# Question-1

**Solve 24x < 100**

**(i) x is a natural numbers (ii) x is an integer.**

**Solution:**

We are given:

24x < 100

or ** < **(Rule 2)

or x < .

(i) When x is a natural number, in this case, the following values of x make the statement true

x = 1, 2, 3, 4.

The solution set of the inequality is {1, 2, 3, 4}.

(ii) When x is an integer, in this case the solutions of the given inequality are â€¦.., -3, -2, -1, 0, 1, 2, 3, 4.

âˆ´ The solution set of the inequality is [â€¦..,-3, -2, -1, 0, 1, 2, 3, ].

# Question-2

**Solve â€“12x > 30, when**

**(i) x is a natural number**

**(ii) x is an integer.**

**Solution:**

We are given â€“12x > 30

or < (Rule 2)

or x <

or x < .

(i) When x is a natural number, there is no value of x which makes the statement true.

The solution set of the inequality is empty i. e.,Ï† .

(ii) When x is an integer, the solutions of the given inequality are .â€¦., -5, -4, -3.

The solution set of the inequality is {.â€¦., -5, -4, -3}.

# Question-3

**Solve 5x â€“ 3 < 7, when**

**(i) x is an integer.**

**(ii) x is a real number**

**Solution:**

We are given 5x â€“ 3 < 7

â‡’ 5x â€“ 3 + 3 < 7 + 3 (Rule 1)

â‡’ 5x < 10

â‡’ < (Rule 2)

â‡’ x < 2.

(i) When x is an integer the solutions of the given inequality are .â€¦.., -3, -2, -1, 0, 1.

âˆ´ The solution set of the inequality is {.â€¦.., -3, -2, -1, 0, 1}.

(ii) When x is a real number, the solutions of the inequality are given by x < 2 i. e., all real numbers x which are less than 2. Therefore, the solution set of the inequalities is x âˆˆ (-âˆž , 2).

# Question-4

**Solve 3x + 8 > 2, when**

**(i) x is an integer**

**(ii) x is a real number.**

**Solution:**

We are given 3x + 8 > 2

â‡’ 3x + 8 â€“ 8 > 2 â€“ 8 (Rule 1)

â‡’ 3x > -6

â‡’ > (Rule 2)

â‡’ x > -3.

(i) When x is an integer, the following values of x make the statement true.

x = -1, -2, 0, 1, 2, 3, â€¦â€¦

âˆ´ The solution set of the inequality is {-2, -1, 0, 1, 2, 3, â€¦â€¦}

(ii) When x is a real number, the solutions of the inequality are given by x > -3 i. e., all real numbers x which are greater than â€“3. Therefore, the solution set of the inequality is (-3, âˆž ).

**Remarks**: Until now we have considered inequalities having solutions as set of natural numbers set of integers as well as the set of real numbers. Henceforth, unless stated other wise, we shall solve the inequalities in the set of real numbers.

# Question-5

**Solve 4x + 3 < 6x + 7**

**Solution:**

4x + 3 < 6x + 7

â‡’ 4x + 3 â€“ 3 < 6x + 7 â€“ 3

â‡’ 4x < 6x + 4

â‡’ 4x â€“ 6x < 6x â€“ 6x + 4 (Rule 1)

â‡’ â€“2x < 4

â‡’ < (Rule 2)

â‡’ x < â€“2.

Solution set = (â€“2, âˆž ).

# Question-6

**Solve 3x â€“7 > 5x â€“ 1**

**Solution:**

3x â€“7 > 5x â€“ 1

â‡’ 3x - 7 + 7 > 5x â€“ 1 + 7 (Rule 1)

â‡’ 3x > 5x + 6

â‡’ 3x â€“ 5x > 5x â€“5x + 6 (Rule 1)

â‡’ -2x > 6

â‡’ > (Rule 2)

â‡’ x > -3.

Solution set = (-âˆž , -3)

# Question-7

**Solve 3(x-1) â‰¤ 2(x-3)**

**Solution:**

3(x-1) â‰¤ 2(x-3)

â‡’ 3x - 3 â‰¤ 2x â€“ 6

â‡’ 3x â€“ 3 + 3 â‰¤ 2x â€“ 6 + 3 (Rule 1)

â‡’ 3x â‰¤ 2x â€“ 3

â‡’ 3x â€“ 2x â‰¤ 2x â€“2x - 3 (Rule 1)

â‡’ x â‰¤ -3

Solution set = (-âˆž , 3).

# Question-8

**Solve 3(2-x) â‰¥ 2(1-x)**

**Solution:**

3(2-x) â‰¥ 2(1-x)

â‡’ 6x - 3 â‰¥ 2 â€“ 2x

â‡’ 6 - 3x +3x â‰¥ 2 â€“ 2x = 3x (Rule 1)

â‡’ 6 â‰¥ 2 + x

â‡’ 6 â€“ 2 â‰¥ 2 + x - 2 (Rule 1)

â‡’ 4 â‰¥ x

â‡’ x â‰¤ 4

Solution set = (-âˆž , 4).

# Question-9

**Solve x + + < 11**

**Solution:**

x + + < 11

â‡’ < 11 (6)

[Multiplying both sides by 6 i. e., L. C. M. of 2 and 3]

â‡’ 6x + 3x + 2x < 66

â‡’ 11x < 66

â‡’ < (Rule 2)

â‡’ x < 6

Solution set = (-âˆž , 6).

