Problem 59 - A Graphical Approach to College Algebra

Question

Solve each rational inequality by hand. Do not use a calculator. $$\frac{3-2 x}{1+x}<0$$

Step-by-Step Solution

Answer

The solution is \( x \in (-\infty, -1) \cup \left(\frac{3}{2}, \infty\right) \).

Step 1: Identify the Critical Points

To find critical points of the rational inequality \( \frac{3-2x}{1+x}<0 \), set the numerator and the denominator equal to zero separately. For the numerator \( 3 - 2x = 0 \), solve to get \( x = \frac{3}{2} \). For the denominator \( 1 + x = 0 \), solve to get \( x = -1 \). These critical points \( x = \frac{3}{2} \) and \( x = -1 \) will help us divide the number line into intervals.

Step 2: Determine Intervals

Use the critical points to divide the number line into intervals: \((-\infty, -1)\), \((-1, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\). These intervals will be tested to see where the inequality \( \frac{3-2x}{1+x} < 0 \) holds.

Step 3: Test Each Interval

Select a test point from each interval and substitute it into \( \frac{3-2x}{1+x} \) to determine the sign of the expression in that interval.- Test point \( x = -2 \) in \((-\infty, -1)\): \( \frac{3 - 2(-2)}{1 + (-2)} = \frac{3 + 4}{-1} = -7 \), which is negative.- Test point \( x = 0 \) in \((-1, \frac{3}{2})\): \( \frac{3 - 2(0)}{1 + 0} = \frac{3}{1} = 3 \), which is positive.- Test point \( x = 2 \) in \((\frac{3}{2}, \infty)\): \( \frac{3 - 2(2)}{1 + 2} = \frac{3 - 4}{3} = -\frac{1}{3} \), which is negative.

Step 4: Conclusion from Test Results

From the test results, the inequality \( \frac{3-2x}{1+x} < 0 \) holds in the intervals \((-\infty, -1)\) and \((\frac{3}{2}, \infty)\). Since the denominator cannot be zero, the point \( x = -1 \) is not included in the solution set, and the numerator does not affect inclusion at \( x = \frac{3}{2} \) as it would make the inequality zero rather than negative. Thus, the solution is \( x \in (-\infty, -1) \cup \left(\frac{3}{2}, \infty\right) \).

Key Concepts

Critical Points

In solving rational inequalities like \( \frac{3-2x}{1+x}<0 \), identifying critical points serves as the foundation. Critical points are values of \( x \) that make the numerator or denominator zero. They help in breaking the inequality into manageable sections.
For our specific inequality, we have:

• Numerator Critical Point: Solve \( 3 - 2x = 0 \) to find \( x = \frac{3}{2} \).
• Denominator Critical Point: Solve \( 1 + x = 0 \) to find \( x = -1 \).

These critical points classify the problem into separate intervals for further analysis. It is important to remember that dividing by zero is undefined, hence denominators should never be zero in viable solutions.

Interval Testing

Interval testing involves using the critical points to divide the number line into distinct intervals. Each interval is then tested to determine where the inequality holds true.
In our scenario, the points \( x = -1 \) and \( x = \frac{3}{2} \) divide the number line into:

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

For each interval, choose a test point and substitute it back into the inequality \( \frac{3-2x}{1+x} \). Analyze whether the resulting value is positive or negative. This determines if the interval satisfies the inequality condition. When testing \(-2\) in \(( -\infty, -1)\), \(0\) in \((-1, \frac{3}{2})\), and \(2\) in \((\frac{3}{2}, \infty)\), we see different behaviors in each distinct section. This systematic testing is crucial in finding the intervals where the inequality is valid.

Solution Set

The solution set represents the range of \( x \) values that satisfy the inequality. Based on interval testing, we compile our observations into a complete understanding of where the inequality holds true.
Upon testing, the inequality \( \frac{3-2x}{1+x} < 0 \) was found to be negative (satisfying the original inequality condition) in the intervals \(( -\infty, -1)\) and \(( \frac{3}{2}, \infty )\).
It is important to note:

• The point \( x = -1 \) is excluded due to undefined division by zero.
• The point \( x = \frac{3}{2} \) is also excluded since it results in zero, not a negative value.

Thus, the solution set is \( x \in ( -\infty, -1) \cup ( \frac{3}{2}, \infty ) \). Each interval distinctly represents where the inequality condition is true, providing a clear picture of the solution geography.