Solution:

The given inequalities are \[4x+3y\le 60\], \[y\ge 2x\], \[x\ge 3\], \[x,y\ge 0\]

For  \[4x+3y\le 60\],

Let us put value of \[x=0\] and \[y=0\] in equation one by one, we get

\[y=20\]and \[x=15\]

We get the required points as \[(0,20)\]and \[(15,0)\]

To check if the origin is included in the line`s graph \[(0,0)\]

\[0\le 60\], which is true, hence the origin would lie in the solution area.

Therefore, the required area would include be on the left of the line`s graph.

We have \[y\ge 2x\],

Let us put value of \[x=0\] and \[y=0\] in equation one by one, we get

\[y=0\] and \[x=0\]

So the line pass through origin.

In order to check which side should be included in the line`s graph solution area, we would check for point \[(15,0)\]

⇒ \[0\ge 15\], which is not true so the required solution area would be to the left of the line’s graph.

Consider, \[x\ge 3\],

Based on \[x\ge 3\]we can say that for any value of y, the value of x would be same.

Also the origin \[(0,0)\]doesn’t satisfies the inequality as \[0\ge 3\]

Therefore, the origin doesn’t lies in the solution area and hence the required solution area would be the right of the line`s graph.

We have \[x,y\ge 0\]

Since given both x and y are greater than \[0\]

Therefore, the solution area would be in the first quadrant only.

In the below graph the shaded area in the graph is the required solution of the given inequalities.