# Question-10

**Solve**

**Solution:**

â‡’

[Multiply both sides by 6]

â‡’ 2x > 3x + 6

â‡’ 2x â€“ 3x > 3x + 6 â€“ 3x (Rule 1)

â‡’ -x > 6

â‡’ x < -6 (Rule 2)

Solution set = (-âˆž , -6).

# Question-11

**Solve**

**Solution:**

â‡’

[Multiply both sides by 15]

â‡’ 9(x-2) â‰¤ 25(2-x)

â‡’ 9x â€“18 â‰¤ 50 â€“ 25x

â‡’ 9x â€“ 18 + 18 â‰¤ 50 â€“ 25x + 18 (Rule 1)

â‡’ 9x â‰¤ 68 â€“ 25x

â‡’ 9x + 25x â‰¤ 68 â€“ 25x + 25x (Rule 1)

â‡’ 34x â‰¤ 68

â‡’ (Rule 2)

â‡’ x â‰¤ 2

Solution set = (-âˆž , 2).

# Question-12

**Solve**

**Solution:**

â‡’

â‡’

[Multiply both sides by 30]

â‡’

â‡’ 3(3x + 20) â‰¥ 10(x â€“ 6)

â‡’ 9x + 60 â‰¥ 10x - 60

â‡’ 9x + 60 â€“ 60 â‰¥ 10x â€“ 60 â€“ 60 (Rule 1)

â‡’ 9x â‰¥ 10x - 120

â‡’ 9x â€“ 10x â‰¥ 10x â€“ 120 â€“ 10x (Rule 1)

â‡’ - x â‰¥ -120

â‡’ x â‰¤ 120 (Rule 2)

Solution set = (-âˆž , 120).

# Question-13

**Solve the inequation: 2(2x + 3) â€“ 10 < 6(x - 2)**

**Solution:**

2(2x + 3) â€“ 10 < 6(x - 2)

4x + 6 â€“ 10 < 6x - 12

4x â€“ 4 < 6x - 12

4x â€“ 4 + 4 < 6x â€“ 12 + 4

4x < 6x - 8

4x â€“ 6x < 6x â€“ 8 â€“ 6x- 2x < â€“ 8

- 2 x/-2 > -8 /-2

x > 4

Solution set = (4, âˆž)

# Question-14

**Solve the inequation: 37 â€“ (3x + 5) â‰¥ 9x â€“ 8(x - 3)**

**Solution:**

37 â€“ 3x â€“ 5 â‰¥ 9x â€“ 8x + 24

32 â€“ 3x â‰¥ x + 24

32 â€“ 3x â€“ 32 â‰¥ x + 24 â€“ 32

â€“ 3x â‰¥ x â€“ 8

â€“ 3x â€“ x â‰¥ x â€“ 8 â€“ x

â€“ 4x â‰¥ â€“ 8

â€“ 4x/â€“4 â‰¥ â€“ 8/â€“4

x â‰¥ 2

Solution set = (2, âˆž).

# Question-15

**Solve**

**Solution:**

Multiplying both sides by 60 i. e., L. C. M. of 3, 4 and 5 we have

â‡’

â‡’ 15x < 20(5x - 2) â€“ 12(7x â€“ 3)

â‡’ 15x < 100x â€“ 40 â€“ 84x + 36

â‡’ 15x < 16x - 4

â‡’ 15x + 4 < 16x â€“ 4 + 4 (Rule 1)

â‡’ 15x + 4 < 16x

â‡’ 15x â€“ 15x + 4 < 16x â€“ 15x

â‡’ 4 < x

â‡’ x > 4

Solution set = (4, âˆž ).

# Question-16

**Solve**

**Solution:**

Multiplying both sides by 60 i. e., L. C. M. of 3, 4 and 5 we have

â‡’

â‡’ 20(2x-1) â‰¥ 15(3x-2) â€“ 12(2-x)

â‡’ 40x â€“ 20 â‰¥ 45x â€“ 30 â€“ 24 + 12x

â‡’ 40x â€“ 20 â‰¥ 57x - 54

â‡’ 40x â€“ 20 + 20 â‰¥ 57x â€“ 54 + 20

â‡’ 40x â‰¥ 57x - 34

â‡’ 40x â€“ 57x â‰¥ 57x â€“ 57x â€“34 (Rule 1)

â‡’ -17x â‰¥ -34

â‡’

â‡’ x â‰¤ 2

Solution set = (-âˆž , 2).

# Question-17

**Solve 3x â€“ 2 < 2x + 1**

**Solution:**

3x â€“ 2 < 2x + 1

â‡’ 3x â€“ 2 + 2 < 2x + 1 + 2

â‡’ 3x < 2x + 3

â‡’ 3x â€“ 2x < 2x + 3 â€“ 2x

â‡’ x < 3

i.e., all real numbers x less than 3 as shown on the number line.

# Question-18

**Solve 5x â€“ 3 â‰¥ 3x â€“ 5**

**Solution:**

5x â€“ 3 â‰¥ 3x â€“ 5

â‡’ 5x â€“ 3 + 3 â‰¥ 3x â€“ 5 + 3

â‡’ 5x â‰¥ 3x - 2

â‡’ 5x â€“ 3x â‰¥ 3x â€“ 2 â€“ 3x

â‡’ 2x â‰¥ -2

â‡’ â‡’ x â‰¥ -1

i.e., all real numbers x greater than or equal to â€“1 as shown on the number line.

