Solve the inequality 3/(x - 2) > 1 and write the solution in interval notation.

Step-by-step solution

1. 1

   Do not multiply both sides by $x-2$. Its sign is unknown: for $x>2$ it is positive (inequality direction preserved) and for $x<2$ it is negative (direction reversed). Multiplying without splitting into cases is the single most common way to get this problem wrong — it silently produces $3>x-2$, i.e. $x<5$, which wrongly includes everything below $2$.

2. 2

   Move everything to one side instead. Subtract 1 and combine over the common denominator $x-2$:

   $\frac{3}{x-2}-1>0\ \Longrightarrow\ \frac{3-(x-2)}{x-2}>0\ \Longrightarrow\ \frac{5-x}{x-2}>0$

   Now the question is purely about the sign of one quotient, and no unknown-sign multiplication was performed.

3. 3

   List the critical points, distinguishing zeros from the undefined point. The numerator vanishes at $5-x=0\Rightarrow x=5$; the denominator vanishes at $x=2$, where the expression is undefined. Both split the line, but only one can ever be part of a solution set.

4. 4

   Test a value in each of the three intervals.

   $x=0:\ \frac{5}{-2}<0\ \times\qquad x=3:\ \frac{2}{1}>0\ \checkmark\qquad x=6:\ \frac{-1}{4}<0\ \times$

   Only the middle interval works.

5. 5

   Assemble the solution with the right endpoints. $x=2$ is excluded because the expression is undefined there, and $x=5$ is excluded because the quotient equals $0$ and the inequality is strict:

   $2<x<5,\qquad (2,5)$

6. 6

   Confirm at the ends. At $x=2.5$: $\dfrac{3}{0.5}=6>1$ ✓. At $x=5$: $\dfrac{3}{3}=1$, not greater than $1$, so $5$ is correctly excluded ✓. At $x=1$: $\dfrac{3}{-1}=-3>1$ is false — exactly the region the careless cross-multiplication would have wrongly admitted.