Chapter 05 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:
$$ \begin{align*} & a x+b<0 \tag{5}\\ & a x+b>0 \tag{6}\\ & a x+b \leq 0 \tag{7}\\ & a x+b \geq 0 \tag{8}\\ & a x+b y<c \tag{9}\\ & a x+b y>c \tag{10}\\ & a x+b y \leq c \tag{11}\\ & a x+b y \geq c \tag{12}\\ & a x^{2}+b x+c \leq 0 \tag{13}\\ & a x^{2}+b x+c>0 \tag{14} \end{align*} $$
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 nonzero 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 $5 x3<3 x+1$ when (i) $x$ is an integer, (ii) $x$ is a real number.
Solution We have, $5 x3<3 x+1$
or $\quad \quad$ $5 x3+3<3 x+1+3$ $\quad \quad \quad$ (Rule 1)
or $\quad \quad$ $5 x<3 x+4$
or $\quad \quad$ $5 x3 x<3 x+43 x$ $\quad \quad \quad \quad$ (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 $\quad 4 x6 x<6 x+46 x$
or $\quad2 x<4 \quad$ 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 $\frac{52 x}{3} \leq \frac{x}{6}5$.
Solution We have $\quad \quad \quad \quad$ $\frac{52 x}{3} \leq \frac{x}{6}5$
or $\quad \quad \quad \quad$ $2(52 x) \leq x30 \text {. }$
or $\quad \quad \quad \quad$ $104 x \leq x30$
or $\quad \quad \quad \quad$ $5 x \leq40 \text {, i.e., } x \geq 8$
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 $\frac{3 x4}{2} \geq \frac{x+1}{4}1$. Show the graph of the solutions on number line.
Solution We have $ \frac{3 x4}{2}\geq\frac{x+1}{4}1$
$ \text{or} \quad \frac{3 x4}{2} \geq \frac{x3}{4} $
$ \text{or} \quad 2(3 x4) \geq(x3) $
or $\quad \quad \quad \quad$ $6 x8 \geq x3$
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 $(11,13),(13,15),(15,17),(17,19)$
EXERCISE 5.1
1. Solve $24 x<100$, when
(i) $x$ is a natural number.
(ii) $x$ is an integer.
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Answer :
The given inequality is $24 x<100$.
$24 x<100$
$\Rightarrow \dfrac{24 x}{24}<\dfrac{100}{24} \quad$ [Dividing both sides by same positive number]
$\Rightarrow x<\dfrac{25}{6}$
(i) It is evident that $1,2,3$, and 4 are the only natural numbers less than $\dfrac{25}{6}$.
Thus, when $x$ is a natural number, the solutions of the given inequality are $1,2,3$, and 4 .
Hence, in this case, the solution set is $ \{1,2,3,4\} $.
(ii) The integers less than $\dfrac{25}{6}$ are …3, 2, 1, $0,1,2,3,4$.
Thus, when $x$ is an integer, the solutions of the given inequality are
…3, 2, 1, 0, 1, 2, 3, 4.
Hence, in this case, the solution set is $ \{\ldots  3 ,  2 , 1, 0,1,2,3,4\} $.
2. Solve $12 x>30$, when
(i) $x$ is a natural number.
(ii) $x$ is an integer.
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Answer :
The given inequality is  $12 x>30$. $12 x>30$
$\Rightarrow \dfrac{12 x}{12}<\dfrac{30}{12} \quad$ [Dividing both sides by same negative number]
$\Rightarrow x<\dfrac{5}{2}$
(i) There is no natural number less than
$ (\dfrac{5}{2}) $
Thus, when xis a natural number, there is no solution of the given inequality.
(ii) The integers less than $(\dfrac{5}{2})$ are …, 5, 4, 3.
Thus, when xis an integer, the solutions of the given inequality are
…, 5, 4, 3.
Hence, in this case, the solution set is $ \{{5, 4, 3} \}$.
3. Solve $ 5 x3<7 $, when
(i) $x$ is an integer.
(ii) $x$ is a real number.
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Answer :
The given inequality is $5 x  3<7$.
