Let's solve both problems by writing the equations in slope-intercept form, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

8) Based on the Table of Values:

We need to calculate the slope $m$ using two points from the table. The formula for slope is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Let's choose the points $(6, 14)$ and $(9, 12)$: $m = \frac{12 - 14}{9 - 6} = \frac{-2}{3} = -\frac{2}{3}$

Now, substitute the slope $m = -\frac{2}{3}$ into the slope-intercept form $y = mx + b$. We can use one point from the table, say $(6, 14)$, to find $b$.

Substitute $x = 6$, $y = 14$, and $m = -\frac{2}{3}$ into the equation $y = mx + b$:

$14 = -\frac{2}{3}(6) + b$ $14 = -4 + b$ $b = 18$

Thus, the equation is: $y = -\frac{2}{3}x + 18$

9) Based on the Graph:

From the graph, we can visually determine the slope and y-intercept. The line passes through the points $(-6, 0)$ and $(0, 4)$.

The slope $m$ is: $m = \frac{4 - 0}{0 - (-6)} = \frac{4}{6} = \frac{2}{3}$

The y-intercept $b$ is where the line crosses the y-axis, which is $b = 4$.

Thus, the equation is: $y = \frac{2}{3}x + 4$

Final Equations:

1. For the table (problem 8): $y = -\frac{2}{3}x + 18$
2. For the graph (problem 9): $y = \frac{2}{3}x + 4$

Would you like further explanation on these solutions or any other questions?

Here are five related questions for deeper understanding:

1. How do you find the slope when given two points?
2. Why does the y-intercept represent the value of $y$ when $x = 0$?
3. Can a line have a slope of zero? If so, what does that line look like?
4. What is the significance of a negative slope compared to a positive slope?
5. How would the equation change if the line passed through the origin?

Tip: Always remember that the slope measures the steepness of a line. A positive slope means the line rises as it moves right, while a negative slope means it falls.