On the grid provided, represent the linear inequalities 2y+x >= 0, x <= 3 and 3x-5y > -15.

First, consider the equation $2y+x = 0$, which can be rewritten as $y = -\frac{1}{2}x$.

• This is a solid line because the inequality includes "equal to" ($\ge$).
• Plot points: If $x=0, y=0$. If $x=2, y=-1$. If $x=-2, y=1$.
• To determine the region, test a point not on the line, e.g., $(1,1)$: $2(1)+1 = 3$. Since $3 \ge 0$ is true, the region containing $(1,1)$ is the feasible region. Therefore, shade the region below the line $y = -\frac{1}{2}x$.

First, consider the equation $x = 3$.

• This is a solid vertical line passing through $x=3$ because the inequality includes "equal to" ($\le$).
• Plot points: $(3,0)$, $(3,1)$, $(3,-1)$, etc.
• To determine the region, test a point not on the line, e.g., $(0,0)$: $0 \le 3$ is true. So, the region containing $(0,0)$ is the feasible region. Therefore, shade the region to the right of the line $x=3$.

First, consider the equation $3x-5y = -15$.

• This is a dashed line because the inequality does not include "equal to" ($>$).
• Plot points: If $x=0, -5y=-15 \Rightarrow y=3$. Point: $(0,3)$. If $y=0, 3x=-15 \Rightarrow x=-5$. Point: $(-5,0)$.
• To determine the region, test a point not on the line, e.g., $(0,0)$: $3(0)-5(0) = 0$. Since $0 > -15$ is true, the region containing $(0,0)$ is the feasible region. Therefore, shade the region above the line $3x-5y = -15$.

The feasible region is the unshaded area on the grid, bounded by these three lines. It is the region that is:

• Above or on the line $y = -\frac{1}{2}x$.
• To the left of or on the line $x = 3$.
• Below the line $y = \frac{3}{5}x+3$.

3 zimekamilika, zimebaki 2 leo. Unapiga hatua.