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This problem involves solving a system of linear inequalities. The inequalities given are:

1. \( x + 2y \geq 6 \)
2. \( x + y \leq 8 \)
3. \( x \geq 0 \) (ensuring \( x \) is non-negative)
4. \( y \geq 0 \) (ensuring \( y \) is non-negative)

### Steps to Find the Solution Region:
1. **Graph the inequalities**:
   - For \( x + 2y \geq 6 \): The boundary line is \( x + 2y = 6 \). Solve for \( y \):
     \[
     y = \frac{6 - x}{2}
     \]
     This is a straight line with intercepts \( (0, 3) \) and \( (6, 0) \). The region \( x + 2y \geq 6 \) lies above this line.
   - For \( x + y \leq 8 \): The boundary line is \( x + y = 8 \). Solve for \( y \):
     \[
     y = 8 - x
     \]
     This is a straight line with intercepts \( (0, 8) \) and \( (8, 0) \). The region \( x + y \leq 8 \) lies below this line.
   - For \( x \geq 0 \) and \( y \geq 0 \): These are constraints limiting the region to the first quadrant.

2. **Identify the feasible region**:
   The solution to the system of inequalities is the intersection of the regions defined by the inequalities. The region must satisfy:
   - Above \( x + 2y = 6 \)
   - Below \( x + y = 8 \)
   - In the first quadrant (\( x \geq 0, y \geq 0 \)).

3. **Find intersection points**:
   - Intersection of \( x + 2y = 6 \) and \( x + y = 8 \):
     Solve the system:
     \[
     x + 2y = 6
     \]
     \[
     x + y = 8
     \]
     Subtract the second equation from the first

Daerah penyelesaian dari sistem pertidaksamaan linear berikut adalah: x + 2y ≥ 6, x + y ≤ 8, x ≥ 0, y ≥ 0.

Understand how to solve the linear inequalities system x + 2y ≥ 6, x + y ≤ 8, x ≥ 0, and y ≥ 0. Explore step-by-step graphing techniques and discover the feasible solution region. Perfect for Grades 9-11 students.

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