NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

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

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 6
Chapter Name	Linear Inequalities
Exercise	Ex 6.1, Ex 6.2, Ex 6.3
Number of Questions Solved	51
Category	NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

NCERT Exercises

Chapter 6 Linear Inequalities Exercise – 6.1

Question 1.
Solve 24x < 100, when
(i) x is a natural number
(ii) x is an integer.
Solution.
Given inequality is 24x < 100
Dividing both sides by 24, we get

This inequality is true when
(i) x is a natural number, {1, 2, 3, 4} satisfies this inequality.
(ii) x is an integer, {…., -4, -3,-2, -1, 0, 1, 2, 3, 4} satisfies this inequality.

Question 2.
Solve – 12x > 30, when
(i) x is a natural number
(ii) x is an integer
Solution.
Given inequality is -12x > 30 Dividing both sides by -12, we get

(i) This inequality is not true for any natural number.
(ii) Integers that satisfy this inequality are (…, -5, -4, -3}.

Question 3.
Solve 5x – 3 < 7, when
(i) x is an integer
(ii) x is a real number
Solution.
Given inequality is 5x – 3 < 7
Transposing 3 to R.H.S., we get
5x < 7 + 3 or 5x < 10
Dividing both sides by 5, we get
x < 2
(i) When x is an integer, {…. -2, -1, 0, 1} satisfies this inequality.
(ii) When x is real number, the solution is (-∞, 2).

Question 4.
Solve 3x + 8 > 2, when
(i) x is an integer
(ii) x is a real number.
Solution.
Given inequality is 3x + 8 > 2
Transposing 8 to R.H.S., we get
3x > 2 – 8 = -6
Dividing both sides by 3, we get
x > -2
(i) When x is an integer, the solution is (-1, 0, 1, 2, 3,…}
(ii) When x is real, the solution is (-2, ∞).

Solve the inequalities in Exercises 5 to 16 for real x.
Question 5.
4x + 3 < 5x + 7
Solution.
The inequality is 4x + 3 < 5x + 7
Transposing 5x to L.H.S. and 3 to R.H.S., we get
4x – 5x < 7 – 3 or -x < 4 Dividing both sides by -1, we get x > -4
∴ The solution is (- 4, ∞).

Question 6.
3x – 7 > 5x – 1
Solution.
The inequality is 3x – 7 > 5x -1
Transposing 5x to L.H.S. and -7 to R.H.S., we get
3x – 5x > -1 + 7 or -2x > 6
Dividing both sides by -2, we get
x < -3
∴ The solution is (-∞, -3).

Question 7.
3(x – 1) < 2(x – 3)
Solution.
The inequality is 3(x – 1) < 2(x – 3) or 3x – 3 < 2x – 6
Transposing 2x to L.H.S. and -3 to R.H.S., we get
3x – 2x < – 6 + 3
⇒ x<-3
∴ The solution is (- ∞, -3],

Question 8.
3(2 -x) > 2(1 -x)
Solution.
The inequality is 3(2 – x) > 2(1 – x) or 6 – 3x > 2 – Zx
Transposing -2x to L.H.S. and 6 to R.H.S., we get
-3x + 2x > 2 – 6 or -x > -4
Multiplying both sides by -1, we get
x ≤ 4
∴ The solution is (- ∞, 4],

Question 9.

Solution.
The inequality is
Simplifying,
Multiplying both sides by , we get
x < 6
∴ The solution is (- ∞, 6),

Question 10.

Solution.
The inequality is
Transposing to L.H.S., we get

Simplifying,
Multiplying both sides by -6, we get
x < -6
∴ The solution is (- ∞, – 6).

Question 11.

Solution.
The inequality is
Multiply both sides by the L.C.M. of 5, 3 i.e., by 15.
3 x 3(x – 2) ≤ 5 x 5(2 – x)
or, 9(x – 2) ≤ 25(2 – x)
Simplifying, 9x – 18 ≤ 50 – 25x
Transposing -25x to L.H.S. and -18 to R.H.S.
∴ 9x + 25x ≤ 50 + 18 or 34x ≤ 68
Dividing both sides by 34, we get
x < 2
∴ Solution is (- ∞, 2].

Question 12.

Solution.
The inequality is
or,
Multiplying both sides by 30,
3(3x + 20) ≥ 10(x – 6) or, 9x + 60 ≥ 10x-60
Transposing 10 x to L.H.S. and 60 to R.H.S., we get
∴ 9x-10x ≥ -60 – 60 or -x ≥-120
Multiplying both sides by -1, we get
x < 120
∴ The solution is (- ∞, 120].

