Problem 41 - Intermediate Algebra

Question

Solve each inequality. $$ \frac{x-3}{x+2}<0 $$

Step-by-Step Solution

Answer

The solution is \((-2, 3)\).

Step 1: Identify Critical Points

The inequality \( \frac{x-3}{x+2}<0 \) can change signs at points where the numerator is zero (\(x = 3\)) or the denominator is zero (\(x = -2\)). These are the critical points.

Step 2: Set Up Intervals

Divide the number line into intervals based on the critical points \(x = -2\) and \(x = 3\): \((-\infty, -2)\), \((-2, 3)\), and \((3, \infty)\).

Step 3: Test Intervals

Choose a test point from each interval to determine where the inequality \( \frac{x-3}{x+2} < 0 \) holds. Use \(x = -3\) for \((-\infty, -2)\), \(x = 0\) for \((-2, 3)\), and \(x = 4\) for \((3, \infty)\).- For \(x = -3\): \( \frac{-3-3}{-3+2} = \frac{-6}{-1} = 6 \) (positive, does not satisfy)- For \(x = 0\): \( \frac{0-3}{0+2} = \frac{-3}{2} \) (negative, satisfies)- For \(x = 4\): \( \frac{4-3}{4+2} = \frac{1}{6} \) (positive, does not satisfy)

Step 4: Conclusion of Solution

The inequality holds in the interval where the test point produces a negative result. Therefore, \( \frac{x-3}{x+2} < 0 \) holds in the interval \((-2, 3)\). Ensure \(x eq -2\) as it makes the expression undefined.

Key Concepts

Understanding Critical Points

Critical points play a key role in solving rational inequalities. They are where the expression under consideration could potentially change signs, affecting the solution to the inequality.

For a rational inequality like \( \frac{x-3}{x+2} < 0 \), critical points emerge from two main components:

• The numerator, \( x-3 \), becomes zero when \( x = 3 \).
• The denominator, \( x+2 \), is zero when \( x = -2 \), which is crucial because division by zero is undefined.

These critical points, \( x = 3 \) and \( x = -2 \), split the number line into different segments where the inequality can be analyzed separately. Understanding these points sets the foundation for solving rational inequalities effectively.

Using the Number Line

The number line is an essential tool in determining where a rational inequality holds true. Once you have identified the critical points (in our example, \( x = -2 \) and \( x = 3 \)), place them on the number line to divide it into distinct intervals.

For \( \frac{x-3}{x+2} < 0 \), the critical points divide the number line into these segments:

• \((-\infty, -2)\)
• \((-2, 3)\)
• \((3, \infty)\)

Visualizing these intervals on a number line helps in the next step, which involves testing intervals to determine where the inequality holds true. By clearly marking these intervals, you ensure each possible case is examined for correctness without overstepping undefined regions.

Conducting Interval Testing

Interval testing is the process of determining which segments of the number line satisfy the inequality. Choose a test point from each interval to see if it satisfies the inequality \( \frac{x-3}{x+2} < 0 \).

Here's how you do it:

• For \((-\infty, -2)\), use \( x = -3 \). This results in a positive value, so it doesn’t satisfy the inequality.
• For \((-2, 3)\), use \( x = 0 \). This results in a negative value, satisfying the inequality.
• For \((3, \infty)\), use \( x = 4 \). This results in a positive value, again not satisfying the inequality.

The interval which produces a negative outcome indicates where the inequality is valid. It’s important to select test points that are easy to calculate with while ensuring they fall within each interval.

Working with Rational Inequalities

Rational inequalities involve ratios of polynomials and often require special steps to solve. It's crucial to handle the equation with care, particularly focusing on the rational expression's sign changes at critical points.

When solving \( \frac{x-3}{x+2} < 0 \), you should remember:

• Identify the critical points by setting the numerator and denominator to zero separately.
• Use a number line and interval testing to explore where the inequality holds true.
• Check and eliminate intervals where the inequality doesn’t meet the condition, such as undefined regions.

Keep in mind rational inequalities might have restrictions on the values of \( x \), particularly where the denominator is zero. Carefully validate each interval to ensure accurate solutions.