$5 x3<7$
$\Rightarrow 5 x3+3<7+3$
$\Rightarrow 5 x<10$
$\Rightarrow \dfrac{5 x}{5}<\dfrac{10}{5}$
$\Rightarrow x<2$
(i) The integers less than 2are …, 4, 3, 2, 1, 0,1 .
Thus, when xis an integer, the solutions of the given inequality are
…, 4, 3, 2, 1, 0, 1.
Hence, in this case, the solution set is $\{4,  3 ,  2,  1,0,1\}$.
(ii) When $x$ is a real number, the solutions of the given inequality are given by $x<2$, that is, all real numbers $x$ which are less than 2.
Thus, the solution set of the given inequality is $x \in( \infty, 2)$.
4. Solve $3 x+8>2$, when
(i) $x$ is an integer.
(ii) $x$ is a real number.
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Answer :
The given inequality is $3 x+8>2$.
$3 x+8>2$
$\Rightarrow 3 x+88>28$
$\Rightarrow 3 x>6$
$\Rightarrow \dfrac{3 x}{3}>\dfrac{6}{3}$
$\Rightarrow x>2$
(i) The integers greater than 2are 1, 0, 1, 2, ..
Thus, when xis an integer, the solutions of the given inequality are
1, $0,1,2 \ldots$
Hence, in this case, the solution set is $\{ 1,0,1,2, \ldots\}$.
(ii) When xis a real number, the solutions of the given inequality are all the real numbers, which are greater than  2 .
Thus, in this case, the solution set is ( $2, \infty$ ).
Solve the inequalities in Exercises 5 to 16 for real $x$.
5. $4 x+3<5 x+7$
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Answer :
$4 x+3<5 x+7$
$\Rightarrow 4 x+37<5 x+77$
$\Rightarrow 4 x4<5 x$
$\Rightarrow 4 x44 x<5 x4 x$
$\Rightarrow4<x$
Thus, all real numbers $x$, which are greater than 4 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is $(4, \infty)$.
6. $3 x7>5 x1$
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Answer :
$ \Rightarrow 3 x7+7>5 x  1+7$
$\Rightarrow 3 x>5 x+6$
$\Rightarrow 3 x  5 x>5 x+6$ $5 x$
$\Rightarrow  2 x>6$
$\Rightarrow \dfrac{2 x}{2}<\dfrac{6}{2}$
$\Rightarrow x<3$
Thus, all real numbers $x$, which are less than  3 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is ($\infty$,  3 ).
7. $3(x1) \leq 2(x3)$
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Answer :
$3(x1) \leq 2(x3)$
$\Rightarrow 3 x3 \leq 2 x6$
$\Rightarrow 3 x3+3 \leq 2 x6+3$
$\Rightarrow 3 x \leq 2 x3$
$\Rightarrow 3 x2 x \leq 2 x3$  $2 x$
$\Rightarrow x \leq 3$
Thus, all real numbers $x$, which are less than or equal to 3 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is $(\infty,3]$.
8. $3(2x) \geq 2(1x)$
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Answer :
$3(2x) \geq 2(1x)$
$\Rightarrow 63 x \geq 22 x$
$\Rightarrow 63 x+2 x \geq 22 x+2 x$
$\Rightarrow 6x \geq 2$
$ \begin{aligned} & \Rightarrow 6x6 \geq 26 \\ & \Rightarrowx \geq4 \\ & \Rightarrow x \leq 4 \end{aligned} $
Thus, all real numbers $x$, which are less than or equal to 4 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is $(\infty,4]$.
9. $x+\dfrac{x}{2}+\dfrac{x}{3}<11$
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Answer :
$ \begin{aligned} & x+\dfrac{x}{2}+\dfrac{x}{3}<11 \\ & \Rightarrow x(1+\dfrac{1}{2}+\dfrac{1}{3})<11 \\ & \Rightarrow x(\dfrac{6+3+2}{6})<11 \\ & \Rightarrow \dfrac{11 x}{6}<11 \\ & \Rightarrow \dfrac{11 x}{6 \times 11}<\dfrac{11}{11} \\ & \Rightarrow \dfrac{x}{6}<1 \\ & \Rightarrow x<6 \end{aligned} $
Thus, all real numbers $x$, which are less than 6 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is ($\infty$, 6).
