91Ó°ÊÓ

The given graph is,

The equation of the boundary line is$3x-y=6$.

The boundary line is $3x-y=6$.

On one side of the boundary line are the points with $3x-y>6$and on another side of the boundary line are the points with$3x-y<6$.

We will consider the point $(4,1)$. Here, $x=4,y=1$

We have to check, which inequality holds at this point.

Consider the inequality, $3x-y>6$

Then,

$$\begin{gathered} 3x-y>6 \\ 3\left(4\right)-1>6 \\ 12-1>6 \\ 11>6 \end{gathered}$$

which is true.

Now, consider the inequality $3x-y<6$

Then we have,

$$\begin{gathered} 3x-y<6 \\ 3\left(4\right)-1<6 \\ 12-1<6 \\ 11<6 \end{gathered}$$

which is false.

Therefore the inequality that is true at the point$(4,1)$is$3x-y>6$

The inequality $3x-y>6$holds at the point $(4,1)$.

Therefore the side of the borderline with the point $(4,1)$ is the side which denotes the inequality $3x-y>6$.

Also, since the borderline is a solid line, we must also include the equal sign in the inequality.

Therefore the required inequality is$3x-yâ‰\yen 6$

The inequality that is shown by the shaded region in the graph with the boundary line$3x-y=6$is$3x-yâ‰\yen 6$.