## Chapter 5 Linear Inequalities

Mathematics is the art of saying many things in many different ways. - MAXWELL

### 5.1 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), ’ $\leq$ ’ (less than or equal) and $\geq$ (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.

### 5.2 Inequalities

Let us consider the following situations:

(i) Ravi goes to market with ₹ 200 to buy rice, which is available in packets of $1 kg$. 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 ₹ $30 x$. Since, he has to buy rice in packets only, he may not be able to spend the entire amount of ₹ 200. (Why?) Hence

$$ 30 x<200 \quad \quad \quad \quad \quad \quad \ldots (1) $$

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 ₹ $(40 x+20 y)$ and we have

$$ 40 x+20 y \leq 120 \quad \quad \quad \quad \quad \quad \ldots (2) $$

Since in this case the total amount spent may be upto ₹ 120 . Note that the statement (2) consists of two statements

$ \text{ and } \quad \begin{aligned} & 40 x+20 y<120 \quad \quad \quad \quad \quad \quad \ldots (3) \\ & 40 x+20 y=120 \quad \quad \quad \quad \quad \quad \ldots (4) \end{aligned} $

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 ’ $<$,’, ‘>’, ’ $\leq$ ’ or ’ $\geq$ ’ form an inequality.

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

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

$x<5 ; y>2 ; x \geq 3, y \leq 4$ are some examples of literal inequalities.

$3<5<7($ read as 5 is greater than 3 and less than 7), $3 \leq x<5($ read as $x$ is greater than or equal to 3 and less than 5) and $2<y \leq 4$ are the examples of double inequalities.

Some more examples of inequalities are:

$ \begin{aligned} & a x+b<0 \quad \quad \quad \quad \quad \quad \ldots (5) \\ & a x+b>0 \quad \quad \quad \quad \quad \quad \ldots (6) \\ & a x+b \leq 0 \quad \quad \quad \quad \quad \quad \ldots (7) \\ & a x+b \geq 0 \quad \quad \quad \quad \quad \quad \ldots (8) \\ & a x+b y<c \quad \quad \quad \quad \quad \quad \ldots (9) \\ & a x+b y>c \quad \quad \quad \quad \quad \quad \ldots (10) \\ & a x+b y \leq c \quad \quad \quad \quad \quad \quad \ldots (11) \\ & a x+b y \geq c \quad \quad \quad \quad \quad \quad \ldots (12) \\ & a x^{2}+b x+c \leq 0 \quad \quad \quad \quad \quad \ldots (13) \\ & a x^{2}+b x+c>0 \quad \quad \quad \quad \quad \ldots (14) \end{aligned} $

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 $a \neq 0$, while inequalities from (9) to (12) are linear inequalities in two variables $x$ and $y$ when $a \neq 0, b \neq 0$.

Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities in one variable $x$ when $a \neq 0)$.

In this Chapter, we shall confine ourselves to the study of linear inequalities in one and two variables only.

### 5.3 Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation

Let us consider the inequality (1) of Section 6.2, viz, $30 x<200$

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

$ \begin{aligned} & \text{ For } x=0 \text{, L.H.S. }=30(0)=0<200(\text{ R.H.S. }) \text{, which is true. } \\ & \text{ For } x=1 \text{, L.H.S. }=30(1)=30<200 \text{ (R.H.S.), which is true. } \\ & \text{ For } x=2 \text{, L.H.S. }=30(2)=60<200 \text{, which is true. } \\ & \text{ For } x=3 \text{, L.H.S. }=30(3)=90<200 \text{, which is true. } \\ & \text{ For } x=4 \text{, L.H.S. }=30(4)=120<200 \text{, which is true. } \\ & \text{ For } x=5 \text{, L.H.S. }=30(5)=150<200 \text{, which is true. } \\ & \text{ For } x=6 \text{, L.H.S. }=30(6)=180<200 \text{, which is true. } \\ & \text{ For } x=7 \text{, L.H.S. }=30(7)=210<200 \text{, which is false. } \end{aligned} $

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 ‘>’, $\leq$ ’ becomes ’ $\geq$ ’ and so on) whenever we multiply (or divide) both sides of an inequality by a negative number. It is evident from the facts that

$ \begin{aligned} & 3>2 \text{ while }-3<-2 \\ & -8<-7 \text{ while }(-8)(-2)>(-7)(-2), \text{ i.e., } 16>14 . \end{aligned} $

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.

### Summary

Two real numbers or two algebraic expressions related by the symbols $<,>, \leq$ or $\geq$ 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 $x \leq a$ ( or $x \geq a$ ) on a number line, put a dark circle on the number $a$ and dark the line to the left (or right) of the number $x$.