Question

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)

Answer 1

Are you sure

Answer 2

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.

Answer 3

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!