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To solve this problem, we need to understand the feasible region defined by the given system of inequalities:

\[
y \geq \frac{10}{9}x + \frac{105}{9}
\]
\[
y \leq 21x + 131
\]
\[
y < -\frac{11}{8}x + \frac{153}{8}
\]

We will find the boundaries where these inequalities intersect to determine the regions in which \(x\) varies and how \(S\) can be divided into two subsets based on those intervals.

### Step 1: Find the intersection points of the boundary lines.

#### 1. Intersection of \( y = \frac{10}{9}x + \frac{105}{9} \) and \( y = 21x + 131 \):

Set the equations equal:
\[
\frac{10}{9}x + \frac{105}{9} = 21x + 131
\]
Multiply through by 9 to clear the fractions:
\[
10x + 105 = 189x + 1179
\]
Rearrange terms:
\[
189x - 10x = 105 - 1179
\]
\[
179x = -1074
\]
\[
x = -\frac{1074}{179} \approx -6.00
\]

Substitute \(x = -6.00\) back into either equation to find \(y\):
Using \(y = 21x + 131\):
\[
y = 21(-6.00) + 131 = -126 + 131 = 5
\]
So, the intersection point is approximately \((-6.00, 5)\).

#### 2. Intersection of \( y = \frac{10}{9}x + \frac{105}{9} \) and \( y = -\frac{11}{8}x + \frac{153}{8} \):

Set the equations equal:
\[
\frac{10}{9}x + \frac{105}{9} = -\frac{11}{8}x + \frac{153}{8}
\]
Multiply through by 72 (the least common multiple of 9 and 8):
\[
8(10x + 105) = 9(-11x + 153)
\]
\[
80x + 840 = -99x + 1377
\]
Rearrange terms:
\[
80x + 99x = 1377 - 840
\]
\[
179x = 537
\]
\[
x = \frac{537}{179} \approx 3.00
\]

Substitute \(x = 3.00\) back into either equation to find \(y\):
Using \(y = \frac{10}{9}x + \frac{105}{9}\):
\[
y = \frac{10}{9}(3.00) + \frac{105}{9} = \frac{30 + 105}{9} = \frac{135}{9} = 15
\]
So, the intersection point is approximately \((3.00, 15)\).

#### 3. Intersection of \( y = 21x + 131 \) and \( y = -\frac{11}{8}x + \frac{153}{8} \):

Set the equations equal:
\[
21x + 131 = -\frac{11}{8}x + \frac{153}{8}
\]
Multiply through by 8:
\[
8(21x + 131) = -11x + 153
\]
\[
168x + 1048 = -11x + 153
\]
Rearrange terms:
\[
168x + 11x = 153 - 1048
\]
\[
179x = -895
\]
\[
x = -\frac{895}{179} \approx -5.00
\]

Substitute \(x = -5.00\) back into either equation to find \(y\):
Using \(y = 21x + 131\):
\[
y = 21(-5.00) + 131 = -105 + 131 = 26
\]
So, the intersection point is approximately \((-5.00, 26)\).

### Step 2: Determine the intervals for \(x\).

The intersection points give us the intervals for \(x\):
1. From \(x \approx -6.00\) to \(x \approx -5.00\)
2. From \(x \approx -5.00\) to \(x \approx 3.00\)

### Step 3: Describe \(S\) as a union of two sets.

Let \( S_1 \) correspond to the interval \([-6.00, -5.00]\), where \(y\) is determined by the upper and lower bounds given by the relevant lines. Let \( S_2 \) correspond to the interval \([-5.00, 3.00]\), similarly defined.

Thus, \(S = S_1 \cup S_2\), where:
- \( S_1 \) is the feasible region for \( -6.00 \leq x \leq -5.00 \)
- \( S_2 \) is the feasible region for \( -5.00 \leq x \leq 3.00 \)

Would you like more details or have any questions?

### Follow-up Questions
1. How do we graphically represent the feasible region \(S\)?
2. What do the slopes of the boundary lines represent in this context?
3. Can we express the feasible region using set notation or interval notation for \(y\)?
4. How do we check if a point lies within the feasible region \(S\)?
5. What are some applications of solving such systems of linear inequalities?

### Tip
When solving inequalities graphically, always plot the boundary lines first and then determine which side of each line satisfies the inequality.

Write S as a union of two sets, where x is a free variable and y is dependent on x. In what intervals does x vary in the two sets?

Learn how to solve a system of linear inequalities to determine the intervals in which x varies. This problem covers solving inequalities and finding the intersection of boundary lines, suitable for students in Grades 10-12.

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