Questions: Use the value method to solve the rational inequality. Write the solution set in interval notation. (x-7)/(x+8) ≥ 3

Solution Steps

To solve the rational inequality \(\frac{x-7}{x+8} \geq 3\), we first need to bring all terms to one side of the inequality to form a single rational expression. Then, we find the critical points by setting the numerator and denominator to zero. These critical points will help us determine the intervals to test for the inequality. Finally, we test each interval to see where the inequality holds true and express the solution in interval notation.

To solve the rational inequality \(\frac{x-7}{x+8} \geq 3\), we will follow a series of steps to find the solution set and express it in interval notation.

First, we move all terms to one side of the inequality to set it to zero: \[ \frac{x-7}{x+8} - 3 \geq 0 \]

To combine the terms into a single fraction, we need a common denominator: \[ \frac{x-7}{x+8} - \frac{3(x+8)}{x+8} \geq 0 \] \[ \frac{x-7 - 3(x+8)}{x+8} \geq 0 \]

Simplify the expression in the numerator: \[ x - 7 - 3x - 24 = -2x - 31 \] Thus, the inequality becomes: \[ \frac{-2x - 31}{x+8} \geq 0 \]

The critical points occur where the numerator or the denominator is zero:

1. Numerator: \(-2x - 31 = 0\) \[ -2x = 31 \quad \Rightarrow \quad x = -\frac{31}{2} \]
2. Denominator: \(x + 8 = 0\) \[ x = -8 \]

The critical points divide the number line into intervals. We test each interval to determine where the inequality holds:

• Interval \((-\infty, -8)\)
• Interval \((-8, -\frac{31}{2})\)
• Interval \((- \frac{31}{2}, \infty)\)

Choose test points from each interval:

1. For \(x = -9\) in \((-\infty, -8)\): \[ \frac{-2(-9) - 31}{-9 + 8} = \frac{18 - 31}{-1} = \frac{-13}{-1} = 13 \quad (\text{positive}) \]
2. For \(x = -7\) in \((-8, -\frac{31}{2})\): \[ \frac{-2(-7) - 31}{-7 + 8} = \frac{14 - 31}{1} = -17 \quad (\text{negative}) \]
3. For \(x = 0\) in \((- \frac{31}{2}, \infty)\): \[ \frac{-2(0) - 31}{0 + 8} = \frac{-31}{8} \quad (\text{negative}) \]

The inequality \(\frac{-2x - 31}{x+8} \geq 0\) holds in the interval \((-\infty, -8]\) because the test point in this interval resulted in a positive value. The critical point \(x = -8\) is included because the inequality is non-strict (\(\geq\)).

Final Answer