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  • Solve the following system of inequalities graphically: 3x + 4y less than or equal to 60, x + 3y less than or equal to 30, x greater than or equal to 0, y greater than or eqaul to 0

Q: Solve the following system of inequalities graphically: 3x + 4y \leq 60,\ x + 3y \leq 30, \ x \geq 0, \ y\geq 0

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3x + 4y \leq 60,\ x + 3y \leq 30, \ x \geq 0, \ y\geq 0

Graphical representation of  3x+4y=60 \, \, ,x+3y=30\, \, \, ,x=0\, \, and\, \, y=0  is given in graph below.

For 3x + 4y \leq 60 

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

 For \ x + 3y \leq 30

The  solution to this inequality is region below the line (x+3y=30)  including points on this line because  points on 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 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, the 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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