a) \( \frac{x^{2}-9}{x-3} \geqslant 0 \)

To solve the inequality \( \frac{x^{2}-9}{x-3} \geqslant 0 \), we first factor the numerator. The expression \( x^2 - 9 \) can be factored as: \[ x^2 - 9 = (x - 3)(x + 3) \] Thus, we can rewrite the inequality as: \[ \frac{(x - 3)(x + 3)}{x - 3} \geqslant 0 \] For \( x \neq 3 \), we can simplify the expression: \[ x + 3 \geqslant 0 \] This simplifies our inequality to: \[ x + 3 \geqslant 0 \] Now, we solve for \( x \): \[ x \geqslant -3 \] However, we must consider the point where the original expression is undefined, which is at \( x = 3 \). Therefore, we need to check the sign of the expression around the critical points \( x = -3 \) and \( x = 3 \). ### Test Intervals 1. **Interval \( (-\infty, -3) \)**: - Choose \( x = -4 \): \[ \frac{(-4 - 3)(-4 + 3)}{-4 - 3} = \frac{(-7)(-1)}{-7} = \frac{7}{-7} = -1 \quad (\text{negative}) \] 2. **Interval \( (-3, 3) \)**: - Choose \( x = 0 \): \[ \frac{(0 - 3)(0 + 3)}{0 - 3} = \frac{(-3)(3)}{-3} = \frac{-9}{-3} = 3 \quad (\text{positive}) \] 3. **Interval \( (3, \infty) \)**: - Choose \( x = 4 \): \[ \frac{(4 - 3)(4 + 3)}{4 - 3} = \frac{(1)(7)}{1} = 7 \quad (\text{positive}) \] ### Summary of Signs - \( (-\infty, -3) \): Negative - \( (-3, 3) \): Positive - \( (3, \infty) \): Positive ### Critical Points - At \( x = -3 \), the expression equals zero. - At \( x = 3 \), the expression is undefined. ### Conclusion The solution to the inequality \( \frac{x^{2}-9}{x-3} \geqslant 0 \) is: \[ [-3, 3) \cup (3, \infty) \] This means \( x \) can take any value from \(-3\) to \(3\) (including \(-3\) but not \(3\)), or any value greater than \(3\).