To determine which system of linear equations the table of values represents, we can analyze the values of \( y_a \) and \( y_b \) based on the given \( x \) values.
From the table:
For \( y_a \):
- When \( x = 0 \), \( y_a = 0 \)
- When \( x = 1 \), \( y_a = 2 \)
- When \( x = 2 \), \( y_a = 4 \)
We can see that \( y_a \) appears to be following a linear relationship with \( x \). We can express this as: \[ y_a = 2x \] This is confirmed by substituting the \( x \) values into the equation:
- For \( x = 0 \): \( 2(0) = 0 \)
- For \( x = 1 \): \( 2(1) = 2 \)
- For \( x = 2 \): \( 2(2) = 4 \)
For \( y_b \):
- When \( x = 0 \), \( y_b = 3 \)
- When \( x = 1 \), \( y_b = 3.5 \)
- When \( x = 2 \), \( y_b = 4 \)
The change in \( y_b \) as \( x \) increases suggests a constant growth. We calculate the slope between points: From \( (0, 3) \) to \( (2, 4) \): \[ \text{slope} = \frac{4 - 3}{2 - 0} = \frac{1}{2} \] This means the equation for \( y_b \) can be represented in the slope-intercept form: \[ y_b = \frac{1}{2}x + 3 \]
Next, let's see if any of the options match these equations:
- \( { y = 12x, y = 2x + 3 } \)
- \( { y = 2x, y = 12x + 3 } \)
- \( { y = 2x, y = 3x + 12 } \)
- \( { y = 2x + 3, y = 12x } \)
From our analysis:
- The first equation \( y_a = 2x \) is found in the second and third systems.
- The second equation \( y_b = \frac{1}{2}x + 3 \) can be rewritten as \( y = 2x + 3 \), fitting into the first option.
- No matching options suggest coefficients.
The closest match is: \[ { y = 2x, y = 12x + 3 } \]
So, the answer is: {y = 2x, y = 12x + 3}.