NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3.

• Linear Inequalities Class 11 Ex 6.1
• Linear Inequalities Class 11 Ex 6.2

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 6
Chapter Name	Linear Inequalities
Exercise	Ex 6.3
Number of Questions Solved	15
Category	NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.3

Solve the following system of inequalities graphically:
Question 1.
x ≥ 3, y ≥ 2
Solution.
x ≥ 3, y ≥ 2
(i) AB represents the line x = 3
Putting x = 0 in x ≥ 3
0 ≥ 3, which is not true.
Therefore, origin does not lie in the region of x ≥ 3
Its graph is shaded on the right side of AB.

(ii) CD represents the line y = 2
Putting y = 0 in y ≥ 2 0 ≥ 2, which is not true.
Therefore, origin does not lie in the region of y ≥ 2.
Its graph is shaded above the CD.
Solution of system x ≥ 3 and y ≥ 2 is shown as the shaded region.

Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution.
Inequalities are 3x + 2y ≤ 12, x ≥ 1, y ≥ 2
(i) The line l₁ : 3x + 2y = 12 passes through (4, 0), (0, 6)
AB represents the line, 3x + 2y = 12.
Consider the inequality 3x + 2y ≤ 12
Putting x = 0, y = 0 in 3x + 2y ≤ 12
0 + 0 ≤ 12, which is true.
Therefore, origin lies in the region 3x + 2y ≤ 12
∴ The region lying below the line AB including all the points lying on it.

(ii) The line l₂ : x = 1 passes through (1, 0). This line is represented by EF. Consider the inequality x ≥ 1
Putting x = 0, 0 ≥ 1, which is not true.
Therefore, origin does not lie in the region of x ≥ 1
The region lies on the right of EF and the points on EF from the inequality x ≥ 1.

(iii) The line l₃ : y = 2 passes through (0, 2). This line is represented by CD.
Consider the inequality y ≥ 2
Putting y = 0 in y ≥ 2, we get 0 ≥ 2 which is false.
∴ Origin does not lie in the region of y ≥ 2. y ≥ 2 is represented by the region above CD and all the points on this line. Hence, the region satisfying the inequalities.
3x + 2y ≥ 12, x ≥ 1, y ≥ 2 is the APQR.

Question 3.
2x + y ≥ 6, 3x + 4y ≤ 12
Solution.
The inequalities are 2x + y ≥ 6, 3x + 4y ≤ 12
(i) The line l₁ : 2x + y = 6 passes through (3, 0), (0, 6)
AB represents the line 2x + y = 6
Putting x = 0, y = 0 in 2x + y ≥ 6 0 ≥ 6, which is false.
∴ Origin does not lie in the region of 2x + y ≥ 6 Therefore, the region lying above the line AB and all the points on AB represents the inequality 2x + y ≥ 6

(ii) The line l₂ : 3x + 4y = 12 passes through (4, 0) and (0, 3).
This line is represented by CD.

Consider the inequality 3x + 4y ≤ 12
Putting x = 0, y = 0 in 3x + 4y ≤ 12, we get 0 ≤ 12, which is true.
∴ 3x + 4y ≤ 12 represents the region below the line CD (towards origin) and all the points lying on it.
The common region is the solution of 2x + 3y ≥ 6 are 3x + 4y ≤ 12 represented by the
shaded region in the graph.

Question 4.
x + y ≥ 4, 2x – y > 0
Solution.
The inequalities are , x + y ≥ 4, 2x – y > 0
(i) The line l₁: x + y = 4 passes through (4, 0) and (0, 4). This line is represented by AB. Consider the inequality x + y ≥ 4
Putting x = 0, y = 0 in x + y ≥ 4, we get 0 ≥ 4, which is false.
Origin does not lie in this region.

Therefore, ,r + y > 4 is represented by the region above the line x + y = 4 and all points lying on it.

(ii) The line l₂ : 2x – y = 0 passes through (0, 0) and (1, 2).
This line is represented by CD.
Consider the inequality 2x – y > 0
Putting x = 1, y = 0, we get 2 > 0, which is true
This shows (1, 0) lies in the region.
i.e. region lying below the line 2x – y = 0
represents 2x — y > 0
∴ The common region to both inequalities is shaded region as shown in the figure.

