6

Chapter

vMathematics is the art of saying many things in many

different ways. – MAXWELLv

6.1 Introduction

In earlier classes, we have studied equations in one variable and two variables and also

solved some statement problems by translating them in the form of equations. Now a

natural question arises: ‘Is it always possible to translate a statement problem in the

form of an equation? For example, the height of all the students in your class is less

than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we

get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘≤’ (less than

or equal) and ≥ (greater than or equal) which are known as inequalities.

In this Chapter, we will study linear inequalities in one and two variables. The

study of inequalities is very useful in solving problems in the field of science, mathematics,

statistics, economics, psychology, etc.

6.2 Inequalities

Let us consider the following situations:

(i) Ravi goes to market with ` 200 to buy rice, which is available in packets of 1kg. The

price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he

buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets

only, he may not be able to spend the entire amount of ` 200. (Why?) Hence

30x < 200

... (1)

Clearly the statement (i) is not an equation as it does not involve the sign of equality.

(ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one

register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of

registers and y, the number of pens which Reshma buys, then the total amount spent by

her is ` (40x + 20y) and we have

40x + 20y ≤ 120 ... (2)

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Since in this case the total amount spent may be upto ` 120. Note that the statement (2)

consists of two statements

40x + 20y < 120 ... (3)

and 40x + 20y = 120 ... (4)

Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation.

Definition 1

Two real numbers or two algebraic expressions related by the symbol

‘<’, ‘>’, ‘

≤’ or ‘≥’ form an inequality

.

Statements such as (1), (2) and (3) above are inequalities.

3 < 5; 7 > 5 are the examples of numerical inequalities while

x < 5; y > 2; x ≥ 3, y ≤ 4 are some examples of literal inequalities.

3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3

< x < 5 (read as x is greater

than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities.

Some more examples of inequalities are:

ax + b < 0 ... (5)

ax + b > 0 ... (6)

ax + b ≤ 0 ... (7)

ax + b ≥ 0 ... (8)

ax + by < c ... (9)

ax + by

> c ... (10)

ax + by ≤ c ... (11)

ax + by ≥ c ... (12)

ax

2

+ bx + c ≤ 0 ... (13)

ax

2

+ bx + c > 0 ... (14)

Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8),

(11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear

inequalities in one variable x when a

≠

0, while inequalities from (9) to (12) are linear

inequalities in two variables x and y when a

≠

0, b

≠

0.

Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities

in one variable x when a

≠

0).

In this Chapter, we shall confine ourselves to the study of linear inequalities in one

and two variables only.

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6.3 Algebraic Solutions of Linear Inequalities in One

Variable and their

Graphical Representation

Let us consider the inequality (1) of Section 6.2, viz, 30x <

200

Note that here x denotes the number of packets of rice.

Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this

inequality is 30x and right hand side (RHS) is 200. Therefore, we have

For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true.

For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true.

For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true.

For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true.

For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true.

For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true.

For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true.

For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false.

In the above situation, we find that the values of x, which makes the above

inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above

inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6}

is called its solution set.

Thus, any solution of an inequality in one variable is a value of the variable

which makes it a true statement.

We have found the solutions of the above inequality by trial and error method

which is not very efficient. Obviously, this method is time consuming and sometimes

not feasible. We must have some better or systematic techniques for solving inequalities.

Before that we should go through some more properties of numerical inequalities and

follow them as rules while solving the inequalities.

You will recall that while solving linear equations, we followed the following rules:

Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation.

Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero

number.

In the case of solving inequalities, we again follow the same rules except with a

difference that in Rule 2, the sign of inequality is reversed (i.e., ‘<‘ becomes ‘>’, ≤’

becomes ‘≥’ and so on) whenever we multiply (or divide) both sides of an inequality by

a negative number. It is evident from the facts that

3 > 2 while – 3 < – 2,

– 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14.

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Thus, we state the following rules for solving an inequality:

Rule 1 Equal numbers may be added to (or subtracted from) both sides of an inequality

without affecting the sign of inequality.

Rule 2 Both sides of an inequality can be multiplied (or divided) by the same positive

number. But when both sides are multiplied or divided by a negative number, then the

sign of inequality is reversed.

Now, let us consider some examples.

Example 1

Solve 30

x < 200 when

(i) x is a natural number,

(ii)

x is an integer.

Solution

We are given 30 x < 200

or

30 200

30 30

x

<

(Rule 2), i.e., x < 20 / 3.

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

statement true.

1, 2, 3, 4, 5, 6.

The solution set of the inequality is {1,2,3,4,5,6}.

(ii) When x is an integer, the solutions of the given inequality are

..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6

The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}

Example 2 Solve 5x – 3 < 3x +1 when

(i) x is an integer, (ii) x is a real number.

Solution We have, 5x –3 < 3x + 1

or 5x –3 + 3 < 3x +1 +3 (Rule 1)

or 5x < 3x +4

or 5x – 3x < 3x + 4 – 3x (Rule 1)

or 2x < 4

or x < 2 (Rule 2)

(i) When x is an integer, the solutions of the given inequality are

..., – 4, – 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 inequality is x ∈ (– ∞, 2).

We have considered solutions of inequalities in the set of natural numbers, set of

integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall

solve the inequalities in this Chapter in the set of real numbers.

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120 MATHEMATICS

Example 3 Solve 4x + 3 < 6x +7.

Solution

We have, 4x + 3 < 6x + 7

or 4x – 6x < 6x + 4 – 6

x

or – 2x < 4 or

x > – 2

i.e., all the real numbers which are greater than –2, are the solutions of the given

inequality. Hence, the solution set is (–2, ∞).

Example 4 Solve

5 2

5

3 6

– x x

–

≤

.

Solution We have

5 2

5

3 6

– x x

–

≤

or 2 (5 – 2x) ≤ x – 30.

or 10 – 4x ≤ x – 30

or – 5x ≤ – 40, i.e., x ≥ 8

Thus, all real numbers x which are greater than or equal to 8 are the solutions of the

given inequality, i.e., x ∈ [8, ∞).

Example 5 Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.

Solution We have 7x + 3 < 5x + 9 or

2

x < 6 or x < 3

The graphical representation of the solutions are given in Fig 6.1.

Fig 6.1

Example 6 Solve

3 4 1

1

2 4

x x

− +

≥ −

. Show the graph of the solutions on number line.

Solution We have

3 4 1

1

2 4

x x− +

≥ −

or

3 4 3

2 4

x x

− −

≥

or 2 (3x – 4) ≥ (x – 3)

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LINEAR INEQUALITIES 121

or 6x – 8 ≥ x – 3

or 5x ≥ 5 or x ≥ 1

The graphical representation of solutions is given in Fig 6.2.

Fig 6.2

Example 7 The marks obtained by a student of Class XI in first and second terminal

examination are 62 and 48, respectively. Find the minimum marks he should get in the

annual examination to have an average of at least 60 marks.

Solution Let x be the marks obtained by student in the annual examination. Then

62 48

60

3

x

+ +

≥

or 110 + x ≥ 180

or x ≥ 70

Thus, the student must obtain a minimum of 70 marks to get an average of at least

60 marks.

Example 8 Find all pairs of consecutive odd natural numbers, both of which are larger

than 10, such that their sum is less than 40.

Solution Let x be the smaller of the two consecutive odd natural number, so that the

other one is x +2. Then, we should have

x > 10 ... (1)

and x + ( x + 2) < 40 ... (2)

Solving (2), we get

2x + 2 < 40

i.e., x < 19 ... (3)

From (1) and (3), we get

10 < x < 19

Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required

possible pairs will be

(11, 13), (13, 15), (15, 17), (17, 19)

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122 MATHEMATICS

EXERCISE 6.1

1. Solve 24

x < 100, when

(i) x is a natural number. (ii) x is an integer.

2. Solve – 12

x > 30, when

(i) x is a natural number. (ii) x is an integer.

3. Solve 5x – 3 < 7, when

(i) x is an integer. (ii) x is a real number.

4. Solve 3

x + 8 >2, when

(i) x is an integer. (ii) x is a real number.

Solve the inequalities in Exercises 5 to 16 for real x.

5. 4x + 3 < 5

x + 7 6. 3x – 7 > 5x – 1

7. 3(x – 1) ≤ 2 (x – 3) 8. 3 (2 –

x) ≥ 2 (1 – x)

9.

11

2 3

x x

x

+ + <

10.

1

3 2

x x

> +

11.

3( 2) 5(2 )

5 3

x x

− −

≤

12.

1 3 1

4 ( 6)

2 5 3

x

x

+ ≥ −

13. 2 (2x + 3) – 10 < 6 (x – 2) 14. 37 – (3x + 5) > 9x – 8 (x – 3)

15.

(5 2) (7 3)

4 3 5

x x x

− −

< −

16.

(2 1) (3 2) (2 )

3 4 5

x x x

− − −

≥ −

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each

case on number line

17. 3x – 2 < 2x + 1 18. 5x – 3 > 3x – 5

19. 3 (1 – x) < 2 (x + 4) 20.

(5 – 2) (7 –3)

–

2 3 5

x x x

≥

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 at least 60 marks.

22. To receive Grade ‘A’ in a 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.

23. Find all pairs of consecutive odd positive integers both of which are smaller than

10 such that their sum is more than 11.

24. Find all pairs of consecutive even positive integers, both of which are larger than

5 such that their sum is less than 23.

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LINEAR INEQUALITIES 123

Fig 6.3 Fig 6.4

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.

26. A man wants to cut three lengths from a single piece of board of length 91cm.

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 of the shortest

board if the third piece is to be at least 5cm 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) + 2

x ≤ 91 and

2x ≥ (x + 3) + 5].

