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)

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)

ax2 + bx + c ≤ 0 … (13)

ax2 + 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.

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.

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 (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.

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

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

or 4x – 6x < 6x + 4 – 6x

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 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)

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)

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).

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

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.

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) + 20y = 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.

We observe that x = 0, y = 0 satisfy the inequality. Thus, we say that the half plane I is the graph (Fig 6.7) of the inequality. Since the points on the line also satisfy the inequality (1) above, the line is also a part of the graph.

Thus, the graph of the given inequality is half plane I including the line itself. Clearly half plane II is not the part of the graph. Hence, solutions of inequality (1) will consist of all the points of its graph (half plane I including the line).

We shall now consider some examples to explain the above procedure for solving a linear inequality involving two variables.

### SUMMARY

• Two real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality.

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

• 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 inequality is reversed.

• The values of x, which make an inequality a true statement, are called solutions of the inequality.

• To represent x < a (or x > a) on a number line, put a circle on the number a and dark line to the left (or right) of the number a.

• To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number a and dark the line to the left (or right) of the number x.

• If an inequality is having ≤ or ≥ symbol, then the points on the line are also included in the solutions of the inequality and the graph of the inequality lies left (below) or right (above) of the graph of the equality represented by dark line that satisfies an arbitrary point in that part.

• If an inequality is having < or > symbol, then the points on the line are not included in the solutions of the inequality and the graph of the inequality lies to the left (below) or right (above) of the graph of the corresponding equality represented by dotted line that satisfies an arbitrary point in that part.

• The solution region of a system of inequalities is the region which satisfies all the given inequalities in the system simultaneously.