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:

$$ \begin{align*} & a x+b<0 \tag{5}\\ & a x+b>0 \tag{6}\\ & a x+b \leq 0 \tag{7}\\ & a x+b \geq 0 \tag{8}\\ & a x+b y<c \tag{9}\\ & a x+b y>c \tag{10}\\ & a x+b y \leq c \tag{11}\\ & a x+b y \geq c \tag{12}\\ & a x^{2}+b x+c \leq 0 \tag{13}\\ & a x^{2}+b x+c>0 \tag{14} \end{align*} $$

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.

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 $\quad \frac{30 x}{30}<\frac{200}{30}$ (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.

$$ x=1,2,3,4,5,6 $$

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

$$ \ldots,-3,-2,-1,0,1,2,3,4,5,6 $$

The solution set of the inequality is $ \{ \ldots,-3,-2,-1,0,1,2,3,4,5,6 \} $

Example 2 Solve $5 x-3<3 x+1$ when (i) $x$ is an integer, (ii) $x$ is a real number.

Solution We have, $5 x-3<3 x+1$

or $\quad \quad$ $5 x-3+3<3 x+1+3$ $\quad \quad \quad$ (Rule 1)

or $\quad \quad$ $5 x<3 x+4$

or $\quad \quad$ $5 x-3 x<3 x+4-3 x$ $\quad \quad \quad \quad$ (Rule 2)

or $\quad \quad$ $2 x<4$

or $\quad \quad$ $x<2$ $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$ (Rule 3)

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

$ \ldots,-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 $x \in(-\infty, 2)$.

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 $4 x+3<6 x+7$.

Solution We have, $\quad 4 x+3<6 x+7$

or $\quad 4 x-6 x<6 x+4-6 x$

or $\quad-2 x<4 \quad$ 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 $(-2, \infty)$.

Example 4 Solve $\frac{5-2 x}{3} \leq \frac{x}{6}-5$.

Solution We have $\quad \quad \quad \quad$ $\frac{5-2 x}{3} \leq \frac{x}{6}-5$

or $\quad \quad \quad \quad$ $2(5-2 x) \leq x-30 \text {. }$

or $\quad \quad \quad \quad$ $10-4 x \leq x-30$

or $\quad \quad \quad \quad$ $-5 x \leq-40 \text {, i.e., } x \geq 8$

Thus, all real numbers $x$ which are greater than or equal to 8 are the solutions of the given inequality, i.e., $x \in[8, \infty)$.

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

Solution We have $7 x+3<5 x+9$ or $2 x<6$ or $x<3$

The graphical representation of the solutions are given in Fig 5.1.

Fig 5.1

Example 6 Solve $\frac{3 x-4}{2} \geq \frac{x+1}{4}-1$. Show the graph of the solutions on number line.

Solution We have $ \frac{3 x-4}{2}\geq\frac{x+1}{4}-1$

$ \text{or} \quad \frac{3 x-4}{2} \geq \frac{x-3}{4} $

$ \text{or} \quad 2(3 x-4) \geq(x-3) $

or $\quad \quad \quad \quad$ $6 x-8 \geq x-3$

or $\quad \quad \quad \quad$ $5 x \geq 5$

or $\quad \quad \quad \quad$ $x \geq 1$

The graphical representation of solutions is given in Fig 5.2.

Fig 5.2

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

$ \frac{62+48+x}{3} \geq 60 $

or $\quad \quad \quad \quad 110+x \geq 180$

or $\quad \quad \quad \quad$ $x \geq 70$

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

$$ \begin{equation*} x>10 \tag{1} \end{equation*} $$

$$ \begin{equation*} \text{ and } \quad \quad \quad x>10 \tag{2} \end{equation*} $$

Solving (2), we get

$$ \begin{equation*} 2 x+2<40 \tag{3} \end{equation*} $$

i.e., $$x<19 \quad \quad \quad \quad \quad \quad \ldots (3) $$

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

Miscellaneous Examples

Example 9 Solve $-8 \leq 5 x-3<7$.

