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