10. $\dfrac{x}{3}>\dfrac{x}{2}+1$
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Answer :
$\dfrac{x}{3}>\dfrac{x}{2}+1$
$\Rightarrow \dfrac{x}{3}\dfrac{x}{2}>1$
$\Rightarrow \dfrac{2 x3 x}{6}>1$
$\Rightarrow\dfrac{x}{6}>1$
$\Rightarrowx>6$
$\Rightarrow x<6$
Thus, all real numbers $x$, which are less than  6 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is ($\infty$, 6).
11. $\dfrac{3(x2)}{5} \leq \dfrac{5(2x)}{3}$
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Answer :
$\dfrac{3(x2)}{5} \leq \dfrac{5(2x)}{3}$
$\Rightarrow 9(x2) \leq 25(2x)$
$\Rightarrow 9 x18 \leq 5025 x$
$\Rightarrow 9 x18+25 x \leq 50$
$\Rightarrow 34 x18 \leq 50$
$\Rightarrow 34 x \leq 50+18$
$\Rightarrow 34 x \leq 68$
$\Rightarrow \dfrac{34 x}{34} \leq \dfrac{68}{34}$
$\Rightarrow x \leq 2$
Thus, all real numbers $x$, which are less than or equal to 2 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is [  $\infty, 2 $].
12. $\dfrac{1}{2}(\dfrac{3 x}{5}+4) \geq \dfrac{1}{3}(x6)$
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Answer :
$\dfrac{1}{2}(\dfrac{3 x}{5}+4) \geq \dfrac{1}{3}(x6)$
$\Rightarrow 3(\dfrac{3 x}{5}+4) \geq 2(x6)$
$\Rightarrow \dfrac{9 x}{5}+12 \geq 2 x12$
$\Rightarrow 12+12 \geq 2 x\dfrac{9 x}{5}$
$\Rightarrow 24 \geq \dfrac{10 x9 x}{5}$
$\Rightarrow 24 \geq \dfrac{x}{5}$
$\Rightarrow 120 \geq x$
Thus, all real numbers $x$, which are less than or equal to 120 , are the solutions of the given inequality. Hence, the solution set of the given inequality is $(\infty ,120]$
13. $2(2 x+3)10<6(x2)$
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Answer :
$ \begin{aligned} & 2(2 x+3)10<6(x2) \\ & \Rightarrow 4 x+610<6 x12 \\ & \Rightarrow 4 x4<6 x12 \\ & \Rightarrow4+12<6 x4 x \\ & \Rightarrow 8<2 x \\ & \Rightarrow 4<x \end{aligned} $
Thus, all real numbers $x$, which are greater than or equal to 4 , are the solutions of the given inequality. Hence, the solution set of the given inequality is $(4, \infty)$.
14. $37(3 x+5) \geq 9 x8(x3)$
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Answer :
$37(3 x+5) \geq 9 x8(x3)$
$\Rightarrow 373 x5 \geq 9 x8 x+24$
$\Rightarrow 323 x \geq x+24$
$\Rightarrow 3224 \geq x+3 x$
$\Rightarrow 8 \geq 4 x$
$\Rightarrow 2 \geq x$
Thus, all real numbers $x$, which are less than or equal to 2 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is $( \infty, 2]$.
15. $\dfrac{x}{4}<\dfrac{(5 x2)}{3}\dfrac{(7 x3)}{5}$
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Answer :
$\dfrac{x}{4}<\dfrac{(5 x2)}{3}\dfrac{(7 x3)}{5}$
$\Rightarrow \dfrac{x}{4}<\dfrac{5(5 x2)3(7 x3)}{15}$
$\Rightarrow \dfrac{x}{4}<\dfrac{25 x1021 x+9}{15}$
$\Rightarrow \dfrac{x}{4}<\dfrac{4 x1}{15}$
$\Rightarrow 15 x<4(4 x1)$
$\Rightarrow 15 x<16 x4$
$\Rightarrow 4<16 x15 x$
$\Rightarrow 4<x$
Thus, all real numbers $x$, which are greater than 4 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is $(4, \infty)$.
16. $\dfrac{(2 x1)}{3} \geq \dfrac{(3 x2)}{4}\dfrac{(2x)}{5}$
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Answer :
$\dfrac{(2 x1)}{3} \geq \dfrac{(3 x2)}{4}\dfrac{(2x)}{5}$
$\Rightarrow \dfrac{(2 x1)}{3} \geq \dfrac{5(3 x2)4(2x)}{20}$
$\Rightarrow \dfrac{(2 x1)}{3} \geq \dfrac{15 x108+4 x}{20}$
$\Rightarrow \dfrac{(2 x1)}{3} \geq \dfrac{19 x18}{20}$
$\Rightarrow 20(2 x1) \geq 3(19 x18)$
$\Rightarrow 40 x20 \geq 57 x54$
$\Rightarrow20+54 \geq 57 x40 x$
$\Rightarrow 34 \geq 17 x$
$\Rightarrow 2 \geq x$
Thus, all real numbers $x$, which are less than or equal to 2 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is $(\infty ,2]$
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line
17. $3 x2<2 x+1$
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Answer :
$3 x2<2 x+1$
$\Rightarrow 3 x2 x<1+2$
$\Rightarrow x<3$
The graphical representation of the solutions of the given inequality is as follows.
18. $3(1x)<2(x+4)$
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Answer :
$3(1x)<2(x+4)$
$\Rightarrow 33 x<2 x+8$
$\Rightarrow 38<2 x+3 x$
$\Rightarrow5<5 x$
$\Rightarrow \dfrac{5}{5}<\dfrac{5 x}{5}$
$\Rightarrow1<x$
The graphical representation of the solutions of the given inequality is as follows.
19. $5 x3 \geq 3 x5$
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Answer :
$5 x3 \geq 3 x5$
$\Rightarrow 5 x3 x \geq5+3$
$\Rightarrow 2 x \geq2$
$\Rightarrow \dfrac{2 x}{2} \geq \dfrac{2}{2}$
$\Rightarrow x \geq1$
The graphical representation of the solutions of the given inequality is as follows.
20. $\dfrac{x}{2} \geq \dfrac{(5 x2)}{3}\dfrac{(7 x3)}{5}$
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Answer :
$\dfrac{x}{2} \geq \dfrac{(5 x2)}{3}\dfrac{(7 x3)}{5}$
$\Rightarrow \dfrac{x}{2} \geq \dfrac{5(5 x2)3(7 x3)}{15}$
$\Rightarrow \dfrac{x}{2} \geq \dfrac{25 x1021 x+9}{15}$
$\Rightarrow \dfrac{x}{2} \geq \dfrac{4 x1}{15}$
$\Rightarrow 15 x \geq 2(4 x1)$
$\Rightarrow 15 x \geq 8 x2$
$\Rightarrow 15 x8 x \geq 8 x28 x$
$\Rightarrow 7 x \geq2$
$\Rightarrow x \geq\dfrac{2}{7}$
The graphical representation of the solutions of the given inequality is as follows.
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.
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Answer :
Let $x$ be the marks obtained by Ravi in the third unit test.
Since the student should have an average of at least 60 marks,
$\dfrac{70+75+x}{3} \geq 60$
$\Rightarrow 145+x \geq 180$
$\Rightarrow x \geq 180145$
$\Rightarrow x \geq 35$
Thus, the student must obtain a minimum of 35 marks 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.
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Answer :
Let $x$ be the marks obtained by Sunita in the fifth examination.
In order to receive grade ‘A’ in the course, she must obtain an average of 90 marks or more in five examinations.
Therefore,
$ \begin{aligned} & \dfrac{87+92+94+95+x}{5} \geq 90 \\ & \Rightarrow \dfrac{368+x}{5} \geq 90 \\ & \Rightarrow 368+x \geq 450 \\ & \Rightarrow x \geq 450368 \\ & \Rightarrow x \geq 82 \end{aligned} $
Thus, Sunita must obtain greater than or equal to 82 marks in the fifth examination.
23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11 .
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Answer :
Let $x$ be the smaller of the two consecutive odd positive integers. Then, the other integer is $x+2$.
Since both the integers are smaller than 10 ,
$x+2<10$
$\Rightarrow x<10$  2
$\Rightarrow x<8 \ldots$ (i)
Also, the sum of the two integers is more than 11 .
