*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 5.
Chapter 6 Linear Inequalities of Class 11 Maths is an important topic which is included in the CBSE Syllabus for 2023-24. The third exercise of this chapter contains 15 questions on the topic Solution of System of Linear Inequalities in Two Variables. The solution of linear inequalities in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are substituted into the inequality. The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. The problems present in Exercise 6.3 of NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities are based on this concept.
The questions in the NCERT textbook are solved by following the steps and methods as per the latest CBSE Syllabus and its guidelines. Hence, students aiming to score high marks for Mathematics subject in the annual examination should practise the problems referring to the NCERT Solutions for Class 11 Maths.
Download the PDF of NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities Exercise 6.3
Solutions of Class 11 Maths Chapter 6 – Exercise 6.3
Solve the following system of inequalities graphically:
1. x ≥ 3, y ≥ 2
Solution:
Given x ≥ 3……… (i)
y ≥ 2…………… (ii)
Since x ≥ 3 means for any value of y, the equation will be unaffected, so similarly, for y ≥ 2, for any value of x, the equation will be unaffected.
Now putting x = 0 in the (i)
0 ≥ 3, which is not true
Putting y = 0 in (ii)
0 ≥ 2, which is not true again
This implies the origin doesn’t satisfy the given inequalities. The region to be included will be on the right side of the two equalities drawn on the graphs.
The shaded region is the desired region.
2. 3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
Given 3x+ 2y ≤ 12
Solving for the value of x and y by putting x = 0 and y = 0 one by one,
We get
y = 6 and x = 4
So the points are (0, 6) and (4, 0)
Now checking for (0, 0)
0 ≤ 12, which is also true
Hence the origin lies in the plane and the required area is toward the left of the equation.
Now checking for x ≥ 1,
The value of x would be unaffected by any value of y
The origin would not lie on the plane
⇒ 0 ≥ 1, which is not true
The required area to be included would be on the left of the graph x ≥1
Similarly, for y ≥ 2
The value of y will be unaffected by any value of x in the given equality. Also, the origin doesn’t satisfy the given inequality.
⇒ 0 ≥ 2, which is not true; hence origin is not included in the solution of the inequality.
The region to be included in the solution would be towards the left of the equality y≥ 2
The shaded region in the graph will give the answer to the required inequalities as it is the region which is covered by all the given three inequalities at the same time while satisfying all the given conditions.
3. 2x + y ≥ 6, 3x + 4y ≤ 12
Solution:
Given 2x + y ≥ 6…………… (i)
3x + 4y ≤ 12 ……………. (ii)
2x + y ≥ 6
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 6 and x = 3
So the point for the (0, 6) and (3, 0)
Now checking for (0, 0)
0 ≥ 6, which is not true; hence the origin does not lie in the solution of equality. The required region is on the right side of the graph.
Checking for 3x + 4y ≤ 12
Putting values of x= 0 and y = 0 one by one in the equation
We get y = 3, x = 4
The points are (0, 3), (4, 0)
Now checking for origin (0, 0)
0 ≤ 12, which is true,
So the origin lies in the solution of the equation.
The region on the right of the equation is the region required.
The solution is the region which is common to the graphs of both inequalities.
The shaded region is the required region.
4. x + y ≥ 4, 2x – y < 0
Solution:
Given x + y ≥ 4
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 4 and x = 4
The points for the line are (0, 4) and (4, 0)
Checking for the origin (0, 0)
0 ≥ 4
This is not true,
So the origin would not lie in the solution area. The required region would be on the right of the line graph.
2x – y < 0
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y= 0 and x = 0
Putting x = 1 we get y = 2
So the points for the given inequality are (0, 0) and (1, 2)
Now that the origin lies on the given equation, we will check for the (4, 0) point to check which side of the line graph will be included in the solution.
⇒ 8 < 0, which is not true, hence the required region would be on the left side of the line 2x-y < 0
The shaded region is the required solution to the inequalities.
5. 2x – y >1, x – 2y < – 1
Solution:
Given 2x – y >1……………… (i)
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = -1 and x = 1/2 = 0.5
The points are (0,-1) and (0.5, 0)
Checking for the origin, putting (0, 0)
0 >1, which is false
Hence the origin does not lie in the solution region. The required region would be on the right of the line graph.
x – 2y < – 1………… (ii)
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = ½ = 0.5 and x = -1
The required points are (0, 0.5) and (-1, 0)
Now checking for the origin, (0, 0)
0 < -1, which is false
Hence, the origin does not lie in the solution area, the required area would be on the left side of the line graph.
