Math Problem Statement

Which point is a solution to the simultaneous inequality y < 1/2 xand y > 1/3 x + 3? A. (10, 5) B. (3, 6) C. (2, 2) D. (25, 12)

Which point is a solution to the simultaneous inequality y < 1/2 x and y > 1/3 x + 3? A. (10, 5) B. (3, 6) C. (2, 2) D. (25, 12)

Solution

To determine which point is a solution to the simultaneous inequalities:

  1. y<12xy < \frac{1}{2}x
  2. y>13x+3y > \frac{1}{3}x + 3

we'll substitute each point into these inequalities and see if both conditions are satisfied.

Checking Points

Option A: (10, 5)

  1. y=5<12×10=5y = 5 < \frac{1}{2} \times 10 = 5 (False, since 55 is not less than 55)
  2. y=5>13×10+3=103+3=1936.33y = 5 > \frac{1}{3} \times 10 + 3 = \frac{10}{3} + 3 = \frac{19}{3} \approx 6.33 (False, since 55 is not greater than 6.336.33)

Conclusion: Not a solution.

Option B: (3, 6)

  1. y=6<12×3=1.5y = 6 < \frac{1}{2} \times 3 = 1.5 (False, since 66 is not less than 1.51.5)
  2. y=6>13×3+3=1+3=4y = 6 > \frac{1}{3} \times 3 + 3 = 1 + 3 = 4 (True)

Conclusion: Not a solution.

Option C: (2, 2)

  1. y=2<12×2=1y = 2 < \frac{1}{2} \times 2 = 1 (False, since 22 is not less than 11)
  2. y=2>13×2+3=23+3=1133.67y = 2 > \frac{1}{3} \times 2 + 3 = \frac{2}{3} + 3 = \frac{11}{3} \approx 3.67 (False)

Conclusion: Not a solution.

Option D: (25, 12)

  1. y=12<12×25=12.5y = 12 < \frac{1}{2} \times 25 = 12.5 (True)
  2. y=12>13×25+3=253+3=34311.33y = 12 > \frac{1}{3} \times 25 + 3 = \frac{25}{3} + 3 = \frac{34}{3} \approx 11.33 (True)

Conclusion: This point satisfies both inequalities.

Answer

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

Would you like further clarification or additional details?


Here are 5 related questions for deeper understanding:

  1. How do you graph linear inequalities on a coordinate plane?
  2. What is the process for finding the solution region for two inequalities?
  3. How can you check if a point lies in a shaded region on a graph?
  4. What does the boundary line of an inequality represent in a graph?
  5. How can simultaneous inequalities be used in real-world applications?

Tip: When checking inequalities, always evaluate each inequality independently to ensure each condition is satisfied.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Inequalities
Coordinate Geometry

Formulas

y < (1/2)x
y > (1/3)x + 3

Theorems

Linear Inequality Theorem

Suitable Grade Level

Grades 9-10