$\therefore x+(x+2)>11$
$\Rightarrow 2 x+2>11$
$\Rightarrow 2 x>11$  2
$\Rightarrow 2 x>9$
$\Rightarrow x>\dfrac{9}{2}$
$\Rightarrow x>4.5$
From (i) and (ii), we obtain
Since $x$ is an odd number, $x$ can take the values, 5 and 7 .
Thus, the required possible pairs are $(5,7)$ and $(7,9)$.
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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Answer
1. Define variables: Let x be the smaller even integer. Since the consecutive even number is next in order, it will be x + 2.
2. Set up the conditions: We know two things:

Both integers must be larger than 5: x > 5 and x>3

Their sum must be less than 23: x + (x + 2) < 23 (This combines x and x + 2 because they represent the consecutive even numbers).
3. Solve the inequality: Combine like terms in the inequality: 2x + 2 < 23 Subtract 2 from both sides:2x < 21 Divide both sides by 2 (since we’re dealing with even integers):x < 10.5
4. Consider the limitations: We found that x must be less than 10.5. However, x also needs to be greater than 5 and an integer (since it represents an even number). Therefore, valid values for x are 6, 8, and 10.
Find the corresponding even integers:
 If x = 6, then the consecutive even number is x + 2 = 8.
 If x = 8, then the consecutive even number is x + 2 = 10.
 If x = 10, then the consecutive even number is x + 2 = 12
Therefore, the valid pairs of consecutive even positive integers are: (6, 8), (8, 10)and (10,12).
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.
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Answer :
Let the length of the shortest side of the triangle be $x cm$.
Then, length of the longest side $=3 x cm$
Length of the third side $=(3 x  2) cm$
Since the perimeter of the triangle is at least $61 cm$,
$x cm+3 x cm+(3 x2) cm \geq 61 cm$
$\Rightarrow 7 x2 \geq 61$
$\Rightarrow 7 x \geq 61+2$
$\Rightarrow 7 x \geq 63$
$\Rightarrow \dfrac{7 x}{7} \geq \dfrac{63}{7}$
$\Rightarrow x \geq 9$
Thus, the minimum length of the shortest side is $9 cm$.
26. A man wants to cut three lengths from a single piece of board of length $91 cm$. The second length is to be $3 cm$ 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 $5 cm$ longer than the second?
[Hint: If $x$ is the length of the shortest board, then $x,(x+3)$ and $2 x$ are the lengths of the second and third piece, respectively. Thus, $x+(x+3)+2 x \leq 91$ and $2 x \geq(x+3)+5$].
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Answer :
Let the length of the shortest piece be $x cm$. Then, length of the second piece and the third piece are $(x+3) cm$ and $2 x cm$ respectively.
Since the three lengths are to be cut from a single piece of board of length $91 cm$,
$x cm+(x+3) cm+2 x cm \leq 91 cm$
$\Rightarrow 4 x+3 \leq 91$ $\Rightarrow 4 x \leq 91$  3
$\Rightarrow 4 x \leq 88$
$\Rightarrow \dfrac{4 x}{4} \leq \dfrac{88}{4}$
$\Rightarrow x \leq 22$
Also, the third piece is at least $5 cm$ longer than the second piece.
$\therefore 2 x \geq(x+3)+5$
$\Rightarrow 2 x \geq x+8$
$\Rightarrow x \geq 8 \ldots(2)$
From (1) and (2), we obtain
$8 \leq x \leq 22$
Thus, the possible length of the shortest board is greater than or equal to $8 cm$ but less than or equal to $22 cm$.
Miscellaneous Examples
Example 9 Solve $8 \leq 5 x3<7$.
Solution In this case, we have two inequalities, $8 \leq 5 x3$ and $5 x3<7$, which we will solve simultaneously. We have $8 \leq 5 x3<7$
or $\quad5 \leq 5 x<10$
$ \text{ or } \quad1 \leq x<2 $
Example 10 Solve $5 \leq \frac{53 x}{2} \leq 8$.