# Question-19

**Solve 3(1 - x) < 2(x + 4)**

**Solution:**

3(1 - x) < 2(x + 4)

â‡’ 3-3x < 2x + 8

â‡’ 3 - 3x â€“3 < 2x + 8 - 3

â‡’ -3x < 2x + 5

â‡’ -3x â€“ 2x < 2x + 5 â€“ 2x

â‡’ -5x < 5

â‡’ (Rule 2)

â‡’ x > -1

i. e., all real numbers x greater than â€“1 as shown on the number line.

# Question-20

**Solve**

**Solution:**

Multiplying on both sides by 30 i. e., L. C. M. of 2, 3 and 5 we have

â‡’

â‡’ 15x < 10(5x-2) â€“ 6(7x-3)

â‡’ 15x < 50x â€“ 20 â€“ 42x + 18

â‡’ 15x < 8x - 2

â‡’ 15x â€“ 8x < 8x â€“ 2 â€“ 8x

â‡’ 7x < -2

â‡’ x <

i.e., all real numbers x less than as shown on the number line.

# Question-21

**Ravi obtained 70 and 75 marks in first two-unit test. Find the minimum marks he should get in the third test to have an average of atleast 60 marks.****Solution:**

Let he should get x marks in the third test such that

â‡’ â‡’ 145 + x â‰¥ 180

â‡’ x â‰¥ 180 â€“ 145 â‡’ x â‰¥ 35

Hence, he should get 35 or more marks in the third test.

# Question-22

**To receive Grade â€˜Aâ€™ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunitaâ€™s marks in first four examination are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get Grade â€˜Aâ€™ in the course.**

**Solution:**

Let Sunita obtain x marks in her fifth examination.

Then,

â‡’ 368 + x â‰¥ 450 â‡’ x â‰¥ 82.

Thus, Sunita must obtain a minimum of 82 marks to get Grade â€˜Aâ€™ in this course.

# 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 the two consecutive odd positive integers be x and x + 2.

x < 10, x + 2 < 10, x + (x + 2) > 11

x + 2 < 10

x < 8

x + (x + 2) > 11

2x + 2 > 11

2x > 9

x > 9/2

i.e 8 > x > 9/2

Thus, the required pairs of consecutive odd positive integers are (5, 7) and (7, 5).

# 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 the two consecutive even positive integers be x and x + 2.

x > 5, x + 2 > 5, x + (x + 2) < 23

x > 5, x + 2 > 5 â‡’x > 5.

2x + 2 < 23

2x < 21

x < 21/2

i.e 21/2 > x > 5

Thus, the required pairs of consecutive odd positive integers are (6, 8) and (8, 10).

# Question-25

**The longest side of a triangle is 3 times the shortest side and the third side is 2cm shorter than the longest side. If the perimeter of the triangle is at least 61cm, find the minimum length of the shortest side.**

**Solution:**

Let the minimum length of the shortest side be x.

Then the longest side of a triangle is 3x and the third side is 3x â€“ 2.

Perimeter of the triangle â‰¥ 61

x + 3x + 3x â€“ 2 â‰¥ 61

7x â€“ 2 â‰¥ 61

7x â‰¥ 61 + 2

7x â‰¥ 63

x â‰¥ 9

Thus, 9 is the required minimum length of the shortest side.

# 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 3cm 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 at least 5cm longer than the second?**

**Solution:**

Let x be the shortest board.

Then x + 3 is the second piece and 2x is the third piece.

x + (x + 3) + 2x â‰¤ 91

and 2x â‰¥ (x + 3) + 5

x + (x + 3) + 2xâ‰¤ 91

4x + 3 â‰¤ 91

4x â‰¤ 91 â€“ 3

4x â‰¤88

x â‰¤ 44/2

2x â‰¥ (x + 3) + 5

2x â‰¥ x + 8

x â‰¥ 8

i.e, 8 â‰¤ x < 22

Thus the possible lengths of the shortest board is at least 8cm long but not more than 22cm long.

# Question-27

**Represent the following inequalities graphically in a two dimensional plane.**

**(1) x + y < 5**

**Solution:**

(i) We draw the graph of the equation x + y = 5 ...(i)

Putting y = 0, x - 5, therefore the point on the x-axis is (5,0). The point on the y-axis is (0, 5). AB is the graph of (i) (See Figure).

Putting x = 0, y = 0 in the given inequality, we have 0 + 0 < 5 or 5 > 0 which is true.

Hence, origin lies in the half plane region I.

Clearly, any point on the line does not satisfy the given inequality.

Hence, the shaded region 1 excluding the points on the line is the solution region of the inequality.

# Question-28

**Represent the following inequalities graphically in a two dimensional plane.**

**2x + y > 6**

**Solution:**

We draw the graph of the equation

2x + y = 6 ...(i)

Putting x = 0, y = 6, therefore the point on y-axis is (0, 6) and the point on x-axis is (3, 0). AB is the graph of (i).

Putting x = 0, y = 0 in the given inequality, we have 2(0) + 0 â‰¥ 6 or 0 â‰¥ 6, which is false.

Hence, origin does not lie in the half plane region I.

Clearly, any point on the line satisfy the given inequality.

Hence, the shaded region II including the points on the line is the solution region of the inequality.

# Question-29

**Represent the following inequalities graphically in a two dimensional plane. 3x + 4y < 12.**

**Solution:**

(iii) We draw the graph of the equation

3x + 4y = 12 ...(i)

Putting x - 0, y = 3, therefore the point on y-axis is (0, 3) and the point on x-axis is (4, 0). AB is the graph of (i).