6.4 Graphical Solution of Linear Inequalities in Two Variables

In earlier section, we have seen that a graph of an inequality in one variable is a visual

representation and is a convenient way to represent the solutions of the inequality.

Now, we will discuss graph of a linear inequality in two variables.

We know that a line divides the Cartesian plane into two parts. Each part is

known as a half plane. A vertical line will divide the plane in left and right half planes

and a non-vertical line will divide the plane into lower and upper half planes

(Figs. 6.3 and 6.4).

A point in the Cartesian plane will either lie on a line or will lie in either of the half

planes I or II. We shall now examine the relationship, if any, of the points in the plane

and the inequalities ax + by <

c or ax + by > c.

Let us consider the line

ax + by = c, a ≠ 0, b ≠ 0 ... (1)

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124 MATHEMATICS

Fig 6.5

There are three possibilities namely:

(i) ax + by = c (ii) ax + by > c (iii) ax + by < c.

In case (i), clearly, all points (x, y) satisfying (i) lie on the line it represents and

conversely. Consider case (ii), let us first assume that b > 0. Consider a point P (α,β)

on the line ax + by = c, b > 0, so that

aα

+ bβ = c.

Take an arbitrary point

Q (α , γ) in the half plane II (Fig 6.5).

Now, from Fig 6.5, we interpret,

γ > β (Why?)

or b

γ

> bβ or aα + b γ > aα + bβ

(Why?)

or aα + b γ > c

i.e., Q(α,

γ

) satisfies the inequality

ax + by > c.

Thus, all the points lying in the half

plane II above the line ax + by = c satisfies

the inequality ax + by > c. Conversely, let (α, β) be a point on line ax + by = c and an

arbitrary point Q(α, γ) satisfying

ax + by > c

so that aα + bγ > c

⇒

aα + b γ > aα + bβ (Why?)

⇒

γ > β (as b > 0)

This means that the point (α,

γ

) lies in the half plane II.

Thus, any point in the half plane II satisfies ax + by > c, and conversely any point

satisfying the inequality ax + by > c lies in half plane II.

In case b < 0, we can similarly prove that any point satisfying ax + by > c lies in

the half plane I, and conversely.

Hence, we deduce that all points satisfying ax + by > c lies in one of the half

planes II or I according as b > 0 or b < 0, and conversely.

Thus, graph of the inequality ax + by > c will be one of the half plane (called

solution region) and represented by shading in the corresponding half plane.

Note 1 The region containing all the solutions of an inequality is called the

solution region.

2. In order to identify the half plane represented by an inequality, it is just sufficient

to take any point (a, b) (not online) and check whether it satisfies the inequality or

not. If it satisfies, then the inequality represents the half plane and shade the region

A

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LINEAR INEQUALITIES 125

Fig 6.6

which contains the point, otherwise, the inequality represents that half plane which

does not contain the point within it. For convenience, the point (0, 0) is preferred.

3. If an inequality is of the type ax + by ≥ c or ax + by ≤ c, then the points on the

line ax + by = c are also included in the solution region. So draw a dark line in the

solution region.

4. If an inequality is of the form ax + by

> c or

ax + by < c, then the points on the

line ax + by = c are not to be included in the solution region. So draw a broken or

dotted line in the solution region.

In Section 6.2, we obtained the following linear inequalities in two variables

x and y: 40x + 20y ≤ 120 ... (1)

while translating the word problem of purchasing of registers and pens by Reshma.

Let us now solve this inequality keeping in mind that x and y can be only whole

numbers,

since the number of articles cannot be a fraction or a negative number. In

this case, we find the pairs of values of x and y, which make the statement (1) true. In

fact, the set of such pairs will be the solution set of the inequality (1).

To start with, let

x = 0. Then L.H.S. of (1) is

40x + 20y = 40 (0) + 20

y = 20y.

Thus, we have

20y ≤ 120 or y ≤ 6 ... (2)

For x = 0, the corresponding values of y can be 0, 1, 2, 3, 4, 5, 6 only. In this case, the

solutions of (1) are (0, 0), (0, 1), (0,2), (0,3), (0,4),

(0, 5) and (0, 6).

Similarly, other solutions of (1), when

x = 1, 2 and 3 are: (1, 0), (1, 1), (1, 2), (1,

3), (1, 4), (2, 0), (2, 1), (2, 2), (3, 0)

This is shown in Fig 6.6.

Let us now extend the domain of x and y

from whole numbers to real numbers, and see

what will be the solutions of (1) in this case.

You will see that the graphical method of solution

will be very convenient in this case. For this

purpose, let us consider the (corresponding)

equation and draw its graph.

40x + 20y = 120 ... (3)

In order to draw the graph of the inequality

(1), we take one point say (0, 0), in half plane I

and check whether values of x and y satisfy the

inequality or not.

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