Question 13.
2(2x + 3) – 10 < 6(x – 2)
Solution.
The inequality is 2(2x + 3) – 10 < 6(x – 2)
Simplifying, 4x + 6 -10 < 6x -12
or, 4x – 4 < 6x – 12
Transposing 6x to L.H.S. and – 4 to R.H.S., we get
∴ 4x – 6x < -12 + 4 or -2x < – 8 Dividing both sides by -2, we get x>4
∴ The solution is (4, ∞).

Question 14.
37 – (3x + 5) ≥ 9x – 8(x – 3)
Solution.
The inequality is 37 – (3x + 5) ≥ 9x – 8(x – 3)
Simplifying, 37 – 3x – 5 ≥ 9x – 8x + 24
or 32 – 3x ≥ x + 24
Transposing x to L.H.S. and 32 to R.H.S., We get
-3x – x ≥ 24 – 32 or -4x ≥ -8
Dividing both sides by – 4, we get
x < 2
∴ The solution is (- ∞, 2].

Question 15.

Solution.
The inequality is
Multiplying each term by the L.C.M. of 4, 3, 5, i.e., by 60, we get
15x < 100x – 40 – 84x + 36
or, 15x < 100x – 84x -40 + 36
or, 15x < 16x – 4
Tranposing 16x to L.H.S., we get
15x – 16x < -4 or -x < -4 Multiplying both sides by -1, we get x >4
∴ The solution is (4, ∞).

Question 16.

Solution.
The inequality is
Multiplying each term by L.C.M. of 3,4, 5, i.e., by 60

or, 20(2x – 1) ≥ (3x – 2) x 15 – (2 – x) x 12
or, 40x – 20 ≥ 45x – 30 – 24 + 12x
or, 40x – 20 ≥ 57x – 54
Transposing 57x to L.H.S. and -20 to R.H.S., we get
40x – 57x ≥ -54 + 20 or -17x ≥ -34
Dividing both sides by -17, we get
x <2
∴ The solution is (- ∞, 2].

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.
Question 17.
3x – 2 < 2x + 1
Solution.
The inequality is 3x – 2 < 2x + 1
Transposing 2x to L.H.S. and -2 to R.H.S, we get
3x- 2x < 1 + 2 or, x <3

Question 18.
5x – 3 ≥ 3x – 5
Solution.
The inequality is 5x – 3 ≥ 3x – 5
Transposing 3x to L.H.S. and -3 to R.H.S., we get
∴ 5x – 3x ≥ -5 + 3 or, 2x ≥ -2
Dividing both sides by 2, we get

Question 19.
3(1 – x) < 2(x + 4)
Solution.
3(1-x) < 2(x + 4)
Simplifying 3 – 3x < 2x + 8
Transposing 2x to L.H.S. and 3 to R.H.S., we get
-3x – 2x < 8 – 3 or -5x < 5
Dividing both sides by -5, we get

Question 20.

Solution.
The inequality is

Question 21.
Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution.
Let Ravi gets x marks in third unit test.
∴ Average marks obtained by Ravi

He has to obtain atleast 60 marks,
∴
Multiplying both sides by 3,
145 + x ≥ 60 x 3 = 180
Transposing 145 to R.H.S., we get
x ≥ 180 – 145 = 35
∴ Ravi should get atleast 35 marks in the third unit test.

Question 22.
To receive Grade ‘A’ in the course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
Solution.
Let Sunita obtained x marks in the fifth examination.
∴ Average marks of 5 examinations

This average must be atleast 90
∴
Multiplying both sides by 5
368 + x ≥ 5 x 90 = 450
Transposing 368 to R.H.S., we get
x ≥ 450 – 368 = 82
∴ Sunita should obtain atleast 82 marks in the fifth examination.

Question 23.
Find all pairs of consecutive odd positive integers both of which are smaller than 10, such that their sum is more than 11.
Solution.
Let x be the smaller of the two odd positive integers. Then the other integer is x + 2. We should havex + 2< 10 and x + (x + 2) > 11 or, 2x + 2 > 11
or, 2x > 11 – 2 or, 2x > 9 or,
Hence, if one number is 5 (odd number), then the other is 7. If the smaller number is 7, then the other is 9. Hence, possible pairs are (5, 7) and (7, 9).

Question 24.
Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
Solution.
Let x be the smaller of the two positive even integers then the other one is x + 2, then we should have x > 5
and x + x + 2 < 23 or, 2x + 2 < 23
or, 2x < 21 or,
Thus, the value of x may be 6,8,10 (even integers) Hence, the pairs may be (6, 8), (8,10), (10,12).

Question 25.
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Solution.
Let the shortest side measures x cm
The longest side will be 3x cm.
Third side will be (3x – 2) cm.
According to the problem, x + 3x + 3x – 2 ≥ 61
or, 7x – 2 ≥ 61 or, 7x ≥ 63 or, x ≥ 9
Hence, the minimum length of the shortest side is 9 cm.

