Problem 15 - Algebra for College Students

Question

Solve each inequality and graph its solution set on a number line. $$\frac{x-3}{x+2}<0$$

Step-by-Step Solution

Answer

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

Step 1: Identify Critical Points

To solve the inequality \( \frac{x-3}{x+2} < 0 \), we first find the values that make the numerator and denominator zero. Set \( x - 3 = 0 \) to find \( x = 3 \), and \( x + 2 = 0 \) to find \( x = -2 \). These are the critical points where the expression could change its sign.

Step 2: Analyze Intervals

The critical points divide the number line into intervals: \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \). We will test each interval to see where the expression \( \frac{x-3}{x+2} \) is less than zero.

Step 3: Test the Interval \((-\infty, -2)\)

Choose a test point from the interval \( (-\infty, -2) \), such as \( x = -3 \). Substituting into the inequality, we have:\[ \frac{-3 - 3}{-3 + 2} = \frac{-6}{-1} = 6 > 0 \]So, \( \frac{x-3}{x+2} \) is not less than zero in this interval.

Step 4: Test the Interval \((-2, 3)\)

Choose a test point from the interval \( (-2, 3) \), such as \( x = 0 \). Substituting into the inequality, we have:\[ \frac{0 - 3}{0 + 2} = \frac{-3}{2} < 0 \]Thus, \( \frac{x-3}{x+2} \) is less than zero in this interval.

Step 5: Test the Interval \((3, \infty)\)

Choose a test point from the interval \( (3, \infty) \), such as \( x = 4 \). Substituting into the inequality, we have:\[ \frac{4 - 3}{4 + 2} = \frac{1}{6} > 0 \]Thus, \( \frac{x-3}{x+2} \) is not less than zero in this interval.

Step 6: Identify Solution Set and Graph

The solution set of the inequality \( \frac{x-3}{x+2} < 0 \) is the interval \((-2, 3)\). This is because the expression is less than zero only in this interval. To graph this, draw a number line and shade the region between \(-2\) and \(3\), using open circles at both endpoints to indicate they are not included in the solution.

Key Concepts

Critical Points

Critical points are where the expression might change its behavior, due to the numerator or denominator equating to zero in the inequality. For example, in the inequality \( \frac{x-3}{x+2}<0 \), the critical points can be discovered by solving \( x-3=0 \) and \( x+2=0 \). This results in critical points at \( x=3 \) and \( x=-2 \). These values must be checked since they are potential places where the inequality transitions between being true and false. Remember, the expression is undefined at \( x=-2 \) because division by zero is not possible.

Number Line

A number line is a visual representation helpful for understanding which intervals satisfy an inequality. When you know the critical points, like \( x=-2 \) and \( x=3 \) from our example, you can divide the number line into segments between these points. For our inequality \( \frac{x-3}{x+2}<0 \), the number line is divided into three intervals:

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

By testing each of these intervals, you can determine in which sections the inequality holds true. The number line will display this clearly by shading the parts of the line where the inequality is satisfied.

Solution Set

The solution set is the collection of values for which the inequality holds true. Once you identify the intervals on the number line and perform interval testing (as explained in the upcoming section), the resulting intervals where the inequality is valid make up the solution set. For the inequality \( \frac{x-3}{x+2} < 0 \), the solution set is the interval \((-2, 3)\). These are the \( x \) values which satisfy the inequality. To show this solution set graphically on a number line, sketch an open interval between the critical points \( -2 \) and \( 3 \), using open circles to indicate that these end points are not included.

Interval Testing

Interval testing involves selecting a test point from each interval divided by your critical points and evaluating the inequality. For the intervals \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \), you pick a number from each, like \( x=-3 \), \( x=0 \), and \( x=4 \) respectively. Plug these into the inequality to check the validity:

• For \( x=-3 \), the inequality yields \( 6 > 0 \), so it doesn't satisfy \( \frac{x-3}{x+2}<0 \).
• For \( x=0 \), it returns \( -\frac{3}{2} < 0 \), satisfying the inequality.
• For \( x=4 \), it evaluates to \( \frac{1}{6} > 0 \), not satisfying the inequality.

From these evaluations, only the interval \( (-2, 3) \) makes \( \frac{x-3}{x+2} < 0 \) true, which forms your solution set.