To solve the inequality $\frac{2x - 1}{x + 3} \geq 0$, we will follow a series of steps.

Step 1: Find the critical points

The critical points are values where the expression equals zero or is undefined. These occur where the numerator or denominator equals zero.

• Numerator: Set $2x - 1 = 0$ $2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}$
• Denominator: Set $x + 3 = 0$ $x = -3$ Thus, the critical points are $x = \frac{1}{2}$ and $x = -3$.

Step 2: Analyze the sign of the expression

We need to examine the sign of $\frac{2x - 1}{x + 3}$ in the intervals formed by the critical points. These intervals are:

1. $(-\infty, -3)$
2. $(-3, \frac{1}{2})$
3. $(\frac{1}{2}, \infty)$

We will test a point in each of these intervals to determine the sign of the expression.

Interval 1: $(-\infty, -3)$

Choose $x = -4$: $\frac{2(-4) - 1}{-4 + 3} = \frac{-8 - 1}{-1} = \frac{-9}{-1} = 9 \quad (\text{positive})$

Interval 2: $(-3, \frac{1}{2})$

Choose $x = 0$: $\frac{2(0) - 1}{0 + 3} = \frac{-1}{3} \quad (\text{negative})$

Interval 3: $(\frac{1}{2}, \infty)$

Choose $x = 1$: $\frac{2(1) - 1}{1 + 3} = \frac{2 - 1}{4} = \frac{1}{4} \quad (\text{positive})$

Step 3: Consider the equality and undefined points

• At $x = \frac{1}{2}$, the expression equals zero: $\frac{2x - 1}{x + 3} = 0$, so $x = \frac{1}{2}$ is part of the solution.
• At $x = -3$, the expression is undefined, so $x = -3$ is not part of the solution.

Step 4: Construct the solution

We are looking for where the expression is greater than or equal to zero. From our analysis:

• The expression is positive in the intervals $(-\infty, -3)$ and $(\frac{1}{2}, \infty)$.
• The expression equals zero at $x = \frac{1}{2}$.

Thus, the solution is: $x \in (-\infty, -3) \cup \left[ \frac{1}{2}, \infty \right)$ However, remember that $x = -3$ is excluded since the expression is undefined there.

Final Solution:

$x \in (-\infty, -3) \cup \left[ \frac{1}{2}, \infty \right)$

Would you like further clarification or a step-by-step breakdown of the solution? Here are some related questions you could explore:

1. How would the solution change if the inequality were strict, i.e., $\frac{2x - 1}{x + 3} > 0$?
2. What happens if we have a similar inequality with different constants in the numerator and denominator?
3. How do we approach solving rational inequalities with higher-degree polynomials?
4. Can the sign analysis method be used for all rational inequalities?
5. How can we graph the inequality $\frac{2x - 1}{x + 3} \geq 0$ to visually verify the solution?

Tip: For rational inequalities, always check where the expression is undefined (denominator zero) and where the numerator equals zero to correctly identify the critical points.