Solution:
The given inequalities are \[2x+y\ge 4\], \[x+y\le 3\], \[2x-3y\le 6\]
For \[2x+y\ge 4\],
Let us put value of \[x=0\] and \[y=0\] in equation one by one, we get
\[y=4\]and \[x=2\]
We get the required points as \[(0,4)\]and \[(2,0)\]
To check if the origin is included in the line`s graph \[(0,0)\]
\[0\ge 4\], which is not true, hence the origin doesn’t lies in the solution area of the line`s graph. The solution area would be given by the right side of the line`s graph.
For \[x+y\le 3\],
Now, put value of \[x=0\] and \[y=0\] in equation one by one, we get
\[y=3\]and \[x=3\]
We get the required points as \[(0,3)\]and \[(3,0)\]
To check if the origin is included in the line`s graph \[(0,0)\]
\[0\le 3\], which is true, hence the solution area would include the origin and hence would be on the left side of the line`s graph.
For \[2x-3y\le 6\]
Let us put value of \[x=0\] and \[y=0\] in equation one by one, we get
\[y=-2\]and \[x=3\]
We get the required points as \[(0,-2)\], \[(3,0)\]
To check if the origin is included in the line`s graph \[(0,0)\]
\[0\le 6\], which is true
Therefore, the origin lies in the solution area and the area would be on the left of the line`s graph.
In the below graph the shaded area in the graph is the required solution of the given inequalities.