∴ the shaded area is the required solution for the given inequalities.
6. x + y ≤ 6, x + y ≥ 4
Solution:
Given x + y ≤ 6,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
Y = 6 and x = 6
The required points are (0, 6) and (6, 0)
Checking further for origin (0, 0)
We get 0 ≤ 6; this is true.
Hence, the origin would be included in the area of the line graph. So the required solution of the equation would be on the left side of the line graph, which will be including the origin.
x + y ≥ 4
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 4 and x = 4
The required points are (0, 4) and (4, 0)
Checking for the origin (0, 0)
0 ≥ 4, which is false
So the origin would not be included in the required area. The solution area will be above the line graph or the area on the right of the line graph.
Hence the shaded area in the graph is the required graph area.
7. 2x + y ≥ 8, x + 2y ≥ 10
Solution:
Given 2x + y ≥ 8
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 8 and x = 4
The required points are (0, 8) and (4, 0)
Checking if the origin is included in the line graph (0, 0)
0 ≥ 8, which is false
Hence, the origin is not included in the solution area, and the required area would be the area to the right of the line graph.
x + 2y ≥ 10
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 5 and x = 10
The required points are (0, 5) and (10, 0)
Checking for the origin (0, 0)
0 ≥ 10, which is false,
Hence, the origin would not lie in the required solution area. The required area would be to the left of the line graph.
The shaded area in the graph is the required solution for the given inequalities.
8. x + y ≤ 9, y > x, x ≥ 0
Solution:
Given x + y ≤ 9,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 9 and x = 9
The required points are (0, 9) and (9, 0)
Checking if the origin is included in the line graph (0, 0)
0 ≤ 9
This is true, so the required area would be including the origin and hence will lie on the left side of the line graph.
y > x,
Solving for y = x
We get x= 0, y = 0, so the origin lies on the line graph.
The other points would be (0, 0) and (2, 2)
Checking for (9, 0) in y > x,
We get 0 > 9, which is false since the area would not include the area below the line graph and hence would be on the left side of the line.
We have x ≥ 0
The area of the required line graph would be on the right side of the line graph.
Therefore the shaded area is the required solution for the given inequalities.
9. 5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
Given 5x + 4y ≤ 20,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 5 and x= 4
The required points are (0, 5) and (4, 0)
Checking if the origin lies in the solution area (0, 0)
0 ≤ 20
This is true; hence the origin would lie in the solution area. The required area of the line graph is on the left side of the graph.
We have x ≥ 1,
For all the values of y, x would be 1,
The required points would be (1, 0), (1, 2) and so on.
Checking for origin (0, 0)
0 ≥ 1, which is not true
So the origin would not lie in the required area. The required area on the graph will be on the right side of the line graph.
Consider y ≥ 2
Similarly, for all the values of x, y would be 2.
The required points would be (0, 2), (1, 2) and so on.
Checking for origin (0, 0)
0 ≥ 2, this is not true
Hence, the required area would be on the right side of the line graph.
The shaded area on the graph shows the required solution for the given inequalities.
10. 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
Given 3x + 4y ≤ 60,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 15 and x = 20
The required points are (0, 15) and (20, 0)
Checking if the origin lies in the required solution area (0, 0)
0 ≤ 60, this is true.
Hence, the origin would lie in the solution area of the line graph.
The required solution area would be on the left of the line graph.
We have x + 3y ≤ 30,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 10 and x = 30
The required points are (0, 10) and (30, 0)
Checking for the origin (0, 0)
0 ≤ 30, this is true.
Hence, the origin lies in the solution area, which is given by the left side of the line graph.
Consider x ≥ 0,
y ≥ 0,
The given inequalities imply the solution lies in the first quadrant only.
Hence, the solution to the inequalities is given by the shaded region in the graph.
11. 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution:
Given 2x + y ≥ 4,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 4 and x =2
The required points are (0, 4) and (2, 0)
Checking for origin (0, 0)
0 ≥ 4, this is not true
Hence, the origin doesn’t lie in the solution area of the line graph. The solution area would be given by the right side of the line graph.
x + y ≤ 3,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 3 and x = 3
The required points are (0, 3) and (3, 0)
Checking for the origin (0, 0)
0 ≤ 3, this is true
Hence, the solution area would include the origin and hence would be on the left side of the line graph.
2x – 3y ≤ 6
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = -2 and x = 3
The required points are (0, -2), (3, 0)
Checking for the origin (0, 0)
0 ≤ 6 this is true
So the origin lies in the solution area, and the area would be on the left of the line graph.
