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Select the points that are part of the solution set for the inequality y § 3a + 1.

Select ALL that apply. (2 points)
(5,3)
(10,5.5)
(10,4)
(10,6)

3 answers

To determine which points are part of the solution set for the inequality \( y \leq 3x + 1 \), we will test each point by substituting the \( x \) and \( y \) values into the inequality.

  1. Point (5, 3):

    • Substitute \( x = 5 \) and \( y = 3 \): \[ 3 \leq 3(5) + 1 \ 3 \leq 15 + 1 \ 3 \leq 16 \quad \text{(True)} \]

  2. Point (10, 5.5):

    • Substitute \( x = 10 \) and \( y = 5.5 \): \[ 5.5 \leq 3(10) + 1 \ 5.5 \leq 30 + 1 \ 5.5 \leq 31 \quad \text{(True)} \]

  3. Point (10, 4):

    • Substitute \( x = 10 \) and \( y = 4 \): \[ 4 \leq 3(10) + 1 \ 4 \leq 30 + 1 \ 4 \leq 31 \quad \text{(True)} \]

  4. Point (10, 6):

    • Substitute \( x = 10 \) and \( y = 6 \): \[ 6 \leq 3(10) + 1 \ 6 \leq 30 + 1 \ 6 \leq 31 \quad \text{(True)} \]

Since all four points satisfy the inequality \( y \leq 3x + 1 \):

The solution set includes the following points:

  • (5, 3)
  • (10, 5.5)
  • (10, 4)
  • (10, 6)

Therefore, all points are part of the solution set.

Are you sure

Let’s clarify the situation regarding the inequality \( y \leq 3x + 1 \) and re-evaluate all the points one by one.

The inequality means the \( y \) value must be less than or equal to \( 3x + 1 \). We can rewrite the inequality in standard form for clarity:

  1. For point (5, 3): \[ y = 3 \quad \text{and} \quad x = 5 \ 3 \leq 3(5) + 1 \ 3 \leq 15 + 1 \ 3 \leq 16 \quad \text{(True)} \]

  2. For point (10, 5.5): \[ y = 5.5 \quad \text{and} \quad x = 10 \ 5.5 \leq 3(10) + 1 \ 5.5 \leq 30 + 1 \ 5.5 \leq 31 \quad \text{(True)} \]

  3. For point (10, 4): \[ y = 4 \quad \text{and} \quad x = 10 \ 4 \leq 3(10) + 1 \ 4 \leq 30 + 1 \ 4 \leq 31 \quad \text{(True)} \]

  4. For point (10, 6): \[ y = 6 \quad \text{and} \quad x = 10 \ 6 \leq 3(10) + 1 \ 6 \leq 30 + 1 \ 6 \leq 31 \quad \text{(True)} \]

Upon reviewing the calculations again, it appears that all points satisfy the condition illustrated by the inequality \( y \leq 3x + 1 \).

To summarize:

  • (5, 3): True
  • (10, 5.5): True
  • (10, 4): True
  • (10, 6): True

Thus, all four points are indeed part of the solution set for the inequality \( y \leq 3x + 1 \). If you have any specific reason to doubt this answer, or if you would like more information, feel free to ask!