Solve the following system of inequalities graphically: \[x-2y\le 3\], \[3x+4y\ge 12\], \[x\ge 0\], \[y\ge 1\]

Class 11 | Exercise 6.3 | Linear Inequalities | Maths | NCERT

Solution:

The given inequalities are \[x-2y\le 3\], \[3x+4y\ge 12\], \[x\ge 0\], \[y\ge 1\]

For \[x-2y\le 3\]

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

\[y=-1.5\]and \[x=3\]

We get the required points as \[(0,-1.5)\]and \[(3,0)\]

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

\[0\le 3\], which is true and hence the solution area would be on the left of the line`s graph

For \[3x+4y\ge 12\],

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

\[y=3\]and \[x=4\]

We get the required points as \[(0,3)\]and \[(0,4)\]

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

\[0\ge 12\], which is not true

Therefore, the solution area would of include the origin and the required solution area would be on the right side of the line`s graph.

For \[x\ge 0\],

We can say that all the values of y, the value of x would be same in the given inequality, which would be the region above the x axis on the graph.

For \[y\ge 1\]

We can say that for all the values of x, the value of y would be same in the given inequality.

As \[y\ge 1\],the origin is not included

Therefore, the solution area would be on the left side of the line`s graph.

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

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