Question 5.
2x – y> 1, x – 2y < -1
Solution.
The inequalities are 2x – y > 1 and x – 2y < -1

(i) Let us draw the graph of line
l₁ : 2x – y = 1, passes through and
(0, -1) which is represented by AB. Consider the inequality 2x – y > 1.
Putting x = y = 0, we get 0 > 1, which is false.
Therefore, origin does not lie in region of 2x – y > 1 i.e., 2x – y > 1 represents the area below the line AB excluding all the points lying on 2x – y = 1.

(ii) Let us draw the graph of the line
l₂ : x – 2y = -1, passes through (-1, 0) and (0,1/2) which is represented by CD.
Consider the inequality x – 2y < -1
Putting x = y = 0, we have 0 < -1, which is false.
Therefore, origin does not lie in region of x – 2y < -1 i.e., x – 2y < -1 represents the area above the line CD excluding all the points lying on x – 2y = -1
⇒ The common region of both the inequality is the shaded region as shown in figure.

Question 6.
x + y ≤ 6, x + y ≥ 4
Solution.
The inequalities are
x + y ≤ 6 and x + y ≥ 4
(i) The line l₁: x + y = 6 passes through (6, 0) and (0, 6). It is represented by AB.
Consider the inequality x + y ≤ 6
Putting x = 0, y = 0 in x + y ≤ 6 0 ≤ 6, which is true.

∴ Origin lies in the region of x + y ≤ 6
∴ x + y ≤ 6 is represented by the region below the line x + y = 6 and all the points lying on it.

(ii) The line l₂ : x + y = 4 passes through (4, 0) and (0, 4). It is represented by CD.
Consider the inequality x + y ≥ 4
Putting x = 0, y = 0 in x + y ≥ 4 or, 0 + 0 ≥ 4, which is false.
∴ Origin does not lie in the region of x + y ≥ 4
∴ x + y ≥ 4 is represented by the region above the line x + y = 4 and all the points lying on it.
∴ The solution region is the shaded region between AB and CD as shown in the figure.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution.
The inequalities are 2x + y ≥ 8 and x + 2y ≥ 10
(i) Let us draw the graph of the line
l₁ : 2x + y = 8 passes through (4, 0) and (0, 8) which is represented by AB.
Consider the inequality 2x + y ≥ 8

Putting x = y = 0, we get 0 ≥ 8, which is false.
∴ Origin does not lie in the region of 2x + y ≥ 8.
i.e., 2x + y ≥ 8 represents the area above the line AB and all the points lying on 2x + y = 8.

(ii) Let us draw the graph of line
l₂ : x + 2y = 10, passes through (10, 0) and (0, 5) which is represented by CD. Consider the inequality x + 2y ≥ 10
Putting x = 0, y = 0, we have 0 ≥ 10, which is false.
∴ The origin does not lie in region of x + 2y ≥ 10
i.e. x + 2y ≥ 10 represents the area above the line CD and all the points lying on x + 2y = 10.
⇒ The common region of both the inequality is the shaded region as shown in the figure.

Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution.
The inequalities are x + y ≤ 9, y > x and x ≥ 0
(i) Consider the inequality x + y ≤ 9
The line l₁ : x + y = 9 passes through (9, 0) and (0, 9). AB represents this line.
Putting x = 0, y = 0 in x + y ≤ 9
0 + 0 = 0 ≤ 9, which is true.

Origin lies in this region. i.e., x + y ≤ 9 represents the area below the line AB and all the points lying on x + y = 9.

(ii) The line l₂ : y = x, passes through the origin and (2, 2).
∴ CD represents the line y = x
Consider the inequality y – x > 0
Putting x = 0,y = 1 in y – x > 0
1 – 0 > 0, which is true.
∴ (0, 1) lies in this region.
The inequality y > x is represented by the region above the line CD, excluding all the points lying on y – x = 0.

(iii) The region x ≥ 0 lies on the right of y-axis.
∴ The common region of the inequalities is the region bounded by ΔPQO is the solution of x + y ≤ 9, y > x, x ≥ 0.

Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution.
The inequalities are 5x + 4y ≤ 20, x ≥ 1, y ≥ 2
(i) The line l₁ : 5x + 4y = 20 passes through (4, 0) and (0, 5). This line is represented by AB.
Consider the inequality 5x + 4y ≤ 20
Putting x = 0, y = 0
0 + 0 = 0 ≤ 20, which is true.
The origin lies in this region, i.e., region below the line 5x + 4y = 20 and all the points lying on it belong to 5x + 4y ≤ 20.

(ii) The line l₂ : y = 2, line is parallel to x-axis at a distance 2 from the origin. It is represented by EF. Putting y = 0, 0 ≥ 2 is not true.
Origin does not lie in this region.
Region above y = 2 represents the inequality y ≥ 2 including the points lying on it.

(iii) The line l₃ : x = 1, line parallel to y-axis at a distance 1 from the origin. It is represented by CD. Putting x = 0in x – 1 ≥ 0
-1 ≥ 0, which is not true.
Origin does not lie in this region.
∴ The region on the right of x = 1 and all the points lying on it belong to x ≥ 1.

∴ Shaded area bounded by ΔPQR is the solution of given inequalities.

Question 10.
3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Solution.
The inequalities are

We first draw the graphs of lines
l₁ : 3x + 4y = 60, l₂ : x + 3y – 30, x = 0 and y = 0.
(i) The line 3x + 4y = 60 passes through (20, 0) and (0,15) which is represented by AB. Consider the inequality 3x + 4y ≤ 60, putting x = 0, y = 0 in 3x + 4y ≤ 60, we get 0 + 0 ≤ 60, which is true.
∴ 3x + 4y ≤ 60 represents the region below AB and all the points on AB.

(ii) Further, x + 3y = 30 passes through (0,10) and (30, 0), CD represents this line.
Consider the inequality x + 3y ≤ 30
Putting x = 0, y = 0 in x + 3y < 30, w’e get 0 < 30 is true.
∴Origin lies in the region x + 3y ≤ 30. This inequality represents the region below it and the line itself.
Thus, we note that inequalities (1) and (2) represent the two regions below the respective lines (including the lines).
Inequality (3) represents the region on the right of y-axis and the i/-axis itself.
Inequality (4) represents the region above x-axis and the x-axis itself.
∴ Shaded area in the figure is the solution area.

Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution.
We have the inequalities :


We first draw the graphs of lines
l₁ : 2x + y = 4, l₂ : x + y = 3 and l₃ : 2x – 3y = 6
(i) 2x + y = 4, passes through (2, 0) and (0, 4) which represents by AB
Consider the inequality 2x + y ≥ 4
Putting x = 0, y = 0 in 2x + y ≥ 4, we get 0 ≥ 4 is false.
∴ Origin does not lie in the region of 2x + y ≥ 4 This inequality represents the region above the line AB and all the points on the line AB.

(ii) Again, x + y = 3 is represented by the line CD, passes through (3, 0) and (0, 3). Consider the inequality x + y ≤ 3, putting x = 0,y = 0 in x + y ≤ 3, we get 0 ≤ 3 is true.
∴ Origin lies in the region of x + y ≤ 3
∴ x + y ≤ 3 represents the region below the line CD and all the points on the line CD.

(iii) Further, 2x – 3y = 6 is represented by EF passes through (0, -2) and (3, 0).
Consider the inequality 2x – 3y ≤ 6, putting x = 0, y = 0 in 2x – 3y ≤ 6, we get 0 ≤ 6, which is true.
∴ Origin lies in it.
∴ 2x – 3y ≤ 6 represents the region above the line EF and all the points on the line EF.
∴ Shaded triangular area in the figure is the solution of given inequalities.

Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution.
The inequalities are
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
(i) The line l₁: x – 2y = 3 passes through (3, 0) and
This is represented by AB. Consider the inequality x – 2y ≤ 3, putting x = 0, y = 0 we get 0 ≤ 3, which is true.
⇒ Origin lies in the region of x – 2y ≤ 3.
Region on the above of this line and including its points represents x – 2y ≤ 3

(ii) The line l₂ : 3x + 4y = 12 passes through (4, 0) and (0, 3). CD represents this line. Consider the inequality 3x + 4y ≥12 putting x = 0, y = 0, we get 0 ≥ 12 which is false.
∴ Origin does not lie in the region of 3x + 4y ≥ 12.
The region above the line CD and including points of the line CD represents 3x + 4y ≥ 12.