Putting x = 0, y = 0 in the given inequality, we have 3(0) + 4(0) â‰¤ 12 or 0 â‰¤ 12, which is true.

Hence, origin lies in the half plane region I.

Clearly, any point on the line satisfy the given inequality.

Hence, the shaded region I including the points on the line is the solution region of the inequality.

# Question-30

**Find the solutions of following inequalities graphically in a two dimensional plane. y + 8 > 2x**

**Solution:**

We draw the graph of the equation

y + 8 = 2x

or y - 2x = -8

or 2x - y = 8 ...(i)

Putting x = 0, y = -8, therefore the point on the y-axis is (0, -8). The point on the x-axis is (4, 0). AB is the graph of (i).

Putting x = 0, y = 0 in the given inequality we have 0 + 8 â‰¥ 2(0) or 8 â‰¥ 0, which is true. Hence origin lies in the half plane region I.

Clearly, any point on the line satisfy the given inequality.

Hence, the shaded region I including the points on the line is the solution region of the inequality.

# Question-31

**Find the solutions of following inequalities graphically in a two dimensional plane. x - y < 2**

**Solution:**

We draw the graph of the equation

x - y = 2 ...(i)

Putting x = 0, y - -2, therefore the point on th y-axis is (0, -2). The point on the x-axis is (-2,0' AB is the graph of (i).

Putting x = 0, y = 0 in the given inequality, w have 0 - 0 < 2 or 0 < 2', which is true. Hence origin lies in the half plane region I.

Clearly, all the points on the line satisfy the given inequality.

Hence, the shaded region I including the points on the line is the solution region of the inequality.

# Question-32

**Find the solutions of following inequalities graphically in a two dimensional plane. 2x -3y>6.**

**Solution:**

We draw the graph of the equation

2x - 3y = 6. ...(i)

Putting x = 0, y = -2, therefore the point on the y-axis is (0, -2).

The point on the jc-axis is (3,0). AB is the graph

Putting x = 0, y = 0 in the given inequality, w< have 2(0) - 3(0) > 6 or 0 > 6, which is false. Hence origin does not lie in the half plane region I.

Clearly any point on the line does not satisfy the given inequality. Hence the shaded region II excluding the points on the line is the solution region of the inequality.

# Question-33

**Solve the following inequalities graphically in two-dimensional plane-3x + 2y > -6.**

**Solution:**

We draw the graph of the equation

-3x + 2y = -6. ...(i)

Putting x = 0, y = -3, therefore the point on the v-axis is (0, -3)

The point on the x-axis is (2,0). AB is the graph of(i).

Putting x - 0, y = 0 in the given inequality, we have -3(0) + 2 (0) > -6 or 0 > -6 or 0 < 6, which is true.

Hence, origin lies in the half plane region I. Clearly any point on the line satisfy the given inequality.

Hence, the shaded region I including the points on line is the solution region of the inequality.

# Question-34

**Solve the following inequalities graphically in two-dimensional plane 3y-5x< 30.**

**Solution:**

We draw the graph of the equation

3y-5x = 30. ...(i)

Putting x' - 0, y - 10, therefore the point on the y-axis is (0,10).

The point on the j-axis is (-6,0). AB is the graph of (i). Putting x = 0, y = 0 in the given inequality, we have 3(0) - 5(0) < 30 or 0 - 0 < 30 or 0 < 30, which is true.

Hence, origin lies in the half-plane region I.

Clearly, no point on the line satisfies the given inequality.

Hence, the shaded region I excluding the points on the line is the solution region of the inequality.

# Question-35

**Solve the following inequalities graphically in two-dimensional plane y < -2**

**Solution:**

We draw the graph of the equation y = -2.

Clearly it is a line parallel to x-axis at a distance of 2 units below it. The line y - -2 divides the xy-plane into two regions one below it and the other above it.

Consider the point 0(0, 0). We find that (0, 0) does not satisfy the inequality y < - 2 because 0 < -2 which is not true.

So, the region represented by the given inequality is the region not containing the origin as shown in the Fig. 6.24, by the shaded region excluding all points on the line.

# Question-36

**Solve the following inequalities graphically in two-dimensional plane x > -3.**

**Solution:**

We draw the graph of the equation x = -3.

Clearly it is a line parallel to v-axis at a distance of 3 units on the left of it.

The line x = -3 divides the xy-plane into two regions - one on the right side and the other on the left side.

Consider the point 0(0, 0).

We find that (0, 0) satisfy the inequality x > -3 because 0 > -3, which is true.

** **

So, the region represented by the given inequality is the region containing the origin as shown in the Figure, by the shaded region excluding all the points on the line.

# Question-37

**Solve the following systems of inequalities graphically x â‰¥ 3, y â‰¥ 2.**

**Solution:**

x â‰¥ 3 â€¦.(i) and y â‰¥ 2 â€¦.(ii)

Graph of inequality (i). Let us draw me graph of the line x = 3. Clearly, x = 3 is a line parallel to y-axis at a distance of 3 units from the origin.

Since (0, 0) does not satisfy the inequality x â‰¥ 3 because 0 â‰¥ 3, which is false. So the portion lying on the right side of x - 3 is the region represented by, x â‰¥ 3.

Graph of inequality (ii). Let us draw the graph of the line v = 2. Clearly, y = 2 is a line parallel to x-axis at a distance of 3 units from it. Since (0. 0) does not satisfy y â‰¥ 2 as 0 â‰¥ 2, which is false. So, the portion not containing the origin is represented by the given inequality.

The common region of the above two regions represents the solution set of the given linear constraints.

# Question-38

**Solve the following systems of inequalities graphically 3x + 2y â‰¤ 12,x â‰¤ l, y â‰¥ 2.**

**Solution:**

3x + 2y â‰¤ 12 (i)

x â‰¥ 1 (ii)

y â‰¥ 2 (iii)

Graph of inequality (i): Let us draw the graph of the line

3x + 2y = 12

at y = 0, x = 4 we get point (4,0) on x-axis

at x = 0, y = 6 we get point (0, 6) on y-axis.

Putting x = y = 0 in (i) we have 0 â‰¤ 12 which is true.

Hence, half plane region containing the origin including all the points on the line is the solution region of the given inequality.

Graph of inequality (ii): Let us draw the graph of the line x = 1. Clearly x = 1 is a line parallel to y-axis at a distance of 1 unit from it. Since (0, 0) does not satisfy x â‰¥ 1, as 0 â‰¥ 1, which is false. So the portion not containing the origin is represented by the given inequality.

Graph of inequality (iii): Let us draw the graph of the line y = 2. Clearly. y = 2 is a line parallel to x - axis at a distance of 2 units from it.

Since (0,0) does not satisfy y â‰¥ 2, as 0 â‰¥ 2, which is false. So the portion not containing the origin is represented by the given inequality.

The common region of the above three regions represents the solution set of the given linear system.

# Question-39

**Solve the following systems of inequalities graphically 2x + y â‰¥ 6, 3x + 4y â‰¤ 12.**

**Solution:**

2x + y â‰¥ 6 ...(i)

3x + 4y â‰¤ 12 â€¦.(ii)

Graph of inequality (i): Let us draw the graph of the line

2x + y = 6

at y = 0, x = 3, we get the point (3, 0) on x - axis

at x = 0, y = 6, we get the point (0, 6) on y - axis

Putting x = y = 0 in (i) we have 0 > 6 which is false.

Hence, half-plane region not containing the origin is the solution region of the given inequality.

Graph of inequality (ii) : Let us draw the graph of the line 3x + Ay = 12. At y = 0, x - 4 we get the point (4, 0) on x-axis.

At x - 0, y = 3 we get the point (0, 3) on v-axis.

Putting X = y - 0 in (ii) we have 0 â‰¤ 12 which is true.

Hence, half-plane containing the origin is the solution region of the given inequality.

The common region of the above two regions represents the solution set of the given linear system

# Question-40

**Solve the following systems of inequalities graphically x + y>4, 2x - y > 0.**

**Solution:**

x + y > 4 ...(i)

2x - y < 0 ...(ii)

Graph of inequality (i): Let us draw the graph o the line

x + y = 4 ...(iii)

At y = 0, x = 4 we get the point (4, 0) on x-axis

At x = 0, y = 4 we get the point (0, 4) on y-axis

Putting x = y - 0 in (i) we get 0 > 4 which is false

Hence, half plane region not containing the origin is the solution region of the given inequality.

Graph of inequality (ii) : Let us draw the graph of the line

2x - y = 0 ...(iv)

On the line (iv)

at x = 0 â‡’ 2(0) -y = 0 â‡’ y = 0

at x = 1 â‡’ 2(1) -y = 0 â‡’ y = 2

(0, 0) and (1, 2) are on the line (iv).

Draw a dotted line joining points (0,0) and (1,2)

To determine the region represented by the given inequality (ii), consider the point not lying on the line (iv), say (2, 1) and it lies in the half-plane of (ii) if 2(2) - 1 < 0, which is not true.

The half-plane region not containing (2, 1) represents the solution set of the given inequality.

The common region of the above two regions represents the solution set of the given linear system.

# Question-41

**Solve the following systems of inequalities graphically**

**2x â€“ y > 1, x â€“ 2y < 1**

**Solution:**

(i) 2x - y > 1 ...(i)

x -2y < -1 â€¦(ii)

Graph of inequality (i). Let us draw the graph of the line 2x - y = 1.

at y = 0, x = we get the point on x - axis

at x - 0, y = -1 we get the point (0, -1) on y-axis.

Putting x = y = 0 in (i) we get 0 > 1 which is false.

Hence, half-plane region not containing the origin is the solution region of the given inequality.

Graph of inequality (ii): Let us draw the graph of the line x - 2y = -1. At y = 0, x = -1, we get the point (-1,0) on x-axis.

At x = 0, y = , we get the point 10, - on y - axis.

Putting x = y = 0 in (ii) we get 0 < -1 which is false.

Hence, half-plane region not containing the origin is the solution region of the given inequality.

The common region of the above two regions represents the solution set of the given linear constraints.

# Question-42

**Solve the following systems of inequalities graphically**

**x + y â‰¤ 6, x + y â‰¥ 4**

**Solution:**

x + y â‰¤ 6 ... (i)

x + y â‰¥ 4 ...(ii)

Graph of inequality (i): Let us draw the graph of the line x + y = 6.

At x = 0, x = 6, we get the point (6, 0) on x-axis.

At x = 0, y = 6, we get the point (0, 6) on y-axis.

Putting x = y = 0 in (0 we get 0 â‰¤ 6 which is true.

Hence, half-plane containing the origin is the solution region of the given inequality.

