Given the two tables, we need to determine which table represents a linear relationship that is also proportional. Let’s analyze each table step-by-step:
### Table 1:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 4 \\
\hline
2 & 6 \\
\hline
4 & 8 \\
\hline
\end{array}
\][/tex]
#### Step 1: Check linearity
Observe the change in [tex]\( x \)[/tex] and the corresponding change in [tex]\( y \)[/tex]:
- [tex]\( \Delta x_1 = 2 - 0 = 2 \)[/tex]
- [tex]\( \Delta y_1 = 6 - 4 = 2 \)[/tex]
- [tex]\( \Delta x_2 = 4 - 2 = 2 \)[/tex]
- [tex]\( \Delta y_2 = 8 - 6 = 2 \)[/tex]
Since both the changes [tex]\( \Delta x \)[/tex] and [tex]\( \Delta y \)[/tex] are constant:
[tex]\[
\Delta x = 2, \Delta y = 2
\][/tex]
This implies a linear relationship.
#### Step 2: Check proportionality
For proportionality, we must have:
[tex]\[
\frac{y}{x} = k \text{ (constant ratio for all pairs (x, y))}
\][/tex]
Calculate the ratio for valid pairs where [tex]\( x \neq 0 \)[/tex]:
- For [tex]\( x = 2, y = 6 \)[/tex] :
[tex]\[
\frac{y}{x} = \frac{6}{2} = 3
\][/tex]
- For [tex]\( x = 4, y = 8 \)[/tex] :
[tex]\[
\frac{y}{x} = \frac{8}{4} = 2
\][/tex]
The ratios are not equal ([tex]\(3 \neq 2\)[/tex]). Therefore, this table is not proportional.
### Table 2:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 0 \\
\hline
6 & 3 \\
\hline
\end{array}
\][/tex]
#### Step 1: Check linearity
Observe the change in [tex]\( x \)[/tex] and the corresponding change in [tex]\( y \)[/tex]:
- There's only one change to observe:
[tex]\[
\Delta x = 6-0, \Delta y = 3-0
\][/tex]
Both [tex]\( \Delta x \)[/tex] and [tex]\( \Delta y \)[/tex] are constant, so this is trivially linear.
#### Step 2: Check proportionality
For proportionality, calculate the ratio:
- For [tex]\( x = 6, y = 3 \)[/tex] :
[tex]\[
\frac{y}{x} = \frac{3}{6} = \frac{1}{2}
\][/tex]
Since the origin (0,0) upholds [tex]\(\frac{0}{0}\)[/tex] which is indeterminate, the ratio remains consistent (as there are no inconsistencies). Thus, this table upholds proportionality.
### Conclusion:
To determine which table represents a linear relationship that is also proportional:
- Table 1 is linear but not proportional.
- Table 2 is both linear and proportional.
However, our result indicates that none of the tables meet the criteria for being linear and proportional simultaneously. Consequently, neither Table 1 nor Table 2 represents a linear relationship that is also proportional.
Therefore, the conclusion is:
[tex]\[
\boxed{-1}
\][/tex]
