Use the image to answer the question.

Question

Which point is a solution to the simultaneous inequalities y<12x
𝑦
<
1
2
𝑥
and y>13x+3
𝑦
>
1
3
𝑥
+
3
?

(1 point)
Responses

(25,12)
(
25
,
12
)
left parenthesis 25 comma 12 right parenthesis

(2,2)
(
2
,
2
)
left parenthesis 2 comma 2 right parenthesis

(3,6)
(
3
,
6
)
left parenthesis 3 comma 6 right parenthesis

(10,5)

Answer 1

To determine which point is a solution to the simultaneous inequalities \( y < 12x \) and \( y > \frac{1}{3}x + 3 \), we need to evaluate each given point against both inequalities.

1. Point (25, 12):

   • For \( y < 12x \):
     • \( 12 < 12(25) \)
     • \( 12 < 300 \) (True)
   • For \( y > \frac{1}{3}x + 3 \):
     • \( 12 > \frac{1}{3}(25) + 3 \)
     • \( 12 > \frac{25}{3} + 3 = \frac{25}{3} + \frac{9}{3} = \frac{34}{3} \approx 11.33 \) (True)
   • Conclusion: (25, 12) satisfies both inequalities.

2. Point (2, 2):

   • For \( y < 12x \):
     • \( 2 < 12(2) \)
     • \( 2 < 24 \) (True)
   • For \( y > \frac{1}{3}x + 3 \):
     • \( 2 > \frac{1}{3}(2) + 3 \)
     • \( 2 > \frac{2}{3} + 3 = \frac{2}{3} + \frac{9}{3} = \frac{11}{3} \approx 3.67 \) (False)
   • Conclusion: (2, 2) does not satisfy both inequalities.

3. Point (3, 6):

   • For \( y < 12x \):
     • \( 6 < 12(3) \)
     • \( 6 < 36 \) (True)
   • For \( y > \frac{1}{3}x + 3 \):
     • \( 6 > \frac{1}{3}(3) + 3 \)
     • \( 6 > 1 + 3 = 4 \) (True)
   • Conclusion: (3, 6) satisfies both inequalities.

4. Point (10, 5):

   • For \( y < 12x \):
     • \( 5 < 12(10) \)
     • \( 5 < 120 \) (True)
   • For \( y > \frac{1}{3}x + 3 \):
     • \( 5 > \frac{1}{3}(10) + 3 \)
     • \( 5 > \frac{10}{3} + 3 = \frac{10}{3} + \frac{9}{3} = \frac{19}{3} \approx 6.33 \) (False)
   • Conclusion: (10, 5) does not satisfy both inequalities.

Among the points evaluated, (25, 12) and (3, 6) both satisfy the inequalities. Since (25, 12) seems higher and still valid, we'll consider it a solution.

So, a solution to the simultaneous inequalities is:

Response: (25, 12).