Graph of inequality (ii)': Let us draw the graph of the line x + y = 4.

At y = 0, x = 4 we get the point (4, 0) on x-axis.

At x = 0, y - 4 we get the point (0, 4) on y-axis.

Putting x - y - 0 in (ii) we get 0 > 4 which is false.

Hence half plane not containing the origin is the solution region of the given inequality.

The common region of the above two regions represent the solution set of the given linear constraints.

# Question-43

**Solve the following systems of inequalities, graphically 2x + y â‰¥ 8, x + 2y â‰¥ 10.**

**Solution:**

(i) 2x + y â‰¥ 8 ...(i)

(ii) x + 2y â‰¥ 10 ...(ii)

Graph of inequality (i): Let us draw the graph of the line 2x + y = 8

At y = 0, x = 4, we get the point (4, 0) on x - axis.

At x = 0, y = 8, we get the point (0, 8) on y-axis.

Putting x = y = 0 in (i) we get 0 > 8 which is false.

Hence, half plane region not containing the origin is the solution region of the given inequality.

Graph of inequality (ii): Let us draw the graph of the line x + 2y = 10

At y = 0, x = 10 we get the point (10, 0) on x - axis

At x = 0, y = 5 we get the point (0, 5) on y - axis

Putting x = y = 0 in (ii) we get 0 â‰¥ 10 which is false.

Hence, half plane region not containing the origin is the solution region of the given inequality.

The common region of the above two regions represent the solution set of the given linear constraints.

# Question-44

**Solve the following systems of inequalities, graphically x + y â‰¤ 9, y > x, x â‰¥ 0**

**Solution:**

x + y â‰¤ 9 ...(i)

y > x ...(ii)

x â‰¤ 0 ...(iii)

Graph of inequality (ii). Let us draw the graph of the line

x + y = 9 ...(iv)

At y = 0, x = 9 we get the point (9, 0) on x-axis.

At x - 0, y = 9 we get the point (0,9) on y-axis.

Putting x = y = 0 in (i) we get 0 â‰¤ 9 which is true.

Hence, half-plane region containing the origin is the solution region of the given inequality.

Graph of inequality (ii). Let us draw the graph of the line y = x

or y - x = 0 ...(v)

on the line (v)

At x = 0â‡’ y-0=0 y = 0

At x=1 â‡’ y - 1 =0â‡’ y=l

âˆ´ (0, 0) and (1, 1) are on the line (v).

Draw a dotted line joining points (0,0) and(1, 1). To determine the region represented by the given inequality (ii) consider the point not lying on the line (v) say (2, 0) and it lies in the half plane of (ii) if 0 > 2, which is not true. Therefore the portion not containing (2, 0) represents the solution set of the given inequality.

Graph of inequality (iii).

Clearly, x â‰¥ 0 represents the region lying on the right side of y - axis.

Triple shaded region is the solution region.

# Question-45

**Solve the following systems of inequalities graphically 5x + 4y â‰¤ 20, x â‰¥ 1, y â‰¥ 2.**

**Solution:**

5x + 4y â‰¤ 20 ...(i)

x â‰¥ 1 , ...(ii)

y â‰¥ 2 . ' ...(iii)

Graph of inequality (i): Let us draw the graph of the line

5x + 4y = 20 ...(iv)

At y = 0 â‡’ 5x + 4(0) = 20 â‡’ x=4

At x = 0 â‡’ 5(0) + 4y = 20 â‡’ x=5

(4,0) and (0, 5) are on the line (iv).

Putting x = y = 0 in (i) we get 5(0) + 4(0) â‰¤ 20 i.e., 0 â‰¤ 20 which is true.

Hence, half plane region containing the origin is the solution region of the given inequality.

Graph of inequality (ii). Let us draw the graph of the line x = 1 which is parallel to the y-axis and is at a unit distance from it.

Putting x = 0 in (ii), we have 0 â‰¥ 1 which is not true. Hence, half-plane region not containing the origin is the solution region of the given inequality.

Graph of inequality (iii). Let us draw the graph of the line y = 2 which is parallel to x-axis and is at a distance of 2 units from it.

Putting y = 0 in (iii), we have 0 > 2 which is not true.

Hence, half plane region not containing the origin is the solution region of the given inequality.

The common region of the above three regions represents the solution set of the given linear system.

# Question-46

**Solve the following systems of inequalities graphically 3x + 4y â‰¤ 60, x + 3y â‰¥ 0, y â‰¥ 0**

**Solution:**

3x + 4y â‰¤ 60 â€¦.(i)

x + 3y â‰¤ 30 â€¦.(ii)

x â‰¥ 0 â€¦.(iii)

y â‰¥ 0. â€¦.(iv)

Graph of inequality (i): Let us draw the graph of the line

3x + 4y = 60 ...(v)

At y = 0 â‡’ 3x + 4(0) = 60

â‡’ x = 20

x = 0 â‡’ 3(0) + 4y = 60

â‡’ y = 15

(20, 0) and (0, 15) are on the line (v)

Putting x = y = 0 in (i) we get 0 â‰¤ 60 which is true.

Hence, half plane region containing the origin is the solution region of the given inequality.

Graph of inequality (ii). Let us draw the graph of the line

x + 3y = 30 ...(vi)

At y = 0 â‡’ x + 3(0) = 30

â‡’ x = 30

x = 0â‡’ 0 + 3y = 30

â‡’ y = 30

(30, 0) and (0, 30) are on the line (vi)

Putting x = y = 0 in (ii) we get 0 < 30 which is true. Hence, half plane region containing the origin is the solution region of the given inequality

Graph of inequality (iii). Clearly, x â‰¥ 0 represents the region lying on the right side of y-axis.

