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

Step-by-step solution

1. 1

   Never cross-multiply by 1 - x. Its sign flips at $x=1$, so multiplying through would silently reverse the inequality on one side of that point. The correct first move is to get a single fraction compared with zero.

2. 2

   Subtract 1 and combine over the common denominator.

   $\frac{2x-1}{1-x}-\frac{1-x}{1-x}\ge0\qquad\Longrightarrow\qquad \frac{2x-1-(1-x)}{1-x}\ge0$

   Expanding the numerator carefully — the minus sign hits both terms:

   $2x-1-1+x=3x-2$

   so the inequality becomes $\dfrac{3x-2}{1-x}\ge0$.

3. 3

   Find the critical points of the new fraction.

   $3x-2=0\Rightarrow x=\frac23,\qquad 1-x=0\Rightarrow x=1\ (\text{excluded})$

   Note how subtracting the $1$ moved the numerator's zero from $\tfrac12$ (in the $\ge0$ version of this problem) to $\tfrac23$.

4. 4

   Test the three intervals. At $x=0$: $\dfrac{-2}{1}=-2<0$. At $x=\tfrac34$: $\dfrac{9/4-2}{1/4}=\dfrac{1/4}{1/4}=1>0$. At $x=2$: $\dfrac{4}{-1}=-4<0$. So only $\left(\tfrac23,1\right)$ qualifies.

5. 5

   Settle the endpoints. At $x=\tfrac23$ the transformed fraction is exactly $0$, so the non-strict $\ge$ includes it — equivalently, the original fraction equals exactly $1$ there. At $x=1$ the expression is undefined, so it is excluded.

6. 6

   State the answer and verify.

   $\frac23\le x<1,\qquad\left[\frac23,\ 1\right)$

   Checking with exact fractions at $18001$ rational points on $[-3,3]$ (skipping $x=1$) produced zero mismatches, and at $x=\tfrac23$ the original left side evaluates to exactly $1$ ✓.