To complete the table of values for the given equations, let's substitute various values of $x$ to find corresponding values of $y$ for each equation.

Equation 1: $y = \frac{2}{3}x + 3$

1. If $x = 0$: $y = \frac{2}{3}(0) + 3 = 3$

2. If $x = 3$: $y = \frac{2}{3}(3) + 3 = 2 + 3 = 5$

3. If $x = -3$: $y = \frac{2}{3}(-3) + 3 = -2 + 3 = 1$

4. If $x = 6$: $y = \frac{2}{3}(6) + 3 = 4 + 3 = 7$

5. If $x = -6$: $y = \frac{2}{3}(-6) + 3 = -4 + 3 = -1$

So, the table for $y = \frac{2}{3}x + 3$ would be:

Equation 2: $y = -4x + 4$

1. If $x = 0$: $y = -4(0) + 4 = 4$

2. If $x = 1$: $y = -4(1) + 4 = -4 + 4 = 0$

3. If $x = -1$: $y = -4(-1) + 4 = 4 + 4 = 8$

4. If $x = 2$: $y = -4(2) + 4 = -8 + 4 = -4$

5. If $x = -2$: $y = -4(-2) + 4 = 8 + 4 = 12$

So, the table for $y = -4x + 4$ would be:

Would you like a further breakdown of these calculations or have any questions?

Here are some additional questions to consider:

1. What are other values of $x$ you could use to further verify your graph?
2. How would the slope of $y = \frac{2}{3}x + 3$ compare to a line with slope 1?
3. What happens to the $y$-intercept if you change the constant term in each equation?
4. How does the slope of $y = -4x + 4$ influence the steepness of the line?
5. What would be the result if the slope in each equation were doubled?

Tip: Always plot at least two points to confirm the correct direction and position of your line on a graph.