Linear Inequalities

Concept:The common region is the intersection of half-planes defined by the given lines and non-negativity constraints.Explanation:1. The lines $x+y=6$ and $2x-y=2$ intersect at $\left(\frac{8}{3}, \frac{10}{3}\right)$, which lies in the first quadrant.2. In the first quadrant ($x \geq 0, y \geq 0$), the two lines and the axes form a bounded triangular region with vertices $(1,0)$, $\left(\frac{8}{3}, \frac{10}{3}\right)$, and $(6,0)$.3. This bounded region lies below the line $x+y=6$ (so $x+y \leq 6$) and below the line $2x-y=2$ (so $2x-y \geq 2$, since $2x-y \geq 2$ rearranges to $y \leq 2x-2$).4. The origin $(0,0)$ does not satisfy $2x-y \geq 2$, confirming the region excludes the origin.Thus, the inequalities are $x+y \leq 6$, $2x-y \geq 2$, $x \geq 0$, $y \geq 0$.Answer:Option C