Solve the inequality 3/(x - 1) - 1 >= 2.

Step-by-step solution

1. 1

   Record the domain restriction first. The denominator $x-1$ cannot be zero, so $x \neq 1$. This value can never be part of the answer no matter what the algebra produces.

2. 2

   Resist cross-multiplying. Multiplying both sides by $x-1$ would require knowing its sign, and $x-1$ is negative for $x<1$ and positive for $x>1$. The safe route is to collect everything on one side and analyse a single quotient.

3. 3

   Move all terms to the left and combine. Adding $1$ to both sides gives $\dfrac{3}{x-1} \ge 3$, so $\frac{3}{x-1}-3 \ge 0 \Longrightarrow \frac{3-3(x-1)}{x-1} \ge 0 \Longrightarrow \frac{6-3x}{x-1} \ge 0.$

4. 4

   Factor out the negative and flip the inequality once, deliberately. $6-3x = -3(x-2)$, and dividing both sides by the constant $-3$ reverses the sign: $\frac{x-2}{x-1} \le 0.$ This is now a clean quotient of two linear factors.

5. 5

   Build the sign chart on the critical values 1 and 2. For $x=0$: $\tfrac{-2}{-1}=2>0$, excluded. For $x=1.5$: $\tfrac{-0.5}{0.5}=-1<0$, included. For $x=3$: $\tfrac{1}{2}>0$, excluded. So only the middle interval works.

6. 6

   Decide the endpoints. $x=2$ makes the numerator zero, and $\le 0$ allows zero, so $x=2$ is included. $x=1$ makes the denominator zero and is excluded by the domain. Hence $1 < x \le 2, \qquad x \in (1,2].$ Check: at $x=2$, $\tfrac{3}{1}-1=2$, which satisfies $\ge 2$ with equality.