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  • Solve the following system of inequalities graphically: 4x plus 3y less than or equal to 60, y greater than or equal to 2x, x is greater than 3, x, y greater than 0

Solve the following system of inequalities graphically:

    Q13.    4x + 3y \leq 60,\ y\geq 2x,\ x\geq 3,\ x,y\geq 0

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4x + 3y \leq 60,\ y\geq 2x,\ x\geq 3,\ x,y\geq 0

Graphical representation of  4x+3y=60 \, \, ,y=2x\, \, \,,x=3\, \, ,x=0\, \, and\, \, y=0  is given in graph below.

For 4x + 3y \leq 60, 

The  solution to this inequality is region below the  line (4x+3y=60) including points on this line because points on the line also satisfy the inequality.

 For y\geq 2x,

The  solution to this inequality is region above the line (y=2x)  including points on this line because  points on the line also satisfy the inequality.

 For x\geq 3,

The  solution to this inequality is region right hand side of  the line (x=3)  including points on this line because  points on the line also satisfy the inequality.

For \ x \geq 0,

The  solution to this inequality is region right hand side of the line (x=0) including points on this line because  points on  the line also satisfy the inequality.

For \ y \geq 0,

The  solution to this inequality is region above  the line (y=0) including points on this line because  points on line also satisfy the inequality.

 

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows: 

Posted by

seema garhwal

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