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

Step-by-step solution

1. 1

   Do not multiply both sides by 2x + 1. Its sign flips at $x=-\tfrac12$, so multiplying through would reverse the inequality on one side of that point without warning. The correct opening move is to compare a single fraction with zero.

2. 2

   Subtract 1 and combine over the common denominator.

   $\frac{3-x}{2x+1}-\frac{2x+1}{2x+1}\ge0\qquad\Longrightarrow\qquad \frac{3-x-(2x+1)}{2x+1}\ge0$

   The subtraction hits both terms of $(2x+1)$:

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

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

3. 3

   Find the critical points.

   $2-3x=0\Rightarrow x=\frac23\quad(\text{numerator zero}),\qquad 2x+1=0\Rightarrow x=-\frac12\quad(\text{excluded})$

   These split the line into $\left(-\infty,-\tfrac12\right)$, $\left(-\tfrac12,\tfrac23\right)$ and $\left(\tfrac23,\infty\right)$.

4. 4

   Test one point per interval. At $x=-1$: $\dfrac{2+3}{-1}=-5<0$. At $x=0$: $\dfrac{2}{1}=2>0$. At $x=1$: $\dfrac{-1}{3}<0$. So only the middle interval qualifies — the fraction is positive exactly between the two critical points.

5. 5

   Decide each endpoint separately, since they behave differently. At $x=\tfrac23$ the numerator is $0$, so the fraction equals $0$ and $\ge0$ holds — include it. At $x=-\tfrac12$ the denominator is $0$, so the expression is undefined — exclude it. That asymmetry gives a half-open interval.

6. 6

   State the solution and verify.

   $-\frac12<x\le\frac23,\qquad\left(-\frac12,\ \frac23\right]$

   Using exact fractions at $18001$ rational test points on $[-3,3]$ (skipping $x=-\tfrac12$), the original inequality holds at exactly the points of this interval and nowhere else ✓.