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(C) $1$. Identify the boundary lines of the shaded region:- Line passing through $(1, 0)$ and $(0, 2)$: The equation is $\frac{x}{1} + \frac{y}{2} = 1

Vedclass · Accepted Answer

$2x + y \geq 2, x - y \leq 1, x + 2y \leq 8, x \geq 0, y \geq 0$

Vedclass · Answer

(C) $1$. Identify the boundary lines of the shaded region:- Line passing through $(1, 0)$ and $(0, 2)$: The equation is $\frac{x}{1} + \frac{y}{2} = 1$, which simplifies to $2x + y = 2$. Since the shaded region is above this line, the inequality is $2x + y \geq 2$.- Line passing through $(1, 0)$ and $(0, -1)$ (implied by the slope): The line passes through $(1, 0)$ and $(0, -1)$, so the equation is $y - 0 = \frac{-1 - 0}{0 - 1}(x - 1)$, which simplifies to $y = x - 1$ or $x - y = 1$. Since the shaded region is to the left of this line, the inequality is $x - y \leq 1$.- Line passing through $(0, 4)$ and $(4, 0)$ (implied by the intersection $B(10/3, 7/3)$): The line passing through $(0, 4)$ and $(4, 0)$ has the equation $\frac{x}{4} + \frac{y}{4} = 1$, which is $x + y = 4$. However, checking the intersection $B(10/3, 7/3)$ with $x + 2y = 8$, we see $10/3 + 2(7/3) = 10/3 + 14/3 = 24/3 = 8$. Thus, the line is $x + 2y = 8$. Since the shaded region is below this line, the inequality is $x + 2y \leq 8$.$2$. Combining these with $x \geq 0, y \geq 0$, we get the system: $2x + y \geq 2, x - y \leq 1, x + 2y \leq 8, x \geq 0, y \geq 0$.$3$. This matches option $C$.

The shaded region in the provided graph represents the solution set for which of the following systems of linear inequalities?

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