NCERT Solutions for Exercise 6.3 Class 11 Maths Chapter 6 - Linear Inequalities

Question:3 Solve the following system of inequalities graphically: $2x +y \geq 6, 3x +4y\leq 12$

Answer:

$2x +y \geq 6, 3x +4y\leq 12$

Graphical representation of $2x +y =6\, \, and\, \, 3x +4y=12$ is given in the graph below.

For $2x +y \geq 6$ ,

The solution to this inequality is region above line $(2x +y =6)$ including points on this line because points on the line also satisfy the inequality.

For $3x +4y\leq 12$ ,

The solution to this inequality is region below the line $( 3x +4y= 12)$ including points on this line because points on the line also satisfy the inequality.

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

This can be represented as follows:

Question:4 Solve the following system of inequalities graphically: $x + y \geq 4, 2x - y <0$

Answer:

$x + y \geq 4, 2x - y <0$

Graphical representation of $x +y =4\, \, and\, \, 2x -y=0$ is given in the graph below.

For $x + y \geq 4,$ ,

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

For $2x - y <0$ ,

The solution to this inequality is half plane corresponding to the line $( 2x -y=0)$ containing point $(1,0)$ excluding points on this line because points on the line does not satisfy the inequality.

Hence, the solution to these linear inequalities is the shaded region as shown in figure including points on line $(x +y =4)$ and excluding points on the line $( 2x -y=0)$ .

This can be represented as follows:

Question:5 Solve the following system of inequalities graphically: $2x - y > 1, \ x -2y < -1$

Answer:

$2x - y > 1, \ x -2y < -1$

Graphical representation of $x -2y =-1\, \, and\, \, 2x -y=1$ is given in graph below.

For $2x - y > 1,$

The solution to this inequality is region below line $( 2x -y=1)$ excluding points on this line because points on line does not satisfy the inequality.

For $\ x -2y < -1$ ,

The solution to this inequality is region above the line $(x -2y =-1)$ excluding points on this line because points on line does not satisfy the inequality.

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

This can be represented as follows:

Question:6 Solve the following system of inequalities graphically: $x + y \leq 6, x + y \geq 4$

Answer:

$x + y \leq 6, x + y \geq 4$

Graphical representation of $x + y = 6,\, \, and\, \, \, x + y = 4$ is given in the graph below.

For $x + y \leq 6,$

The solution to this inequality is region below line $( x+y=6)$ in cluding points on this line because points on the line also satisfy the inequality.

For $x + y \geq 4$ ,

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

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

This can be represented as follows:

Question:7 Solve the following system of inequalities graphically: $2x + y \geq 8 , x + 2y \geq 10$

Answer:

$2x + y \geq 8 , x + 2y \geq 10$

Graphical representation of $2x + y = 8\, \, and\, \, x + 2y =10$ is given in graph below.

For $2x + y \geq 8 ,$

The solution to this inequality is region above line $(2x + y = 8)$ including points on this line because points on line also satisfy the inequality.

For $x + 2y \geq 10$ ,

The solution to this inequality is region above the line $( x + 2y =10)$ 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 lines.

This can be represented as follows:

Question:8 Solve the following system of inequalities graphically: $x + y \leq 9, y > x, x\geq 0$

Answer:

$x + y \leq 9, y > x, x\geq 0$

Graphical representation of $x+y=9,x=y$ and $x=0$ is given in graph below.

For $x + y \leq 9$ ,

The solution to this inequality is region below line $(x+y=9)$ including points on this line because points on line also satisfy the inequality.

For $y > x$ ,

The solution to this inequality represents half plane corresponding to the line $(x=y)$ containing point $(0,1)$ excluding points on this line because points on line does not satisfy the inequality.

For $x\geq 0$ ,

The solution to this inequality is region on right hand side of the line $(x=0)$ 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.

This can be represented as follows:

Question:9 Solve the following system of inequalities graphically: $5x+4y\leq20, \ x\geq 1, \ y\geq 2$

Answer:

$5x+4y\leq20, \ x\geq 1, \ y\geq 2$

Graphical representation of $\, ,5x+4y=20,\, \, \, x=1\, \, and \, \, y=2$ is given in graph below.

For $5x+4y\leq20,$ ,

The solution to this inequality is region below the line $(5x+4y=20)$ including points on this line because points on line also satisfy the inequality.

For $\ x\geq 1,$ ,

The solution to this inequality is region right hand side of the line $(x=1)$ including points on this line because points on line also satisfy the inequality.

For $\ y\geq 2,$

The solution to this inequality is region above the line $(y=2)$ 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:

Question:10 Solve the following system of inequalities graphically: $3x + 4y \leq 60,\ x + 3y \leq 30, \ x \geq 0, \ y\geq 0$

Answer:

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

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:

Question:11 Solve the following system of inequalities graphically: $2x +y \geq 4, \ x + y \leq 3, \ 2x - 3y \leq 6$

Answer:

$2x +y \geq 4, \ x + y \leq 3, \ 2x - 3y \leq 6$

Graphical representation of $2x+y=4 \, \, ,x+y=3$ and $2x-3y=6$ is given in graph below.

For $2x +y \geq 4,$ ,

The solution to this inequality is region above the line $(2x+y=4)$ including points on this line because points on line also satisfy the inequality.

For $\ x + y \leq 3,$ ,

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

For $\ 2x - 3y \leq 6,$

The solution to this inequality is region above the line $(2x-3y= 6)$ 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:

Question:12 Solve the following system of inequalities graphically: $x -2y \leq 3, 3x + 4y \geq 12, x \geq 0, y\geq 1$

Answer:

$x -2y \leq 3, 3x + 4y \geq 12, x \geq 0, y\geq 1$

Graphical representation of $x-2y=3 \, \, ,3x+4y=12\, \, \, ,x=0\, \, and\, \, y=1$ is given in graph below.

For $x -2y \leq 3$ ,

The solution to this inequality is region above the line $(x-2y=3)$ including points on this line because points on line also satisfy the inequality.

For $3x + 4y \geq 12$ ,

The solution to this inequality is region above the line $(3x+4y=12)$ 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 $\ y \geq 1,$

The solution to this inequality is region above the line $(y=1)$ 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:

Question:13 Solve the following system of inequalities graphically: $4x + 3y \leq 60,\ y\geq 2x,\ x\geq 3,\ x,y\geq 0$

Answer:

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

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:

Question:14 Solve the following system of inequality graphically: $3x + 2y \leq 150, \ x +4y \leq 80,\ x\leq 15 \ y\geq 0, \ x\geq 0$

Answer:

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

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:

Question:15 Solve the following system of inequality graphically: $x+2y \leq 10, \ x +y \geq 1, \ x-y\leq 0, x\geq 0, \ y\geq 0$

Answer:

$x+2y \leq 10, \ x +y \geq 1, \ x-y\leq 0, x\geq 0, \ y\geq 0$

Graphical representation of $x+2y=10 \, \, ,x+y=1\, \, \,,x-y=0\, \, ,x=0\, \, and\, \, y=0$ is given in graph below.

For $x+2y \leq 10,$

The solution to this inequality is region below the line $(x+2y=10)$ including points on this line because points on line also satisfy the inequality.

For $\ x +y \geq 1,$ ,

The solution to this inequality is region above the line $(x+y=1)$ including points on this line because points on line also satisfy the inequality.

For $\ x-y\leq 0,$ ,

The solution to this inequality is region above the line $(x-y=0)$ 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, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows: