Question

The linear inequations for which the shaded area in the following figure is the solution set, are

• A. $x+y \leq 1,\, 2 x +y \geq 2,\, x-2 y \geq 8,\, x \leq 0,\, y \geq 0$
• B. $x-y \geq 1,\, 2 x +y \geq 2,\, x+2 y \geq 8,\, x \geq 0,\, y \geq 0$
• C. $x-y \leq 1,\, 2 x +y \geq 2,\, x+2 y \leq 8,\, x \geq 0,\, y \geq 0$
• D. $x+y \geq 1,\, 2 x +y \leq 2,\, x+2 y \geq 8,\, x \geq 0,\, y \geq 0$

Answer 1

Answer: (C)

Let $L_{1}: x+2 y=8;$
$L _{2} 2 x + y =2$
$L _{3}: x - y =1$
ince the shaded area is below the line $L _{1}$, we have $x+2 y \leq 8$.
Since the shaded area is above the line $L_{2}$, we have $2 x +y \geq 2$.
Since the common region is to the left of the line $L _{3}$, we have $x - y \leq 1$