1. ## Basic Inequality question.

$\displaystyle \frac{2}{x-1} > 3$

Now I know the answer is 1 < X < 5/3 but I'm trying to work through it.

I can work it out for 5/3 but I'm not sure how I'm meant to solve for 1.

$\displaystyle 2 > 3x - 3$

$\displaystyle 5 > 3x$

$\displaystyle \frac{5}{3} > x$

Just got a mind blank as to how I'm meant to solve it to get 1 as the other case.

2. Originally Posted by Peleus
$\displaystyle \frac{2}{x-1} > 3$

Now I know the answer is 1 < X < 5/3 but I'm trying to work through it.

I can work it out for 5/3 but I'm not sure how I'm meant to solve for 1.

$\displaystyle 2 > 3x - 3$

$\displaystyle 5 > 3x$

$\displaystyle \frac{5}{3} > x$

Just got a mind blank as to how I'm meant to solve it to get 1 as the other case.
What you've done is fine. The only observation you have to make is that the denominator in your original expression MUST be greater than 0, or else the LHS will be negative. And obviously, the inequality can't be true if the LHS is negative.

$\displaystyle \frac{2}{x-1} > 3$

Multiply through by $\displaystyle x-1$

$\displaystyle 2x - 2 < 3x - 3$

$\displaystyle 1 < x$