(iii) x ≥ 0 is the region on the right of Y-axis and all the points lying on it.

(iv) The line l₃ : y = 1 is the line parallel to X-axis at a distance 1 from it. Consider y ≥ 1 or y – 1 ≥ 0, putting y = 0 in y -1 ≥ 0
We get -1 ≱ 0, origin does not lie in the region.
y ≥ 1 is the region above y = 1 and the points lying on it.

∴ The shaded region shown in figure represents the solution of the given inequalities.

Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution.
The inequalities are 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
(i) The line l₁ : 4x + 3y = 60 passes through (15, 0), (0, 20) and it is represented by AB. Consider the inequality 4x + 3y ≤ 60
Putting x = 0, y = 0.
0 + 0 = 0 ≤ 60 which is true,
therefore, origin lies in this region.
Thus, region is below the line AB and the points lying on the line AB represents the inequality 4x + 3y ≤ 60.

(ii) The line l₂ : y = 2x passes through (0, 0). It is represented by CD.
Consider the inequality y ≥ 2x. Putting x = 0, y = 5 in y – 2x ≥ 0 5 ≥ 0 is true.
∴ (0, 5) lies in this region.

Region lying above the line CD and including the points on the line CD represents y ≥ 2x

(iii) x ≥ 3 is the region lying on the right of line l₃ : x = 3 and points lying on x = 3 represents the inequality x ≥ 3.
∴ The shaded area APQR in which x ≥ 0 and y ≥ 0 is true for each point, is the solution of given inequalities.

Question 14.
3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution.
The inequalities are 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0
(i) The line l₁ : 3x + 2y = 150 passes through the points (50,0) and (0, 75). AB represents the line. Consider the inequality 3x + 2y ≤ 150.
Putting x = 0, y = 0 in 3x + 2y ≤ 150
⇒ 0 ≤ 150 which is true, shows that origin lies in this region.

The region lying below the line AB and the points lying on AB represents the inequality 3x + 2y < 150.

(ii) The line l₂ : x + 4y = 80 passes through the points (80, 0), (0, 20). This is represented by CD.
Consider the inequality x + 4y ≤ 80 putting x = 0, y = 0, we get 0 ≤ 80, which is true.
⇒ Region lying below the line CD and the points on the line CD represents the inequality x + 4y ≤ 80
(iii) x ≤ 15 is the region lying on the left to
l₃ : x = 15 represented by EF and the points lying on EF.

(iv) x ≥ 0 is the region lying on the right side of Y-axis and all the points on Y-axis.

(v) y ≥ 0 is the region lying above the X-axis and all the points on X-axis.
Thus, the shaded region in the figure is the solution of the given inequalities.

Question 15.
x+2y ≤ 10 , x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution.
The inequalities are x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
(i) l₁ : x + 2y = 10 passes through (10, 0) and (0, 5). The line AB represents this equation. Consider the inequality x + 2y ≤ 10 putting x = 0, y = 0, we get 0 ≤ 10 which is true.
∴ Origin lies in the region of x + 2y ≤ 10.
∴ Region lying below the line AB and the points lying on it represents x + 2y ≤ 10

(ii) l₂ : x + y = 1 passes through (1, 0) and (0, 1). Thus line CD represents this equation. Consider the inequality x + y ≥ 1 putting x = 0, y = 0, we get 0 ≥ 1, which is not true. Origin does not lie in the region of x + y ≥ 1.
∴ The region lying above the line CD and the points lying on it represents the inequality x + y ≥1

(iii) l₃ : x – y = 0, passes through (0, 0). This is being represented by EF.
Consider the inequality x – y ≤0, putting x = 0, y = 1, We get 0 – 1 ≤ 0 which is true
⇒ (0,1) lies on x – y ≤ 0
The region lying above the line EF and the points lying on it represents the inequality x – y ≤ 0.

(iv) x ≥ 0 is the region lying on the right of Y-axis and the points lying on x = 0.

(v) y ≥ 0 is the region above X-axis, and the points lying on y = 0.
∴ The shaded area in the figure represents the given inequalities.

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