Solution:

Given inequalities  \[3x+2y\le 12,x\ge 1,y\ge 2\]

Let us solve values of x and y by putting \[x=0\]and \[y=0\]one by one we get,

\[y=6\]and \[x=4\]

We get the points as \[(0,6)\] and \[(4,0)\]

Now let us check for \[(0,0)\]

\[0\le 12\]satisfies the condition

Therefore, the origin lies in the plane and the required area is toward the left of the equation.

Now let us check for x ≥ 1,

Now the value of x would be unaffected by any value of y

The origin would not lie on the plane

⇒ \[0\le 1\]which is not true

We know that the required area to be included would be on the left of the graph \[x\ge 1\]

again, for \[y\ge 2\]

we can see that the value of y will be unaffected by any value of x in the given equality. Also, the origin doesn’t satisfy the given inequality.

⇒ \[0\ge 2\]which is not true, hence origin should not include in the solution of the inequality.

Therefore, region to be included in the solution would be towards the left of the equality \[y\ge 2\]

The shaded region in the graph is the answer for the inequalities, as it satisfies all the inequalities.