The Linear Inequality problem is relatively simple:
Solve for $X$: $$\quad \frac{1}{X-2} < 0$$
Although the solution is clearly $X \in (-\infty, 2)$, I was curious to find a problem in my method: \begin{align*}\frac{1}{X-2} &< 0 \\ \frac{1 + X - 2}{X- 2} &< 1 &\text{[Adding 1 on Both Sides]} \\ X - 1 &< X -2 \\ -1 &< -2, \end{align*} which is false. Where is the error?
The error occurs when multiplying $X - 2$ to both sides. Remember, when multiplying both sides by a negative quantity, you must reverse the inequality. By failing to reverse the inequality, you are effectively assuming that $X - 2 > 0$, or in other words, $X > 2$. But, the solution to this inequality was $(-\infty, 2)$, which means that this assumption is always false, hence the absurd conclusion.
If you wanted to fix this method, there are two common ways to do this. First, you can separate into cases. What if $X < 2$, what if $X > 2$, and what if $X = 2$? Obviously the last case can't happen, as $X - 2$ appears in the denominator. If $X > 2$, then as you've proven, $-1 < -2$, which is absurd, hence $X > 2$ cannot be true either. Thus, there are no solutions such that $X \ge 2$, and the only possible case is $X < 2$.
In this case, we have $X - 2 < 0$, so $$\frac{1 + X - 2}{X - 2} < 1 \iff X - 1 > X - 2 \iff -1 > -2.$$ Note that this is always true, and logically equivalent to the original inequality, so under the overarching assumption that $X - 2 < 0$, there are no other restrictions. Hence, the solution is $X \in (-\infty, 2)$.
The second way to fix this is to multiply by $(X - 2)^2$. Since $X \neq 2$, this has the advantage of being strictly positive. Thus, \begin{align*} \frac{1 + X - 2}{X - 2} < 1 &\iff (X - 1)(X - 2) < (X - 2)^2 \\ &\iff (X - 2)^2 - (X - 1)(X - 2) > 0 \\ &\iff (X - 2)(X - 2 - (X - 1)) > 0 \\ &\iff (X - 2)(-1) > 0 \\ &\iff X - 2 < 0, \end{align*} since, again $-1 < 0$, and dividing both sides by a negative number flips signs.