To determine whether [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in each table, we need to see if there exists a constant [tex]\( k \)[/tex] such that [tex]\( y = kx \)[/tex] for all pairs [tex]\( (x, y) \)[/tex] in each table. Specifically, [tex]\( k \)[/tex] should be the same value for each pair in the table.
### Table 1:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & 3 \\
\hline
3 & 9 \\
\hline
6 & 18 \\
\hline
7 & 21 \\
\hline
\end{array}
\][/tex]
Calculate [tex]\( k \)[/tex] for each pair:
[tex]\[
k = \frac{y}{x}
\][/tex]
[tex]\[
k_1 = \frac{3}{1} = 3
\][/tex]
[tex]\[
k_2 = \frac{9}{3} = 3
\][/tex]
[tex]\[
k_3 = \frac{18}{6} = 3
\][/tex]
[tex]\[
k_4 = \frac{21}{7} = 3
\][/tex]
Since [tex]\( k \)[/tex] is consistently 3 for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 1.
### Table 2:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-4 & -2 \\
\hline
-2 & -1 \\
\hline
2 & 1 \\
\hline
6 & 3 \\
\hline
\end{array}
\][/tex]
Calculate [tex]\( k \)[/tex] for each pair:
[tex]\[
k_1 = \frac{-2}{-4} = \frac{1}{2}
\][/tex]
[tex]\[
k_2 = \frac{-1}{-2} = \frac{1}{2}
\][/tex]
[tex]\[
k_3 = \frac{1}{2} = \frac{1}{2}
\][/tex]
[tex]\[
k_4 = \frac{3}{6} = \frac{1}{2}
\][/tex]
Since [tex]\( k \)[/tex] is consistently [tex]\( \frac{1}{2} \)[/tex] for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 2.
### Table 3:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-2 & 2 \\
\hline
-1 & 1 \\
\hline
2 & -2 \\
\hline
5 & -5 \\
\hline
\end{array}
\][/tex]
Calculate [tex]\( k \)[/tex] for each pair:
[tex]\[
k_1 = \frac{2}{-2} = -1
\][/tex]
[tex]\[
k_2 = \frac{1}{-1} = -1
\][/tex]
[tex]\[
k_3 = \frac{-2}{2} = -1
\][/tex]
[tex]\[
k_4 = \frac{-5}{5} = -1
\][/tex]
Since [tex]\( k \)[/tex] is consistently [tex]\( -1 \)[/tex] for all pairs, [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] in Table 3.
### Table 4:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-1 & 0 \\
\hline
0 & -1 \\
\hline
1 & -2 \\
\hline
2 & -3 \\
\hline
\end{array}
\][/tex]
Calculate [tex]\( k \)[/tex] for each pair:
Note that direct variation requires [tex]\( \frac{y}{x} \)[/tex] to be constant, and for [tex]\( x = 0 \)[/tex], [tex]\( y \)[/tex] should also be 0. However, when [tex]\( x=0 \)[/tex], [tex]\( y \)[/tex] is -1 which does not hold for direct variation.
[tex]\[
k_1 = \frac{0}{-1} = 0 / -1 = 0 \quad \text{(undefined)}
\][/tex]
\]
This inconsistency indicates [tex]\( y \)[/tex] does not vary directly with [tex]\( x \)[/tex] in Table 4.
### Conclusion:
The tables in which [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex] are:
[tex]\[
\boxed{1, 2, 3}
\][/tex]
