Which point is a solution to the simultaneous inequalities y<1/2x

To determine which point is a solution to the simultaneous inequalities \( y < \frac{1}{2}x \) and \( y > \frac{1}{3}x + 3 \), we will substitute each point into the inequalities and check if both conditions are satisfied.

1. Point (2, 2):

   • For \( y < \frac{1}{2}x \):
     • \( 2 < \frac{1}{2}(2) \) ⇒ \( 2 < 1 \) (False)
   • This point does not satisfy the first inequality.

2. Point (25, 12):

   • For \( y < \frac{1}{2}x \):
     • \( 12 < \frac{1}{2}(25) \) ⇒ \( 12 < 12.5 \) (True)
   • For \( y > \frac{1}{3}x + 3 \):
     • \( 12 > \frac{1}{3}(25) + 3 \)
     • \( 12 > \frac{25}{3} + 3 \) ⇒ \( 12 > \frac{25}{3} + \frac{9}{3} \)
     • \( 12 > \frac{34}{3} \) ⇒ \( 12 > 11.33 \) (True)
   • This point satisfies both inequalities.

3. Point (3, 6):

   • For \( y < \frac{1}{2}x \):
     • \( 6 < \frac{1}{2}(3) \) ⇒ \( 6 < 1.5 \) (False)
   • This point does not satisfy the first inequality.

4. Point (10, 5):

   • For \( y < \frac{1}{2}x \):
     • \( 5 < \frac{1}{2}(10) \) ⇒ \( 5 < 5 \) (False)
   • This point does not satisfy the first inequality.

The only point that satisfies both inequalities is (25, 12).