Hence, the shaded area in the graph is the required solution area for the given inequalities.
12. x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0 , y ≥ 1
Solution:
Given, x – 2y ≤ 3
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = -3/2 = -1.5 and x = 3
The required points are (0, -1.5) and (3, 0)
Checking for the origin (0, 0)
0 ≤ 3, this is true.
Hence, the solution area would be on the left of the line graph
3x + 4y ≥ 12,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 3 and x = 4
The required points are (0, 3) and (4, 0)
Checking for the origin (0, 0)
0 ≥ 12, this is not true
So the solution area would include the origin, and the required solution area would be on the right side of the line graph.
We have x ≥ 0,
For all the values of y, the value of x would be the same in the given inequality, which would be the region above the x-axis on the graph.
Consider, y ≥ 1
For all the values of x, the value of y would be the same in the given inequality.
The solution area of the line would not include the origin, as 0 ≥ 1 is not true.
The solution area would be on the left side of the line graph.
The shaded area in the graph is the required solution area which satisfies all the given inequalities at the same time.
13. 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution:
Given, 4x + 3y ≤ 60,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 20 and x = 15
The required points are (0, 20) and (15, 0)
Checking for the origin (0, 0)
0 ≤ 60, this is true.
Hence, the origin would lie in the solution area. The required area would be included on the left of the line graph.
We have y ≥ 2x,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 0 and x = 0
Hence, the line would pass through the origin.
To check which side would be included in the line graph solution area, we would check for the point (15, 0)
⇒ 0 ≥ 15, this is not true, so the required solution area would be to the left of the line graph.
Consider, x ≥ 3,
For any value of y, the value of x would be the same.
Also, the origin (0, 0) doesn’t satisfy the inequality as 0 ≥ 3
So the origin doesn’t lie in the solution area; hence the required solution area would be the right of the line graph.
We have x, y ≥ 0
Since given both x and y are greater than 0
∴ the solution area would be in the first Ist quadrant only.
The shaded area in the graph shows the solution area for the given inequalities
14. 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution:
Given, 3x + 2y ≤ 150
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 75 and x = 50
The required points are (0, 75) and (50, 0)
Checking for the origin (0, 0)
0 ≤ 150, this is true
Hence, the solution area for the line would be on the left side of the line graph, which would include the origin too.
We have x + 4y ≤ 80,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 20 and x = 80
The required points are (0, 20) and (80, 0)
Checking for the origin (0, 0)
0 ≤ 80, this is also true, so the origin lies in the solution area.
The required solution area would be toward the left of the line graph.
Given x ≤ 15,
For all the values of y, x would be the same
Checking for the origin (0, 0)
0 ≤ 15, this is true, so the origin would be included in the solution area. The required solution area would be towards the left of the line graph.
Consider y ≥ 0, x ≥ 0
Since x and y are greater than 0, the solution would lie in the 1st quadrant.
The shaded area in the graph satisfies all the given inequalities and hence is the solution area for given inequalities.
15. x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution:
Given, x + 2y ≤ 10,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 5 and x = 10
The required points are (0, 5) and (10, 0)
Checking for the origin (0, 0)
0 ≤ 10, this is true.
Hence, the solution area would be toward the origin, including the same. The solution area would be toward the left of the line graph.
We have x + y ≥ 1,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 1 and x = 1
The required points are (0, 1) and (1, 0)
Checking for the origin (0, 0)
0 ≥ 1, this is not true.
Hence, the origin would not be included in the solution area. The required solution area would be toward the right of the line graph.
Consider x – y ≤ 0,
Putting values of x = 0 and y = 0 in the equation one by one, we get values of
y = 0 and x = 0
Hence, the origin would lie on the line.
To check which side of the line graph would be included in the solution area, we would check for the (10, 0)
10 ≤ 0, which is not true hence the solution area would be on the left side of the line graph.
Again we have x ≥ 0, y ≥ 0
Since both x and y are greater than 0, the solution area would be in the 1st quadrant.
Hence, the solution area for the given inequalities would be the shaded area of the graph satisfying all the given inequalities.
Access Other Exercise Solutions of Class 11 Maths Chapter 6 – Linear Inequalities
Exercise 6.1 Solutions 26 Questions
Exercise 6.2 Solutions 10 Questions
Miscellaneous Exercise on Chapter 6 Solutions 14 Questions
Also explore – NCERT Class 11 Solutions
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