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Question

(A) Given inequality: $y + 8 > 2x$. First,we draw the boundary line $y = 2x - 8$. To draw this line,we find two points: If $x = 0$,$y = -8$. Point: $(

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The region above the line $y = 2x - 8$.

Vedclass · Answer

(A) Given inequality: $y + 8 > 2x$. First,we draw the boundary line $y = 2x - 8$. To draw this line,we find two points: If $x = 0$,$y = -8$. Point: $(0, -8)$. If $y = 0$,$2x = 8 \implies x = 4$. Point: $(4, 0)$. Since the inequality is strict $(>)$,we draw the line as a dashed line. Now,test the origin $(0, 0)$ in the inequality: $0 + 8 > 2(0) \implies 8 > 0$,which is true. Therefore,the solution region is the half-plane containing the origin $(0, 0)$,which is the region above the line $y = 2x - 8$.

Solve the following inequality graphically in a two-dimensional plane: $y + 8 > 2x$.

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