To determine which system of linear equations the table represents, we need to find the equations for both \( Y_A \) and \( Y_B \) based on the provided values in the table.
Table of Values
X | YA | YB
0 | 0 | 3
1 | 2 | 3.5
2 | 4 | 4
For \( Y_A \):
- \( Y_A \) values are (0, 0), (1, 2), (2, 4).
- The pattern shows that \( Y_A \) increases by 2 when \( X \) increases by 1.
- Thus, we can see that: \[ Y_A = 2X \]
For \( Y_B \):
- \( Y_B \) values are (0, 3), (1, 3.5), (2, 4).
- Let's calculate the slope between the points:
- From (0, 3) to (1, 3.5): \[ \text{slope} = \frac{3.5 - 3}{1 - 0} = 0.5 \]
- From (1, 3.5) to (2, 4): \[ \text{slope} = \frac{4 - 3.5}{2 - 1} = 0.5 \]
- This indicates a slope of \( \frac{1}{2} \), so we can express \( Y_B \) as: \[ Y_B = \frac{1}{2}X + 3 \] (We use the point (0, 3) to find the y-intercept; when \( X = 0 \), \( Y_B = 3 \))
Summary of the equations:
- \( Y_A = 2X \)
- \( Y_B = \frac{1}{2}X + 3 \)
Matching with the provided options:
- \( Y = 2X \) \ \ and \ \ \( Y = 3X + \frac{1}{2} \) (Not a match)
- \( Y = 2X + 3 \) \ and \ \ \( Y = \frac{1}{2}X \) (Not a match)
- \( Y = 2X \) \ and \ \ \( Y = \frac{1}{2}X + 3 \) (This is a match!)
- \( Y = \frac{1}{2}X \) \ and \ \ \( Y = 2X + 3 \) (Not a match)
Conclusion:
The system of linear equations represented by the table of values is Option 3: \[ Y = 2X \quad \text{and} \quad Y = \frac{1}{2}X + 3 \]