Solution We have $\quad5 \leq \frac{53 x}{2} \leq 8$
or $\quad10 \leq 53 x \leq 16 \quad$ or $\quad15 \leq3 x \leq 11$
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:
$$ \begin{aligned} & 3 x7<5+x \quad\quad\quad\quad\quad\quad\ldots(1) \\ & 115 x \leq 1 \quad\quad\quad\quad\quad\quad\ldots(2) \end{aligned} $$
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
$$ 115 x \leq 1 $$
or $ \quad  5 x \leq10 \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 $C=\frac{5}{9} \quad(F32)$, 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}(F32), \text{ we get } $ $ 30<\frac{5}{9}(F32)<35 $
or $\quad\quad\quad$ $ \frac{9}{5} \times(30)<(F32)<\frac{9}{5} \times(35) $
$ \begin{matrix} \text{ or } & 54<(F32)<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.
Miscellaneous Exercise on Chapter 5
Solve the inequalities in Exercises 1 to 6.
1. $2 \leq 3 x4 \leq 5$
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Answer :
$2 \leq 3 x4 \leq 5$
$\Rightarrow 2+4 \leq 3 x4+4 \leq 5+4$
$\Rightarrow 6 \leq 3 x \leq $ 9
$\Rightarrow 2 \leq x \leq 3$
Thus, all the real numbers, $x$, which are greater than or equal to 2 but less than or equal to 3 , are the solutions of the given inequality. The solution set for the given inequalityis $[2,3]$.
2. $6 \leq3(2 x4)<12$
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Answer :
$6 \leq A a3(2 x4)<12$
$\Rightarrow 2 \leq (2 x4) < 4$
$\Rightarrow2 \geq 2 x4>4$
$\Rightarrow 4$  $2 \geq 2 x>4$  4
$\Rightarrow 2 \geq 2 x>0$
$\Rightarrow 1 > = x>0$
Thus, the solution set for the given inequalityis $(0,1]$.
3. $3 \leq 4\dfrac{7 x}{2} \leq 18$
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Answer :
$3 \leq 4\dfrac{7 x}{2} \leq 18$
$\Rightarrow34 \leq\dfrac{7 x}{2} \leq 184$
$\Rightarrow7 \leq\dfrac{7 x}{2} \leq 14$
$\Rightarrow 7 \geq \dfrac{7 x}{2} \geq14$
$\Rightarrow 1 \geq \dfrac{x}{2} \geq2$
$\Rightarrow 2 \geq x \geq4$
Thus, the solution set for the given inequalityis [4, 2].
4. $15<\dfrac{3(x2)}{5} \leq 0$
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Answer :
$15<\dfrac{3(x2)}{5} \leq 0$
$\Rightarrow$  $75<3(x  2) \leq 0$
$\Rightarrow$  $25<x  ~ 2 \leq 0$ $\Rightarrow  25+2<x \leq 2$
$\Rightarrow 23<x \leq 2$
Thus, the solution set for the given inequalityis (23, 2]
5. $12<4\dfrac{3 x}{5} \leq 2$
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Answer :
$12<4\dfrac{3 x}{5} \leq 2$
$\Rightarrow124<\dfrac{3 x}{5} \leq 24$
$\Rightarrow16<\dfrac{3 x}{5} \leq2$
$\Rightarrow80<3 x \leq10$
$\Rightarrow \dfrac{80}{3}<x \leq \dfrac{10}{3}$
Thus, the solution set for the given inequalityis $(\dfrac{80}{3}, \dfrac{10}{3}]$.
6. $7 \leq \dfrac{(3 x+11)}{2} \leq 11$.
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Answer :
$7 \leq \dfrac{(3 x+11)}{2} \leq 11$
$\Rightarrow 14 \leq 3 x+11 \leq 22$
$\Rightarrow 1411 \leq 3 x \leq 2211$
$\Rightarrow 3 \leq 3 x \leq 11$
$\Rightarrow 1 \leq x \leq \dfrac{11}{3}$
Thus, the solution set for the given inequalityis $[1, \dfrac{11}{3}]$.
Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.
