1. **State the problem:** Solve the system of inequalities for problem 7 parts a) and b). 2. **Problem 7a:** \[ \frac{9 - x^2}{x} \geq 0 \quad \text{and} \quad 2x - 1 \geq 0 \] 3. **Analyze the first inequality:** \[ \frac{9 - x^2}{x} \geq 0 \] The numerator is \(9 - x^2 = (3 - x)(3 + x)\). The denominator is \(x\). 4. **Find critical points:** Numerator zeros: \(x = 3, -3\) Denominator zero: \(x = 0\) (excluded from domain) 5. **Sign analysis intervals:** - \(( -\infty, -3 )\) - \(( -3, 0 )\) - \(( 0, 3 )\) - \(( 3, \infty )\) 6. **Test signs in each interval:** - For \(x < -3\), numerator \(9 - x^2 < 0\), denominator \(x < 0\), fraction positive. - For \(-3 < x < 0\), numerator \(9 - x^2 > 0\), denominator \(x < 0\), fraction negative. - For \(0 < x < 3\), numerator \(9 - x^2 > 0\), denominator \(x > 0\), fraction positive. - For \(x > 3\), numerator \(9 - x^2 < 0\), denominator \(x > 0\), fraction negative. 7. **Include points where fraction is zero:** At \(x = -3\) and \(x = 3\), numerator zero, fraction zero, satisfies \(\geq 0\). At \(x = 0\), undefined. 8. **Solution for first inequality:** \[ (-\infty, -3] \cup (0, 3] \] 9. **Second inequality:** \[ 2x - 1 \geq 0 \implies x \geq \frac{1}{2} \] 10. **Combine both inequalities:** Intersection of \( (-\infty, -3] \cup (0, 3] \) and \( [\frac{1}{2}, \infty) \) is \( [\frac{1}{2}, 3] \). --- 11. **Problem 7b:** \[ \frac{(x + 5)(x - 1)}{x} \geq 0 \quad \text{and} \quad 10x - 1 < 0 \] 12. **Analyze first inequality:** Zeros of numerator: \(x = -5, 1\) Zero of denominator: \(x = 0\) (excluded) 13. **Intervals for sign analysis:** \( (-\infty, -5), (-5, 0), (0, 1), (1, \infty) \) 14. **Test signs:** - For \(x < -5\): numerator positive (both factors negative), denominator negative, fraction negative. - For \(-5 < x < 0\): numerator negative (one factor positive, one negative), denominator negative, fraction positive. - For \(0 < x < 1\): numerator negative, denominator positive, fraction negative. - For \(x > 1\): numerator positive, denominator positive, fraction positive. 15. **Include zeros where fraction is zero:** At \(x = -5\) and \(x = 1\), numerator zero, fraction zero, satisfies \(\geq 0\). At \(x = 0\), undefined. 16. **Solution for first inequality:** \[ [-5, 0) \cup [1, \infty) \] 17. **Second inequality:** \[ 10x - 1 < 0 \implies x < \frac{1}{10} \] 18. **Combine both inequalities:** Intersection of \( [-5, 0) \cup [1, \infty) \) and \( (-\infty, \frac{1}{10}) \) is \( [-5, 0) \). --- 19. **Final answers:** - For 7a: \( x \in [\frac{1}{2}, 3] \) - For 7b: \( x \in [-5, 0) \) 20. **Summary:** We solved each inequality by finding critical points, testing signs, and combining with the second inequality in each system.