Question

7D%2B1" id="TexFormula1" title="|\frac{x+1}{x-1}+1|\ \textgreater \ \frac{x+1}{x-1}+1" alt="|\frac{x+1}{x-1}+1|\ \textgreater \ \frac{x+1}{x-1}+1" align="absmiddle" class="latex-formula">

Answer 1

The inequality boils down to

|y| > y

By definition of absolute value, we have

• |y| = y if y ≥ 0

• |y| = -y if y < 0

So if y ≥ 0, we have

y > y

but this is a contradiction.

On the other hand, if y < 0, we have

-y > y   ==>   2y < 0   ==>   y < 0

and no contradiction.

Now replace y with (x + 1)/(x - 1) + 1. Then you're left with solving

(x + 1)/(x - 1) + 1 < 0

(x + 1 + x - 1)/(x - 1) < 0

2x/(x - 1) < 0

The left side is negative if either 2x > 0 and x - 1 < 0, or 2x < 0 and x - 1 > 0. The first case reduces to x > 0 and x < 1, or 0 < x < 1. In the second case, we get x < 0 and x > 1, but x cannot satisfy both conditions, so we throw this case out.