Questions: Which point is a solution to the simultaneous inequalities (y < frac12 x) and (y > frac13 x + 3)? (1 point) ((10,5)) ((2,2)) ((25,12)) ((3,6))

Solution Steps

To determine which point is a solution to the simultaneous inequalities \( y < \frac{1}{2} x \) and \( y > \frac{1}{3} x + 3 \), we need to check each given point to see if it satisfies both inequalities. For each point, substitute the \( x \) and \( y \) values into both inequalities and verify if both conditions are true.

We need to evaluate each point against the inequalities \( y < \frac{1}{2} x \) and \( y > \frac{1}{3} x + 3 \).

For the point \( (10, 5) \):

• Check \( 5 < \frac{1}{2} \cdot 10 \): \[ 5 < 5 \quad \text{(False)} \]
• Check \( 5 > \frac{1}{3} \cdot 10 + 3 \): \[ 5 > \frac{10}{3} + 3 = \frac{10}{3} + \frac{9}{3} = \frac{19}{3} \quad \text{(False)} \] This point does not satisfy both inequalities.

For the point \( (2, 2) \):

• Check \( 2 < \frac{1}{2} \cdot 2 \): \[ 2 < 1 \quad \text{(False)} \]
• Check \( 2 > \frac{1}{3} \cdot 2 + 3 \): \[ 2 > \frac{2}{3} + 3 = \frac{2}{3} + \frac{9}{3} = \frac{11}{3} \quad \text{(False)} \] This point does not satisfy both inequalities.

For the point \( (25, 12) \):

• Check \( 12 < \frac{1}{2} \cdot 25 \): \[ 12 < 12.5 \quad \text{(True)} \]
• Check \( 12 > \frac{1}{3} \cdot 25 + 3 \): \[ 12 > \frac{25}{3} + 3 = \frac{25}{3} + \frac{9}{3} = \frac{34}{3} \quad \text{(True)} \] This point satisfies both inequalities.

For the point \( (3, 6) \):

• Check \( 6 < \frac{1}{2} \cdot 3 \): \[ 6 < 1.5 \quad \text{(False)} \]
• Check \( 6 > \frac{1}{3} \cdot 3 + 3 \): \[ 6 > 1 + 3 = 4 \quad \text{(True)} \] This point does not satisfy both inequalities.

Final Answer