\( \begin{array}{l}\text { 9. Which point is a solution to the system } \\ \text { below? } \\ \qquad \)\begin{tabular}{l|l|} & \text { A) }\( (1,1 / 2) \) \\ \( y<-12 x+4 \) & \text { B) }\( (0,6) \) \\ \text { y<-6x+4} & \text { C) }\( (-1 / 2,5) \) \\ \text { D) }\( (-3,2) \)\end{tabular}\end{array} \( \begin{array}{l}\end{array} \)

To determine which point is a solution to the given system of inequalities, we need to check each point against both inequalities: 1. \( y < -12x + 4 \) 2. \( y < -6x + 4 \) We will evaluate each point one by one. ### Point A: \( (1, \frac{1}{2}) \) 1. Check the first inequality: \[ \frac{1}{2} < -12(1) + 4 \implies \frac{1}{2} < -12 + 4 \implies \frac{1}{2} < -8 \quad \text{(False)} \] 2. Check the second inequality: \[ \frac{1}{2} < -6(1) + 4 \implies \frac{1}{2} < -6 + 4 \implies \frac{1}{2} < -2 \quad \text{(False)} \] ### Point B: \( (0, 6) \) 1. Check the first inequality: \[ 6 < -12(0) + 4 \implies 6 < 0 + 4 \implies 6 < 4 \quad \text{(False)} \] 2. Check the second inequality: \[ 6 < -6(0) + 4 \implies 6 < 0 + 4 \implies 6 < 4 \quad \text{(False)} \] ### Point C: \( (-\frac{1}{2}, 5) \) 1. Check the first inequality: \[ 5 < -12(-\frac{1}{2}) + 4 \implies 5 < 6 + 4 \implies 5 < 10 \quad \text{(True)} \] 2. Check the second inequality: \[ 5 < -6(-\frac{1}{2}) + 4 \implies 5 < 3 + 4 \implies 5 < 7 \quad \text{(True)} \] ### Point D: \( (-3, 2) \) 1. Check the first inequality: \[ 2 < -12(-3) + 4 \implies 2 < 36 + 4 \implies 2 < 40 \quad \text{(True)} \] 2. Check the second inequality: \[ 2 < -6(-3) + 4 \implies 2 < 18 + 4 \implies 2 < 22 \quad \text{(True)} \] ### Conclusion Both Point C \( (-\frac{1}{2}, 5) \) and Point D \( (-3, 2) \) satisfy both inequalities. However, since the question asks for which point is a solution to the system, we can conclude that: - **Point C: \( (-\frac{1}{2}, 5) \)** is a solution. - **Point D: \( (-3, 2) \)** is also a solution. If only one point is to be selected, we can choose either Point C or Point D as a valid solution.