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We are asked to find the area of the region enclosed by the graphs of three lines:

1. \( y = 8 - 3x \)
2. \( y = 6 - x \)
3. \( y = 3 \)

### Steps to solve:

1. **Find the points of intersection** of these three lines, which will help us define the region to integrate.
   
   - Set \( y = 8 - 3x \) equal to \( y = 6 - x \) to find the first intersection:
     \[
     8 - 3x = 6 - x
     \]
     Solving this equation:
     \[
     8 - 6 = 3x - x \implies 2 = 2x \implies x = 1
     \]
     Substituting \( x = 1 \) back into one of the equations, \( y = 8 - 3(1) = 5 \).
     So, the point of intersection is \( (1, 5) \).

   - Set \( y = 8 - 3x \) equal to \( y = 3 \) to find the second intersection:
     \[
     8 - 3x = 3 \implies 5 = 3x \implies x = \frac{5}{3}
     \]
     Thus, the point of intersection is \( \left( \frac{5}{3}, 3 \right) \).

   - Set \( y = 6 - x \) equal to \( y = 3 \) to find the third intersection:
     \[
     6 - x = 3 \implies x = 3
     \]
     So, the third intersection is \( (3, 3) \).

2. **Integrate to find the area** between the curves. We will integrate the difference between the two upper curves and the lower curve from \( x = 1 \) to \( x = 3 \).

   The area can be expressed as:
   \[
   A = \int_1^{\frac{5}{3}} \left( (8 - 3x) - 3 \right) \, dx + \int_{\frac{5}{3}}^3 \left( (6 - x) - 3 \right) \, dx
   \]

Let's calculate this.The area of the region enclosed by the graphs of \( y = 8 - 3x \), \( y = 6 - x \), and \( y = 3 \) is approximately:

\[
A = \frac{14}{9}
\]

Would you like a detailed explanation of any step, or do you have any other questions?

Here are some related questions for further exploration:

1. How would you graph the three lines to visualize the enclosed region?
2. How can you find the exact area using numerical methods?
3. What happens to the area if one of the lines changes?
4. How can you apply the same technique to find areas between curves that aren't linear?
5. How would the problem change if the curves intersected in more complex ways?

**Tip:** When integrating between curves, always ensure that you carefully set up the bounds and correctly subtract the lower function from the upper function to avoid negative areas.

Find the area of the region enclosed by the graphs of y = 8 - 3x, y = 6 - x, and y = 3.

Calculate the area of the region enclosed by the graphs of y = 8 - 3x, y = 6 - x, and y = 3. This problem uses concepts from calculus, including finding points of intersection and using definite integration to calculate the area between curves. Suitable for undergraduate students.

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