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  • Solve the following system of inequality graphically: 3x + 2 y less than or equal to 150, x + 4y less than or equal to 80, x less than or equal to 15, y greater than or equal to 0, x greater than or equal to 0

Solve the following system of inequality graphically:

    Q14.    3x + 2y \leq 150, \ x +4y \leq 80,\ x\leq 15 \ y\geq 0, \ x\geq 0

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3x + 2y \leq 150, \ x +4y \leq 80,\ x\leq 15 \ y\geq 0, \ x\geq 0

Graphical representation of  3x+2y=150 \, \, ,x+4y=80\, \, \,,x=15\, \, ,x=0\, \, and\, \, y=0  is given in graph below.

For 3x + 2y \leq 150, 

The  solution to this inequality is region below the  line (3x+2y=150) including points on this line because points on the line also satisfy the inequality.

 For x+4y\leq 80,

The  solution to this inequality is region below the line (x+4y=80)  including points on this line because points on the line also satisfy the inequality.

 For x\leq 15,

The  solution to this inequality is region left hand side of  the line (x=15)  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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