91Ó°ÊÓ

Convert the inequality$yâ‰\yen 2x+3$into equality that is convert the inequality into equation.

Therefore, it is obtained that: $y=2x+3$

Therefore, the equation of the boundary is $y=2x+3$. The inequality $yâ‰\yen 2x+3$has equal to sign, therefore the boundary is included in the solution. Therefore, the boundary is denoted by solid lines.

Draw the graph of the boundary line $y=2x+3$.

Substitute 0 for x and find the value of y.

$$\begin{gathered} y=2x+3 \\ y=2\left(0\right)+3 \\ y=3 \end{gathered}$$

Therefore, one of the point is$\left(0,3\right)$

Substitute 0 for y and find the value of x.

$$\begin{gathered} y=2x+3 \\ 0=2x+3 \\ −3=2x \\ \frac{−3}{2}=x \end{gathered}$$

Therefore the other point is$\left(\frac{−3}{2},0\right)$

Therefore, draw the graph of the boundary line$y=2x+3$by drawing a line passing through the points$\left(0,3\right)$and $\left(\frac{−3}{2},0\right)$.

Now to draw the graph of the inequality $yâ‰\yen 2x+3$, take any point which is not on the line $y=2x+3$in the inequality $yâ‰\yen 2x+3$. If the condition obtained is true, then shade the region towards that point and if the condition obtained is false, then shade the region away from the point.

Let the point be$\left(0,0\right)$and the point$\left(0,0\right)$is not on the line $y=2x+3$.

Substitute the point in the inequality $yâ‰\yen 2x+3$.

$$\begin{gathered} yâ‰\yen 2x+3 \\ 0â‰\yen 2\left(0\right)+3 \\ 0â‰\yen 3 \end{gathered}$$

As, the condition obtained is $0â‰\yen 3$, which is false. Therefore, to draw the graph of the inequality $yâ‰\yen 2x+3$, shade the region away from the point width="34">0,0.

The graph of the given inequality$yâ‰\yen 2x+3$ is:

Convert the inequality$−4x−3y>12$ into equality that is convert the inequality into equation.

Therefore, it is obtained that:$−4x−3y=12$

Therefore, the equation of the boundary is $−4x−3y=12$. The inequality$−4x−3y>12$ has no equal to sign, therefore the boundary is not included in the solution. Therefore, the boundary is denoted by dashed lines.

Draw the graph of the boundary line $−4x−3y=12$.

Substitute 0 for x and find the value of y.

$$\begin{gathered} −4x−3y=12 \\ −4\left(0\right)−3y=12 \\ −3y=12 \\ y=−4 \end{gathered}$$

Therefore, one of the point is$\left(0,−4\right)$

Substitute 0 for y and find the value of x.

$$\begin{gathered} −4x−3y=12 \\ −4x−3\left(0\right)=12 \\ −4x=12 \\ x=−3 \end{gathered}$$

Therefore the other point is$\left(−3,0\right)$

Therefore, draw the graph of the boundary line$−4x−3y=12$ by drawing a line passing through the points$\left(0,−4\right)$ and $\left(−3,0\right)$.

Now to draw the graph of the inequality width="99" height="20" role="math">−4x−3y>12, take any point which is not on the line $−4x−3y=12$in the inequality $−4x−3y>12$. If the condition obtained is true, then shade the region towards that point and if the condition obtained is false, then shade the region away from the point.

Let the point be$\left(0,0\right)$ and the point$\left(0,0\right)$ is not on the line $−4x−3y=12$.

Substitute the point in the inequality $−4x−3y>12$.

$$\begin{gathered} −4x−3y>12 \\ −4\left(0\right)−3\left(0\right)>12 \\ 0>12 \end{gathered}$$

As, the condition obtained is $0>12$, which is false. Therefore, to draw the graph of the inequality $−4x−3y>12$, shade the region away from the point $\left(0,0\right)$.

The graph of the given inequality$−4x−3y>12$ is:

The graph of the inequalities$yâ‰\yen 2x+3$ and$−4x−3y>12$ in one graph is:

Shade the common region of both the inequalities$yâ‰\yen 2x+3$ and$−4x−3y>12$ to find the solution of the given system of inequalities.