Problem 54 - Intermediate Algebra

Question

Solve each inequality. $$ \frac{2 x-1}{x+2} \geq-1 $$

Step-by-Step Solution

Answer

The solution is \((-\infty, -2) \cup [-\frac{1}{3}, \infty)\)."

Step 1: Understand the Inequality

The given inequality is \( \frac{2x - 1}{x + 2} \geq -1 \). This is a rational inequality, meaning that it involves a fraction with a variable in the denominator.

Step 2: Move Constant to One Side

To simplify the inequality, move the \(-1\) to the other side by adding \(1\) to both sides: \[ \frac{2x - 1}{x + 2} + 1 \geq 0 \]

Step 3: Combine Terms into a Single Fraction

Turn the expression into a single fraction: \( \frac{2x - 1}{x + 2} + \frac{x + 2}{x + 2} \geq 0 \). Simplifying, we have: \[ \frac{2x - 1 + x + 2}{x + 2} \geq 0 \] which further reduces to \( \frac{3x + 1}{x + 2} \geq 0 \).

Step 4: Find Critical Points

The critical points occur where the numerator or denominator is zero. Set \(3x + 1 = 0\) to find the root: \(x = -\frac{1}{3}\). Set \(x + 2 = 0\): \(x = -2\). These critical points split the number line into intervals.

Step 5: Test the Intervals

Test values from each interval to determine where \( \frac{3x + 1}{x + 2} \geq 0 \). Use \(x = -3\) for \((-\infty, -2)\), \(x = -1\) for \((-2, -\frac{1}{3})\), and \(x = 0\) for \((-\frac{1}{3}, \infty)\). Calculate: - For \(x = -3\), \(\frac{3(-3) + 1}{-3 + 2} = \frac{-9 + 1}{-1} = 8\). Positive.- For \(x = -1\), \(\frac{3(-1) + 1}{-1 + 2} = \frac{-3 + 1}{1} = -2\). Negative.- For \(x = 0\), \(\frac{3(0) + 1}{0 + 2} = \frac{1}{2}\). Positive.

Step 6: Evaluate Critical Points

Check critical points \(x = -\frac{1}{3}\) and \(x = -2\). Plug these into \( \frac{3x + 1}{x + 2}\). At \(x = -\frac{1}{3}\), the numerator is zero, so the inequality holds as \(\geq\). At \(x = -2\), the expression is undefined.

Step 7: Determine Solution Interval

From the above tests, the intervals satisfying the inequality are \((-\infty, -2)\) and \([-\frac{1}{3}, \infty)\). Exclude \(-2\) since it's undefined. Thus, the solution is \((-\infty, -2) \cup [-\frac{1}{3}, \infty)\).

Key Concepts

Rational Inequalities

A rational inequality involves a fraction in which both the numerator and the denominator are polynomials. When you encounter an inequality such as \( \frac{2x - 1}{x + 2} \geq -1 \), you're dealing with a rational inequality. This can be challenging because of the need to handle variables in the denominator.

To solve rational inequalities:

• First, aim to rewrite the inequality in a "greater than or equal to zero" format by moving all terms to one side.
• For instance, in our exercise, we transform: \( \frac{2x - 1}{x + 2} \geq -1 \) to \( \frac{3x + 1}{x + 2} \geq 0 \).

Rational inequalities often require manipulation into a single fraction to facilitate further analysis.

Critical Points in Inequalities

Critical points occur where the fraction's numerator or denominator is zero. They break the number line into different regions that need to be tested separately.

To find these points in our example:

• Set the numerator \(3x + 1 = 0\) giving \(x = -\frac{1}{3}\).
• Set the denominator \(x + 2 = 0\) resulting in \(x = -2\).

These points are crucial because they help identify changes in the sign of the inequality, indicating potential shifts between valid and invalid inequality solutions.

Inequality Solution Intervals

To obtain a solution for a rational inequality, you need intervals on the number line where the inequality holds true. Once you've identified the critical points:

• Divvy up the line into intervals based on those points:
  • \(( -\infty, -2 )\)
  • \((-2, -\frac{1}{3})\)
  • \(\left[-\frac{1}{3}, \infty \right)\)
• Test values within each interval to see if they satisfy the inequality.

We found:

• \(x = -3\) validated in \(( -\infty, -2 )\)
• \(x = 0\) validated in \(\left[-\frac{1}{3}, \infty \right)\)

Ultimately, you exclude any undefined critical points like \(x = -2\). This results in the solution: \(( -\infty, -2 )\cup [-\frac{1}{3}, \infty)\), which is where the inequality is true.