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

\[
y = 8 - 3x, \quad y = 6 - x, \quad \text{and} \quad y = 1.
\]

### Step 1: Find the points of intersection
We first need to determine the x-values where the graphs intersect, which will define the limits for integration.

#### Intersection of \(y = 8 - 3x\) and \(y = 6 - x\):
\[
8 - 3x = 6 - x
\]
Solving this:
\[
8 - 6 = 3x - x
\]
\[
2 = 2x \quad \Rightarrow \quad x = 1.
\]

#### Intersection of \(y = 6 - x\) and \(y = 1\):
\[
6 - x = 1
\]
Solving this:
\[
x = 5.
\]

#### Intersection of \(y = 8 - 3x\) and \(y = 1\):
\[
8 - 3x = 1
\]
Solving this:
\[
7 = 3x \quad \Rightarrow \quad x = \frac{7}{3}.
\]

### Step 2: Set up the integrals for the area
The region is bounded by these three lines, and we need to split it into two parts based on the intersection points.

1. From \(x = 1\) to \(x = \frac{7}{3}\), the upper curve is \(y = 8 - 3x\) and the lower curve is \(y = 1\).
2. From \(x = \frac{7}{3}\) to \(x = 5\), the upper curve is \(y = 6 - x\) and the lower curve is \(y = 1\).

The total area is the sum of the areas between the curves in these intervals.

\[
\text{Area} = \int_1^{\frac{7}{3}} \left( (8 - 3x) - 1 \right) \, dx + \int_{\frac{7}{3}}^5 \left( (6 - x) - 1 \right) \, dx.
\]

### Step 3: Solve the integrals
#### First integral:
\[
\int_1^{\frac{7}{3}} (7 - 3x) \, dx = \left[ 7x - \frac{3x^2}{2} \right]_1^{\frac{7}{3}}.
\]
Evaluating this:
\[
\left( 7 \times \frac{7}{3} - \frac{3 \times \left(\frac{7}{3}\right)^2}{2} \right) - \left( 7 \times 1 - \frac{3 \times 1^2}{2} \right).
\]
Simplifying:
\[
\left( \frac{49}{3} - \frac{3 \times 49}{18} \right) - \left( 7 - \frac{3}{2} \right).
\]
This gives:
\[
\frac{49}{3} - \frac{49}{6} - 7 + \frac{3}{2} = \frac{98}{6} - \frac{49}{6} - \frac{11}{2} = \frac{49}{6} - \frac{33}{6} = \frac{16}{6} = \frac{8}{3}.
\]

#### Second integral:
\[
\int_{\frac{7}{3}}^5 (5 - x) \, dx = \left[ 5x - \frac{x^2}{2} \right]_{\frac{7}{3}}^5.
\]
Evaluating this:
\[
\left( 5 \times 5 - \frac{5^2}{2} \right) - \left( 5 \times \frac{7}{3} - \frac{\left( \frac{7}{3} \right)^2}{2} \right).
\]
Simplifying:
\[
\left( 25 - \frac{25}{2} \right) - \left( \frac{35}{3} - \frac{49}{18} \right).
\]
This gives:
\[
\left( \frac{50}{2} - \frac{25}{2} \right) - \left( \frac{210}{18} - \frac{49}{18} \right) = \frac{25}{2} - \frac{161}{18}.
\]
Converting to a common denominator:
\[
\frac{225}{18} - \frac{161}{18} = \frac{64}{18} = \frac{32}{9}.
\]

### Step 4: Total area
Adding the two areas:
\[
\frac{8}{3} + \frac{32}{9} = \frac{24}{9} + \frac{32}{9} = \frac{56}{9}.
\]

Thus, the area of the region enclosed by the curves is \(\frac{56}{9}\).

---

Would you like any additional details or clarification? Here are some related questions you might find interesting:

1. How would the solution change if one of the lines were shifted vertically?
2. Can this area be verified geometrically or using a different method like geometry?
3. What would happen if we added a curve like \(y = x^2\) to this region?
4. How can you visualize the region to better understand the integral setup?
5. How do you generalize the approach to find the area between more complicated curves?

**Tip:** Always sketch the region when solving area problems. It helps to visualize which curve is on top and ensures proper limits for integration.

Find the area of the region enclosed by the graphs of y = 8 - 3x, y = 6 - x, and y = 1.

This page explains how to find the area of the region enclosed by the lines y = 8 - 3x, y = 6 - x, and y = 1. Learn the steps involving solving for intersection points and using integrals to compute the area between the curves.

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