Solution In this case, we have two inequalities, $-8 \leq 5 x-3$ and $5 x-3<7$, which we will solve simultaneously. We have $-8 \leq 5 x-3<7$

or $\quad-5 \leq 5 x<10$

$ \text{ or } \quad-1 \leq x<2 $

Example 10 Solve $-5 \leq \frac{5-3 x}{2} \leq 8$.

Solution We have $\quad-5 \leq \frac{5-3 x}{2} \leq 8$

or $\quad-10 \leq 5-3 x \leq 16 \quad$ or $\quad-15 \leq-3 x \leq 11$

or $\quad 5 \geq x \geq-\frac{11}{3}$

which can be written as $\frac{-11}{3} \leq x \leq 5$

Example 11 Solve the system of inequalities:

$$ \begin{aligned} & 3 x-7<5+x \quad\quad\quad\quad\quad\quad\ldots(1) \\ & 11-5 x \leq 1 \quad\quad\quad\quad\quad\quad\ldots(2) \end{aligned} $$

and represent the solutions on the number line.

Solution From inequality (1), we have

$$ 3 x - 7 < 5 + x $$

or $ \quad x < 6 \quad\quad\quad\quad\quad\quad\ldots(3)$

Also, from inequality (2), we have

$$ 11-5 x \leq 1 $$

or $ \quad - 5 x \leq-10 \quad \text{ i.e., } x \geq 2 \quad\quad\quad\quad\quad\quad\ldots(4)$

If we draw the graph of inequalities (3) and (4) on the number line, we see that the values of $x$, which are common to both, are shown by bold line in Fig 5.3.

Thus, solution of the system are real numbers $x$ lying between 2 and 6 including 2, i.e., $2 \leq x<6$

Example 12 In an experiment, a solution of hydrochloric acid is to be kept between $30^{\circ}$ and $35^{\circ}$ Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by $C=\frac{5}{9} \quad(F-32)$, where $C$ and $F$ represent temperature in degree Celsius and degree Fahrenheit, respectively.

Solution It is given that $30<C<35$.

Putting $ C=\frac{5}{9}(F-32), \text{ we get } $ $ 30<\frac{5}{9}(F-32)<35 $

or $\quad\quad\quad$ $ \frac{9}{5} \times(30)<(F-32)<\frac{9}{5} \times(35) $

$ \begin{matrix} \text{ or } & 54<(F-32)<63 \\ \text{ or } & 86<F<95 . \end{matrix} $

Thus, the required range of temperature is between $86^{\circ} F$ and $95^{\circ} F$.

Example 13 A manufacturer has 600 litres of a $12\%$ solution of acid. How many litres of a $30 \%$ acid solution must be added to it so that acid content in the resulting mixture will be more than $15 \%$ but less than $18 \%$ ?

Solution Let $x$ litres of $30 \%$ acid solution is required to be added. Then Total mixture $=(x+600)$ litres

Therefore $30 \% x+12 \%$ of $600>15 \%$ of $(x+600)$

and $\quad \quad \quad 30 \% x+12 \%$ of $600<18 \%$ of $(x+600)$

$ \begin{array}{ll} \text{or} & \frac{30 x}{100}+\frac{12}{100}(600)>\frac{15}{100}(x+600) \\ \\ \text{and} & \frac{30 x}{100}+\frac{12}{100}(600)<\frac{18}{100}(x+600) \\ \\ \text{or}& 30 x+7200>15 x+9000 \\ \text{and} & 30 x+7200<18 x+10800 \\ \text{or} & 15 x>1800 \text{ and } 12 x<3600 \\ \text{or} & x>120 \text{ and } x<300, \\ \text{i.e.} & 120<x<300 \end{array} $

Thus, the number of litres of the $30 %$ solution of acid will have to be more than 120 litres but less than 300 litres.

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 $x$.