Solve the inequality graphically: y + 3 ≥ 0 .

Given y + 3 ≥ 0 Subtract both side by 3 ⇒ y ≥ - 3 x 0 - 1 4 y - 3 - 3 - 3 Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in the given inequality y + 3 ≥ 0 we get 3 ≥ 0 Which is true, Hence, origin lies in plane. So, we shade upward region of line.

Solve the inequality Graphically: 5 x + 9 ≥ 0 .

Given 5 x + 9 ≥ 0 Subtract both sides by 9 . ⇒ 5 x ≥ - 9 x ≥ - 9 5 x ≥ - 1 . 8 x - 1 . 8 - 1 . 8 - 1 . 8 - 1 . 8 y 0 - 1 1 4 ,Draw a line x = - 1 . 8 in X Y coordinate system. Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in the given inequality 5 x + 9 ≥ 0 ⇒ 9 ≥ 0 Which is true, Therefore, origin lies in plane. Hence, we shade the solution region as right side of the line.

Solve the inequality graphically: x + y ≤ 5 .

Given x + y ≤ 5 x 0 5 3 y 5 0 2 Using the above table, draw a line x + y = 5 on rectangular coordinate system. Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in x + y ≤ 5 ⇒ 0 ≤ 5 Which is true, Hence, origin lies in plane. So, we shade the solution region as downward region of line.

Solve the inequality Graphically: 2 x + y ≥ 6 .

Given 2 x + y ≥ 6 Putting x = 0 in ( 1 ) 2 ( 0 ) + y = 6 0 + y = 6 y = 6 Putting y = 0 in ( 1 ) 2 x + ( 0 ) = 6 x = 6 2 x = 3 x 0 3 y 6 0 Using the above table, draw a line 2 x + y = 6 Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in 2 x + y ≥ 6 ⇒ 0 ≥ 6 Which is not true, So, origin does not lies in plane. Hence, the solution is upward region of the line.

Solve the Inequality Graphically: 3 y - 2 x ≤ 4 .

Given 3 y - 2 x ≤ 4 x 0 - 2 2 y 1 . 3 0 8 Using the above table, draw a line 3 y - 2 x = 4 in X Y co-ordinate system Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in 3 y - 2 x ≤ 4 ⇒ 0 ≤ 4 Which is true, Hence, origin lies in plane. So, the solution is downward region of the line.

Solve the inequality graphically: y + 8 ≥ 2 x .

Given y + 8 ≥ 2 x x 0 4 y - 8 0 Using the above table, draw a line y + 8 = 2 x on rectangular coordinate system Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in y + 8 ≥ 2 x ⇒ 8 ≥ 0 Which is true, So, origin lies in plane. Hence, the solution is upward region of the line.

Solve the inequality graphically: 0 ≤ 2 x - 5 y + 10

Given 0 ≤ 2 x - 5 y + 10 x 0 - 5 y 2 0 Using the above table, draw a line 2 x - 5 y = - 10 on rectangular coordinate system. Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in 0 ≤ 2 x - 5 y + 10 ⇒ 0 ≤ 10 Which is true, So, origin lies in plane. Hence, the solution is downward region of the line.

Solve the inequality graphically: x ≤ 8 - 4 y .

Given x ≤ 8 - 4 y x 0 8 y 2 0 Using the above table, draw a line x = 8 - 4 y on X Y rectangular co-ordinate system Checking for ( 0 , 0 ) Putting x = 0 , y = 0 in x ≤ 8 - 4 y ⇒ 0 ≤ 8 Which is true, So, origin lies in plane. Hence, the solution is downward region of the line.

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 7 - Linear Inequalities>EXERCISE 7.3

M L Aggarwal Solutions for Chapter: Linear Inequalities, Exercise 3: EXERCISE 7.3

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Linear Inequalities, Exercise 3: EXERCISE 7.3

Attempt the practice questions on Chapter 7: Linear Inequalities, Exercise 3: EXERCISE 7.3 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 1 solutions are prepared by Experienced Embibe Experts.