Solve the inequality (2x + 1)/(x - 4) < 0.

Step-by-step solution

1. 1

   Understand what makes a quotient negative. A fraction is negative exactly when its numerator and denominator have opposite signs. So instead of cross-multiplying — which is invalid, because $x-4$ changes sign — track the sign of each part separately.

2. 2

   Find the critical points.

   $2x+1=0\Rightarrow x=-\frac12\quad(\text{numerator zero}),\qquad x-4=0\Rightarrow x=4\quad(\text{denominator zero})$

   The fraction can only change sign at these two values, so they split the line into $\left(-\infty,-\tfrac12\right)$, $\left(-\tfrac12,4\right)$ and $(4,\infty)$.

3. 3

   Test one point in each interval. At $x=-1$: $\dfrac{-1}{-5}=0.2>0$ (both negative). At $x=0$: $\dfrac{1}{-4}=-0.25<0$ (signs differ). At $x=5$: $\dfrac{11}{1}=11>0$ (both positive). Only the middle interval is negative.

4. 4

   Handle both endpoints, which are excluded for different reasons. At $x=-\tfrac12$ the fraction equals $0$, and the inequality is strict ($<0$, not $\le0$), so $0$ does not qualify. At $x=4$ the fraction is undefined. Both ends are therefore open.

5. 5

   State the solution.

   $-\frac12<x<4,\qquad\left(-\frac12,\ 4\right)$

   Unlike many rational inequalities, both brackets here happen to be open — but for two entirely different reasons, which is worth noticing.

6. 6

   Verify with exact arithmetic. Evaluating the fraction with fractions (not floats) at $12001$ rational points on $[-10,10]$, skipping $x=4$, the inequality holds at exactly the points of $\left(-\tfrac12,4\right)$ and nowhere else ✓.