Worksheet displaying coordinate pairs and linear equations for plotting graphs.

To solve the problem, we need to match each table of values (labeled as $ A, B, C, D, E, F, G, H, I, J $) with the corresponding equation that generates those values. Let's go through each table step by step.

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Step 1: Analyze Table $ A $

| $ x $ | $ y $ |
|--------|-------|
| 3 | 6 |
| 4 | 8 |
| 5 | 10 |

- The relationship between $ x $ and $ y $ appears linear.
- Calculate the slope:
$$
\text{slope} = \frac{\Delta y}{\Delta x} = \frac{8 - 6}{4 - 3} = 2
$$
- The equation is of the form $ y = 2x + c $. Using the point $ (3, 6) $:
$$
6 = 2(3) + c \implies c = 0
$$
- Therefore, the equation is $ y = 2x $.

Match: $ A $ corresponds to $ y = 2x $ (Option $ A $).

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Step 2: Analyze Table $ B $

| $ x $ | $ y $ |
|--------|-------|
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |

- The relationship is clearly $ y = x $.

Match: $ B $ corresponds to $ y = x $ (Option $ I $).

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Step 3: Analyze Table $ C $

| $ x $ | $ y $ |
|--------|-------|
| 1 | 3 |
| 2 | 6 |
| 3 | 9 |

- The relationship is $ y = 3x $.

Match: $ C $ corresponds to $ y = 3x $ (Option $ C $).

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Step 4: Analyze Table $ D $

| $ x $ | $ y $ |
|--------|-------|
| 0 | 0 |
| 1 | 2 |
| 2 | 4 |

- The relationship is $ y = 2x $.

Match: $ D $ corresponds to $ y = 2x $ (Option $ A $).

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Step 5: Analyze Table $ E $

| $ x $ | $ y $ |
|--------|-------|
| 0 | 1 |
| 1 | 5 |
| 2 | 9 |

- The relationship is linear. Calculate the slope:
$$
\text{slope} = \frac{\Delta y}{\Delta x} = \frac{5 - 1}{1 - 0} = 4
$$
- The equation is of the form $ y = 4x + c $. Using the point $ (0, 1) $:
$$
1 = 4(0) + c \implies c = 1
$$
- Therefore, the equation is $ y = 4x + 1 $.

Match: $ E $ corresponds to $ y = 4x + 1 $ (Option $ D $).

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Step 6: Analyze Table $ F $

| $ x $ | $ y $ |
|--------|-------|
| 1 | 2 |
| 2 | 4 |
| 3 | 6 |

- The relationship is $ y = 2x $.

Match: $ F $ corresponds to $ y = 2x $ (Option $ A $).

---

Step 7: Analyze Table $ G $

| $ x $ | $ y $ |
|--------|-------|
| 0 | 5 |
| 1 | 3 |
| 2 | 1 |

- The relationship is linear. Calculate the slope:
$$
\text{slope} = \frac{\Delta y}{\Delta x} = \frac{3 - 5}{1 - 0} = -2
$$
- The equation is of the form $ y = -2x + c $. Using the point $ (0, 5) $:
$$
5 = -2(0) + c \implies c = 5
$$
- Therefore, the equation is $ y = -2x + 5 $.

Match: $ G $ corresponds to $ y = -2x + 5 $ (Option $ E $).

---

Step 8: Analyze Table $ H $

| $ x $ | $ y $ |
|--------|-------|
| 0 | 6 |
| 1 | 4 |
| 2 | 2 |

- The relationship is linear. Calculate the slope:
$$
\text{slope} = \frac{\Delta y}{\Delta x} = \frac{4 - 6}{1 - 0} = -2
$$
- The equation is of the form $ y = -2x + c $. Using the point $ (0, 6) $:
$$
6 = -2(0) + c \implies c = 6
$$
- Therefore, the equation is $ y = -2x + 6 $.

Match: $ H $ corresponds to $ y = -2x + 6 $ (Option $ H $).

---

Step 9: Analyze Table $ I $

| $ x $ | $ y $ |
|--------|-------|
| 0 | 8 |
| 1 | 8 |
| 2 | 8 |

- The relationship is constant: $ y = 8 $.

Match: $ I $ corresponds to $ y = 8 $ (Option $ I $).

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Step 10: Analyze Table $ J $

| $ x $ | $ y $ |
|--------|-------|
| 2 | 1 |
| 4 | 2 |
| 6 | 3 |

- The relationship is $ y = \frac{x}{2} $.

Match: $ J $ corresponds to $ y = \frac{x}{2} $ (Option $ J $).

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Final Matches:

- $ A \to A $
- $ B \to I $
- $ C \to C $
- $ D \to A $
- $ E \to D $
- $ F \to A $
- $ G \to E $
- $ H \to H $
- $ I \to I $
- $ J \to J $

Boxed Answer:

$$
\boxed{
A \to A, B \to I, C \to C, D \to A, E \to D, F \to A, G \to E, H \to H, I \to I, J \to J
}
$$