To determine which system of linear equations the table of values represents, let's analyze the values in the table for \( y_a \) and \( y_b \):
For \( y_a \)
Looking at the values for \( y_a \):
- When \( x = 0 \), \( y_a = 0 \)
- When \( x = 1 \), \( y_a = 2 \)
- When \( x = 2 \), \( y_a = 4 \)
This corresponds to the equation \( y_a = 2x \) (it passes through the origin and has a slope of 2).
For \( y_b \)
Now for \( y_b \):
- When \( x = 0 \), \( y_b = 3 \)
- When \( x = 1 \), \( y_b = 3.5 \)
- When \( x = 2 \), \( y_b = 4 \)
To find the equation, we can observe the differences. The difference between \( y_b \) values from \( x = 0 \) to \( x = 1 \) is \( 0.5 \) (3.5 - 3) and from \( x = 1 \) to \( x = 2 \) is also \( 0.5 \). This indicates a constant slope of \( 0.5 \).
The slope of the line for \( y_b \) can be calculated as:
- Slope = (Change in \( y \)) / (Change in \( x \)) = \( (4 - 3) / (2 - 0) = 1/2 \)
The y-intercept is 3, so the equation for \( y_b \) can be written as: \[ y_b = \frac{1}{2}x + 3 \]
Combining the Equations
Now we have the following equations:
- For \( y_a \): \( y_a = 2x \)
- For \( y_b \): \( y_b = \frac{1}{2}x + 3 \)
Choosing the System
Looking at the given systems:
- {y = 2x, y = 12x}
- {y = \frac{1}{2}x, y = 2x + 3}
- {y = 2x, y = 3x + \frac{1}{2}}
- {y = 2x, y = 12x + 3}
The correct choice based on our calculations would be: {y = 2x, y = \frac{1}{2}x + 3}
So, the system that represents the table of values is not listed, as the answer choices you provided do not contain the linear equations we derived. However, the equations representing the data from the table are \( y_a = 2x \) and \( y_b = \frac{1}{2}x + 3 \).