11) Solve the inequality \( \frac{x-7}{x+8} \geq 0 \)

To solve the inequality \( \frac{x-7}{x+8} \geq 0 \), we first determine the critical points where the expression is equal to zero or undefined. The numerator \( x - 7 = 0 \) gives us the critical point \( x = 7 \), while the denominator \( x + 8 = 0 \) gives \( x = -8 \). These points divide the number line into intervals: \( (-\infty, -8) \), \( (-8, 7) \), and \( (7, \infty) \). Next, we test each interval to find where the expression is non-negative: - For \( x < -8 \) (e.g., \( x = -9 \)): \( \frac{-9 - 7}{-9 + 8} = \frac{-16}{-1} = 16 \geq 0\) (True) - For \( -8 < x < 7 \) (e.g., \( x = 0 \)): \( \frac{0 - 7}{0 + 8} = \frac{-7}{8} < 0\) (False) - For \( x > 7 \) (e.g., \( x = 8 \)): \( \frac{8 - 7}{8 + 8} = \frac{1}{16} > 0\) (True) Now, we consider the sign of the original expression at the critical points: - At \( x = -8 \), the expression is undefined. - At \( x = 7 \), the expression is zero. Thus, the solution to the inequality is \( (-\infty, -8) \cup [7, \infty) \).