To match the table of values with the correct system of linear equations, we can analyze the data provided:
The table shows values for \( x \), \( ya \), and \( yb \):
\[ \begin{array}{|c|c|c|} \hline x & ya & yb \ \hline 0 & 0 & 3 \ 1 & 2 & 3.5 \ 2 & 4 & 4 \ \hline \end{array} \]
Let's analyze the values for \( ya \) and \( yb \) to determine the equations.
For \( ya \):
- When \( x = 0 \), \( ya = 0 \).
- When \( x = 1 \), \( ya = 2 \).
- When \( x = 2 \), \( ya = 4 \).
The points are (0, 0), (1, 2), and (2, 4). This suggests a linear relationship, and we can find the slope:
\[ \text{slope} = \frac{2 - 0}{1 - 0} = 2 \]
Using point-slope form \( y = mx + b \):
- The slope \( m = 2 \) and it passes through the origin (0,0). Thus, the equation for \( ya \) is:
\[ ya = 2x \]
For \( yb \):
- When \( x = 0 \), \( yb = 3 \).
- When \( x = 1 \), \( yb = 3.5 \).
- When \( x = 2 \), \( yb = 4 \).
The points are (0, 3), (1, 3.5), and (2, 4).
Calculating the slope:
\[ \text{slope} = \frac{3.5 - 3}{1 - 0} = 0.5 \] \[ \text{slope} = \frac{4 - 3.5}{2 - 1} = 0.5 \]
Using point-slope form \( y = mx + b \):
- The slope \( m = 0.5 \) and it has a y-intercept \( b = 3 \). Thus, the equation for \( yb \) is:
\[ yb = 0.5x + 3 \]
Now, putting these two results together, we can express the system of equations for \( ya \) and \( yb \):
\[ \begin{align*} ya &= 2x \ yb &= 0.5x + 3 \ \end{align*} \]
Comparing this to the given options, we identify:
The correct representation from the responses is:
\[ {y = 2x, y = \frac{1}{2}x + 3} \]
Thus, the last option is:
\[ {y = 2x, y = \frac{1}{2}x + 3} \]