Im sorry I´m not too familiar with the english expression?
What did you mean?

7. Instead of solving the one for $\displaystyle <,$ you solved it for $\displaystyle >.$

8. ## Any other way?

Any other way of solving it?
Ill be more then happy to learn it...

9. I'll do it when I get back.

I have to leave now.

10. Originally Posted by Henryt999
$\displaystyle 2/x<x-1$
Since Im not allowed to multiply by x I can´t work it out.
Never done this kind of stuff before.

$\displaystyle 2/x<x-1$ -> $\displaystyle 2/x -x +1< 0$
I can wright it as (2-x^2+x)/x <0

Multiplying $\displaystyle (-1)$ gives me $\displaystyle (x^2-x-2)/x>0$

Whatever I do I can´t get it right....
All help apprecieted.
1. Multiply both sides by X
$\displaystyle (2/x)(x) < (X)(X-1) \implies 2 < X^2 -1x$

Domain: All real numbers greater than 2