Graph of inequality (iv). Clearly, y â‰¥ 0 represents the region lying above the x-axis.

The common region of the above four regions is the solution set.

# Question-47

**Solve the following systems of inequalities graphically**

**2x + y â‰¥ 4, x + y â‰¤ 3, 2x â€“ 3y â‰¤ 6.**

**Solution:**

2x + y â‰¥ 4 â€¦â€¦â€¦â€¦ (1)

x + y â‰¤ 3 â€¦â€¦â€¦â€¦ (2)

2x â€“ 3y â‰¤ 6 â€¦â€¦â€¦â€¦ (3)

Graph of inequality (i): Let us draw the graph of the line 2x + y = 4.

y = 0 â‡’ 2x + 0 = 4 â‡’ x = 2.

x = 0 â‡’ 2(0) + y = 4 â‡’ y = 4.

(2, 0) and (0, 4) are on the line 2x + y â‰¥ 4.

Putting x = y = 0 in (i) we get 0 â‰¥ 4 which is false.

Hence, half-plane region not containing the origin is the solution region.

Graph of inequality (ii): Let us draw the graph of the line.

x + y = 3

y = 0 â‡’ x + 0 = 3 â‡’ x = 3.

x = 0 â‡’ 0 + y = 3 â‡’ y = 3.

Putting x = y = 0 in (ii) we get 0 â‰¤ 3 which is true.

Hence, half-plane region containing the origin is the solution region.

Graph of inequality (iii): Let us draw the graph of the line.

2x â€“ 3y = 6

y = 0 â‡’ 2x + 3(0) = 6 â‡’ x = 3.

x = 0 â‡’ 2(0) â€“ 3y = 6 â‡’ y = -2.

(3, 0) and (0, -2) are the points on the line 2x â€“ 3y = 6.

Putting x = y = 0 in (ii) we get 0 â‰¤ 6 which is true.

Hence, half-plane region containing the origin is the solution region.

The common region of the above three regions is the solution set.

# Question-48

**Solve the following systems of inequalities graphically**

**x - 2y â‰¤ 3, 3x +4y â‰¥ 12, x â‰¥ 0, y â‰¥ 1.**

**Solution:**

x - 2y â‰¤ 3 â€¦â€¦â€¦â€¦â€¦ (1)

3x +4y â‰¥ 12 â€¦â€¦â€¦â€¦â€¦ (2)

x â‰¥ 0 â€¦â€¦â€¦â€¦â€¦ (3)

y â‰¥ 1 â€¦â€¦â€¦â€¦â€¦ (4)

Graph of inequality (i): Let us draw the graph of the line

x - 2y = 3

y = 0 â‡’ x â€“ 2(0) = 3 â‡’ x = 3.

x = 0 â‡’ 0 â€“ 2y = 3 â‡’ y = .

(3, 0) and (0, ) are on the line x - 2y = 3.

Putting x = y = 0 in (i) we get 0 â‰¤ 3 which is true.

Hence, half-plane region containing the origin is the solution region.

Graph of inequality (ii): Let us draw the graph of the line.

3x +4y = 12

y = 0 â‡’ 3x + 4(0) = 12 â‡’ x = 4.

x = 0 â‡’ 3(0) + 4y = 12 â‡’ y = 3.

âˆ´ (4, 0) and (0, 3) are the points on the line 3x + 4y = 12.

Putting x = y = 0 in (ii) we get 0 â‰¥ 12 which is false.

Hence, half-plane region not containing the origin is the solution region.

Graph of inequality (iii): Clearly, x â‰¥ 0 represents the region lying on the right side of y-axis.

Graph of inequality (iv): Let us draw the graph of the line y = 1 which is parallel to x â€“ axis and is at a distance of 1 unit from it.

Putting y = 0 in (iv), we have 0 â‰¥ 1 which is false.

Hence, half-plane region not containing the origin is the solution region of the given inequality.

The common region of the above four regions represents the solution set of the given linear constraints.

# Question-49

**Solve the following systems of inequalities graphically 4x + 3y â‰¤ 60, y â‰¥ 2x, x â‰¥ 3, x, y â‰¥ 0.**

**Solution:**

4x + 3y â‰¤ 60 â€¦â€¦â€¦â€¦â€¦ (1)

y â‰¥ 2x â€¦â€¦â€¦â€¦â€¦ (2)

x â‰¥ 3 â€¦â€¦â€¦â€¦â€¦ (3)

x â‰¥ 0 â€¦â€¦â€¦â€¦â€¦ (4)

y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦ (5)

Graph of inequality (i): Let us draw the graph of the line

4x + 3y = 60

y = 0 â‡’ 4x + 3(0) = 60 â‡’ x = 15.

x = 0 â‡’ 4(0) + 3y = 60 â‡’ y =20.

(15, 0) and (0, 20) are on the line 4x + 3y = 60.

Putting x = y = 0 in (i) we get 0 â‰¤ 60 which is true.

Hence, half-plane region containing the origin is the solution region.

Graph of inequality (ii): Let us draw the graph of the line.

y = 2x

y = 0 â‡’ 0 = 2x â‡’ x = 0.

x = 0 â‡’ y = 2(1) â‡’ y = 2.