Question 26.
A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths for the shortest board if the third piece is to be atleast 5 cm longer than the second?
[Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].
Solution.
Let x be the length of the shortest board, then x + 3 is the second length and 2x is the third length. Thus, x + (x + 3) + 2x ≤ 91
or 4x + 3 ≤ 91 or 4x ≤ 88 or x ≤ 22
According to the problem, 2x ≥ (x + 3) + 5 or x ≥ 8
∴ Atleast 8 cm but not more than 22 cm are the possible lengths for the shortest board.

Chapter 6 Linear Inequalities Exercise – 6.2

Solve the following inequalities graphically in two-dimensional plane
Question 1.
x + y < 5
Solution.
Consider the equation x + y = 5. It passes through the points (0, 5) and (5, 0). The line x + y = 5 is represented by AB. Consider the inequality x + y < 5

Put x = 0, y = 0
0 + 0 = 0 < 5, which is true. So, the origin O lies in the plane x + y < 5
∴ Shaded region represents the inequality x + y < 5

Question 2.
2x + y ≥ 6
Solution.
Consider the equation 2x + y = 6
The line passes through (0, 6), (3, 0).
The line 2x + y = 6 is represented by AB.

Now, consider 2x + y ≥ 6
Put x = 0, y = 0
0 + 0 ≥ 6, which does not satisfy this inequality.
∴ Origin does not lie in the region of 2x + y ≥ 6.
The shaded region represents the inequality 2x + y ≥ 6

Question 3.
3x + 4y ≤ 12
Solution.
We draw the graph of the equation 3x + 4y = 12. The line passes through the points (4, 0), (0, 3). This line is represented by AB. Now consider the inequality 3x + 4y ≤ 12
Putting x = 0, y = 0 0 + 0 = 0 ≤ 12, which is true

∴ Origin lies in the region of 3x + 4y ≤ 12 The shaded region represents the inequality 3x + 4y ≤ 12

Question 4.
y + 8 ≥ 2x
Solution.
Given inequality is y + 8 ≥ 2x
Let us draw the graph of the line, y+ 8 = 2x
The line passes through the points (4, 0), (0, -8).
This line is represented by AB.

Now, consider the inequality y + 8 ≥ 2x.
Putting x = 0, y = 0
0 + 8 ≥ 0, which is true
∴ Origin lies in the region of y + 8 ≥ 2x
The shaded region represents the inequality y + 8 ≥ 2x.

Question 5.
x – y ≤ 2
Solution.
Given inequality is x – y ≤ 2
Let us draw the graph of the line x – y = 2
The line passes through the points (2, 0), (0, -2)
This line is represented by AB.

∴ Origin lies in the region of x – y ≤ 2
The shaded region represents the inequality x – y ≤ 2.

Question 6.
2x – 3y > 6
Solution.
We draw the graph of line 2x – 3y = 6.
The line passes through (3, 0), (0, -2)
AB represents the equation 2x – 3y = 6
Now consider the inequality 2x – 3y > 6
Putting x = 0, y = 0
0 – 0 > 6, which is not true
∴ Origin does not lie in the region of 2x – 3y > 6.

The shaded region represents the inequality 2x – 3y > 6

Question 7.
-3x + 2y ≥ -6.
Solution.
Let us draw the line -3x + 2y = -6
The line passes through (2, 0), (0, -3)
The line AB represents the equation -3x + 2y = -6
Now consider the inequality -3x+ 2y ≥ -6
Putting x = 0, y = 0
0 + 0 ≥ -6, which is true.
∴ Origin lies in the region of -3x + 2y ≥ -6

The shaded region represents the inequality -3x + 2y ≥ – 6

Question 8.
3y- 5x < 30
Solution.
Given inequality is 3y – 5x < 30
Let us draw the graph of the line 3y – 5x = 30
The line passes through (-6, 0), (0, 10)
The line AB represents the equation 3y – 5x = 30
Now, consider the inequality 3y – 5x < 30
Putting x = 0, y = 0
0 – 0 < 30, which is true.
∴ Origin lies in the region of 3y – 5x < 30

The shaded region represents the inequality 3y – 5x < 30

Question 9.
y<- 2
Solution.
Given inequality is y < -2 ………(1)
Let us draw the graph of the line y = -2
AB is the required line.
Putting y = 0 in (1), we have
0 < -2, which is not true.

The solution region is the shaded region below the line.
Hence, every point below the line (excluding the line) is the solution area.

Question 10.
x > -3
Solution.
Let us draw the graph of x = -3
∴ AB represents the line x = -3
By putting x = 0 in the inequality x > -3
We get, 0 > -3, which is true.
∴ Origin lies in the region of x > -3.

Graph of the inequality x > -3 is shown in the figure by the shaded area

Chapter 6 Linear Inequalities Exercise – 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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