The region formed by the inequalities 2x + 3y | vedclass

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Question

(B) To determine the region,we analyze the given inequalities:$1$. $2x + 3y \leq 5$: This represents a region on or below the line $2x + 3y = 5$.$2$.

Vedclass · Accepted Answer

lies in the first quadrant and is bounded

Vedclass · Answer

(B) To determine the region,we analyze the given inequalities:$1$. $2x + 3y \leq 5$: This represents a region on or below the line $2x + 3y = 5$.$2$. $4x - 3y \leq -2$ or $3y \geq 4x + 2$: This represents a region on or above the line $3y = 4x + 2$.$3$. $x \geq 0$: This restricts the region to the right of the $y$-axis (first and fourth quadrants).Intersection points:- Intersection of $2x + 3y = 5$ and $3y = 4x + 2$: Substituting $3y$,we get $2x + (4x + 2) = 5 \implies 6x = 3 \implies x = 0.5$. Then $3y = 4(0.5) + 2 = 4 \implies y = 4/3$.- When $x = 0$,$2x + 3y = 5 \implies y = 5/3$ and $3y = 4x + 2 \implies y = 2/3$.Since the region includes $x = 0$ and values of $y$ where $x$ can be negative for the second inequality,but the constraint $x \geq 0$ limits it,we observe the region spans across the $y$-axis boundary. Specifically,for $x \geq 0$,the region exists in the first quadrant,but the boundary $4x - 3y + 2 \leq 0$ allows $x$ to be $0$ while $y \geq 2/3$,and $2x + 3y - 5 \leq 0$ allows $x=0$ while $y \leq 5/3$. Thus,the region lies in the first quadrant.

The region formed by the inequalities $2x + 3y - 5 \leq 0$,$4x - 3y + 2 \leq 0$,and $x \geq 0$ is:

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