âˆ´ (0, 0) and (1, 2) are the points on the line y = 2x.

To determine the region represented by the given inequality (ii) consider the point not lying on the line y = 2x, say (2, 0) and it lies in the half plane of (ii) if 0 â‰¥ 4, which is not true. Therefore, the portion not containing (2, 0) represents the solution set of the given inequality.

Graph of inequality (iii): Clearly, x â‰¥ 0 represents the region lying on the right side of y-axis.

Graph of inequality (iv): Let us draw the graph of the line x = 3 which is parallel to the y â€“ axis and is at a distance of 3 unit from it.

Putting x = 0 in (iii), we have 0 â‰¥ 3 which is not true.

Hence, half-plane region not containing the origin is the solution region of the given inequality.

Graph of inequality (iv): Clearly x â‰¥ 0 represents the region lying on the right side of y-axis.

Graph of inequality (v): Clearly y â‰¥ 0 represents the region lying above the x-axis.

The common region of the above five regions represent the solution set of the given linear system.

Triple shaded triangular area is the solution area in the solution region.

# Question-50

**Solve the system of inequalities graphically**

**3x + 2y â‰¤ 150, x + 4y â‰¤ 80, x â‰¤ 15, x â‰¥ 0.**

**Solution:**

3x + 2y â‰¤ 150 â€¦â€¦â€¦â€¦â€¦â€¦. (1)

x + 2y â‰¤ 80 â€¦â€¦â€¦â€¦â€¦â€¦. (2)

x â‰¤ 15 â€¦â€¦â€¦â€¦â€¦â€¦ (3)

x â‰¥ 0 â€¦â€¦â€¦â€¦â€¦â€¦ (4)

Graph of inequality (i). Let us draw the graph of the line 3x + 2y = 150.

At y = 0 â‡’ 3x + 2(0) = 150 â‡’ x = 50

and at x = 0 â‡’ 3(0) + 2y = 150 â‡’ y = 75

(50, 0) and (0, 75) are the points on the line 3x + 2y = 150.

Putting x = y = 0 in (i) we have 0 â‰¤ 150 which is true.

Hence half-plane region containing the origin is the solution region of this inequalities.

Graph of inequality (ii). Let us draw the graph of the line x + 4y = 80.

At y = 0 â‡’ x + 4(0) = 80 â‡’ x = 80

x = 0 â‡’ 0 + 4y = 80 â‡’ y = 20

âˆ´ (80, 0) and (0, 20) are the points on the line x + 4y = 80.

Putting x = y = 0 in (ii) we have 0 â‰¤ 80 which is true.

Hence half-plane region containing the origin is the solution region of this inequality.

Graph of inequality (iii). Let us draw the graph of the line x = 15 which is parallel to the y â€“ axis and is at a distance of 15 units from it.

Putting x = 0 in (iii), we have 0 â‰¤ 15 which is true, the solution region of the given this inequality.

Graph of inequality (iv). Clearly x â‰¥ 0 represents the region lying on the right side of y axis.

The common region of the above four regions represent the solution set of the given linear system.

# Question-51

**Solve the system of inequalities graphically**

**x + 2y â‰¤ 10, x + y â‰¥ 1, x â€“ y â‰¤ 0, x â‰¥ 0, y â‰¥ 0.**

**Solution:**

x + 2y â‰¤ 10 â€¦â€¦â€¦â€¦â€¦â€¦. (1)

x + y â‰¥ 1 â€¦â€¦â€¦â€¦â€¦â€¦. (2)

x - y â‰¤ 0 â€¦â€¦â€¦â€¦â€¦â€¦ (3)

x â‰¥ 0 â€¦â€¦â€¦â€¦â€¦â€¦ (4)

y â‰¥ 0 â€¦â€¦â€¦â€¦â€¦â€¦ (5)

Graph of inequality (i). Let us draw the graph of the line x + 2y â‰¤ 10.

y = 0 â‡’ x + 2(0) = 10 â‡’ x = 10

x = 0 â‡’ 0 + 2y â‡’ y = 5

(10, 0) and (0, 5) are the points on the line x + 2y = 10.

Putting x = y = 0 in (i) we have 0 â‰¤ 10 which is true. Hence half-plane region containing the origin is the solution region.

Graph of inequality (ii). Let us draw the graph of the line x + y = 1.

At y = 0 â‡’ x + (0) = 1 â‡’ x = 1

x = 0 â‡’ 0 + y = 1 â‡’ y = 1

âˆ´ (1, 0) and (0, 1) are the points on the line x + y = 1.

Putting x = y = 0 in (ii) we have 0 â‰¥ 1 which is false.

Hence half-plane region not containing the origin is the solution region.

Graph of inequality (iii). Let us draw the graph of the line x - y = 0.

On the line

At x = 0 â‡’ 0 â€“ y = 0 â‡’ y = 0

at x = 1 â‡’ 1 â€“ y = 0 â‡’ y = 1

âˆ´ (0, 0) and (1, 1) are on the line x â€“ y = 0.

To determine the region represented by the given inequality (iii) consider the point not lying on the line x â€“ y = 0 say (2, 0) and it his in the half-plane of (iii) if 2 â‰¤ which is not true. Therefore the portion not containing (2, 0) represents the solution set of the given inequality.

Graph of inequality (iv) and (v).

Clearly, x â‰¥ 0, represents the region lying on the right side of y-axis and y â‰¥ 0 represents the region lying above the x-axis.

The common region of the above five regions is the solution sets.