The question:
Solve $\displaystyle |\frac{x - 1}{x + 1}| < 1$
My attempt:
$\displaystyle |x - 1| < |x + 1|$
$\displaystyle x - 1 < x + 1$
$\displaystyle x < x + 2$
Help! I'm clearly doing something wrong here. :/
This doesn't follow. If x- 1> 0 and x+ 1> 0 (which happens for x> 1), then |x- 1|= x- 1 and |x+ 1|= x+ 1 so x- 1< x+ 1 from which -1< 1 which is always true.
But if x- 1 and x+ 1 are both negative (which happens for x< -1), then |x- 1|= -(x- 1) and |x+ 1|= -(x+ 1) so -(x- 1)< -(x+ 1) from which x- 1> x+ 1, -1> 1 which is always false.
If -1< x< 1, then x- 1< 0 and x+ 1> 0 so |x- 1|< |x+ 1| becomes -(x- 1)< x+ 1 or -x+ 1< x+ 1 which reduces to -x< x, 0< 2x, x> 0.
The inequality is true for all positive x except x= 1. (If the problem had been $\displaystyle \frac{|x- 1|}{|x+ 1|}\le 1$ then it would be true for all positive x.)
$\displaystyle x < x + 2$
Help! I'm clearly doing something wrong here. :/
Ahh, thanks. I'll remember that.
I'm still having trouble simplifying it. I've done:
$\displaystyle -1 < |\frac{x - 1}{x + 1}| < 1$
$\displaystyle -(x + 1) < (x - 1) < (x + 1)$
$\displaystyle (-x - 1) < (x - 1) < (x + 1)$
$\displaystyle -x < x < x + 2$
$\displaystyle 0 < 2x < 2x + 2$
$\displaystyle 0 < x < x + 1$
So x > 0 (which is correct).
But am I doing this correctly? Or was it just a coincidence I got the answer?