Solve for the solution set of the following inequalities using case method. (x+3)/(2x -1) <0

To solve the inequality \(\frac{x + 3}{2x - 1} < 0\) using the case method, we'll analyze the sign of the expression based on the values of \(x\). The fraction will be negative when the numerator and denominator have opposite signs. We'll consider different cases based on the signs of the numerator and denominator.

### Step 1: Identify the critical points
First, find the values of \(x\) that make the numerator or denominator equal to zero:
- Numerator: \(x + 3 = 0 \Rightarrow x = -3\)
- Denominator: \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)

These points divide the number line into intervals that we can test for the inequality.

### Step 2: Determine the sign of the expression in each interval
We consider the following intervals based on the critical points:
1. \(x < -3\)
2. \(-3 < x < \frac{1}{2}\)
3. \(x > \frac{1}{2}\)

#### Case 1: \(x < -3\)
- The numerator \(x + 3\) is negative.
- The denominator \(2x - 1\) is negative.

Since both numerator and denominator are negative, their quotient is positive:
\[
\frac{x + 3}{2x - 1} > 0
\]
This case does not satisfy the inequality.

#### Case 2: \(-3 < x < \frac{1}{2}\)
- The numerator \(x + 3\) is positive.
- The denominator \(2x - 1\) is negative.

Since the numerator is positive and the denominator is negative, their quotient is negative:
\[
\frac{x + 3}{2x - 1} < 0
\]
This case satisfies the inequality.

#### Case 3: \(x > \frac{1}{2}\)
- The numerator \(x + 3\) is positive.
- The denominator \(2x - 1\) is positive.

Since both numerator and denominator are positive, their quotient is positive:
\[
\frac{x + 3}{2x - 1} > 0
\]
This case does not satisfy the inequality.

### Step 3: Combine the solution set
From the analysis above, the inequality \(\frac{x + 3}{2x - 1} < 0\) is satisfied in the interval:
\[
\boxed{(-3, \frac{1}{2})}
\]

### Summary
The solution set is \((-3, \frac{1}{2})\), meaning \(x\) can take any value in this interval.

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Would you like to go over any details, or do you have any questions? Here are five related questions to consider:

1. How would the solution change if the inequality was \(\frac{x + 3}{2x - 1} > 0\)?
2. What happens if \(x = -3\) or \(x = \frac{1}{2}\)?
3. How would you solve \(\frac{x + 3}{2x - 1} \leq 0\) using the case method?
4. What if the inequality was \(\frac{x + 3}{2x - 1} < -1\)?
5. How can the solution be verified using a graphical method?

**Tip:** Always check for critical points where the numerator or denominator equals zero; they are key to determining the sign changes in an inequality.

Learn how to solve the rational inequality (x+3)/(2x-1) < 0 using the case method. Understand the critical points, analyze sign changes, and determine the solution set. Suitable for advanced high school students or college-level algebra.

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