7. $5 x+1>24,5 x1<24$
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Answer :
$5 x+1>24$
$\Rightarrow 5 x>25$
$\Rightarrow x>5$
$5 x1<24$
$\Rightarrow 5 x<25$
$\Rightarrow x<5$
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is $(5,5)$. The solution of the given system of inequalities can be represented on number line as
8. $2(x1)<x+5,3(x+2)>2x$
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Answer :
$2(x1)<x+5$
$\Rightarrow 2 x2<x+5$
$\Rightarrow 2 xx<5+2$
$\Rightarrow x<7 \ldots$ (1)
$3(x+2)>2x$
$\Rightarrow 3 x+6>2x$
$\Rightarrow 3 x+x>26$
$\Rightarrow 4 x>4$
$\Rightarrow x>1 \ldots(2)$
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (1, 7). The solution of the given system of inequalities can be represented on number line as
9. $3 x7>2(x6), 6x>112 x$
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Answer :
$3 x$  $7>2 (x $  6 $ ) $
$\Rightarrow 3 x$  $7>2 x$  12
$\Rightarrow 3 x$  $2 x>$  $12+7$
$\Rightarrow x>$ 5
6  $x>11$  $2 x$
$\Rightarrow  x+2 x>11 6$
$\Rightarrow x>5$.
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is $(5, \infty)$. The solution of the given system of inequalities can be represented on number line as
10. $5(2 x7)3(2 x+3) \leq 0,2 x+19 \leq 6 x+47$.
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Answer :
$5(2 x7)3(2 x+3) \leq 0$
$\Rightarrow 10 x356 x9 \leq 0$
$\Rightarrow 4 x44 \leq 0$
$\Rightarrow 4 x \leq 44$
$\Rightarrow x \leq 11 \ldots$ (1)
$2 x+19 \leq 6 x+47$
$\Rightarrow 1947$ $\leq x2 x$
$\Rightarrow28 \leq x $
$\Rightarrow7 \leq x$…
From (1) and (2), it can be concluded that the solution set for the given system of inequalities is [7, 11]. The solution of the given system of inequalities can be represented on number line as
11. A solution is to be kept between $68^{\circ} F$ and $77^{\circ} F$. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by $ F=\dfrac{9}{5} C+32 ? $
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Answer :
Since the solution is to be kept between $68^{\circ} F$ and $77^{\circ} F$,
$68<F<77$
Putting $F=\dfrac{9}{5} C+32$, we obtain
$68<\dfrac{9}{5} C+32<77$
$\Rightarrow 6832<\dfrac{9}{5} C<7732$
$\Rightarrow 36<\dfrac{9}{5} C<45$
$\Rightarrow 36 \times \dfrac{5}{9}<C<45 \times \dfrac{5}{9}$
$\Rightarrow 20<C<25$
Thus, the required range of temperature in degree Celsius is between $20^{\circ} C$ and $25^{\circ} C$.
12. A solution of $8 %$ boric acid is to be diluted by adding a $2 %$ boric acid solution to it. The resulting mixture is to be more than $4 %$ but less than $6 %$ boric acid. If we have 640 litres of the $8 %$ solution, how many litres of the $2 %$ solution will have to be added?
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Answer :
Let $x$ litres of $2 %$ boric acid solution is required to be added.
Then,total mixture $=(x+640)$ litres
This resulting mixture is to be more than $4 %$ but less than $6 %$ boric acid.
$\therefore 2 % x+8 %$ of $640>4 %$ of $(x+640)$
And, $2 % x+8 %$ of $640<6 %$ of $(x+640)$
$2 % x+8 %$ of $640>4 %$ of $(x+640)$
$\Rightarrow \dfrac{2}{100} x+\dfrac{8}{100}(640)>\dfrac{4}{100}(x+640)$
$\Rightarrow 2 x+5120>4 x+2560$
$\Rightarrow 5120$  $2560>4 x  2 x$
$\Rightarrow 5120$  $2560>2 x$
$\Rightarrow 2560>2 x$
$\Rightarrow 1280>x$
$2 % x+8 %$ of $640<6 %$ of $(x+640)$
$\dfrac{2}{100} x+\dfrac{8}{100}(640)<\dfrac{6}{100}(x+640)$
$\Rightarrow 2 x+5120<6 x+3840$
$\Rightarrow 5120$  $3840<6 x  2 x$
$\Rightarrow 1280<4 x$
$\Rightarrow 320<x$
$\therefore 320<x<1280$
Thus, the number of litres of $2 %$ of boric acid solution that is to be added will have to be more than 320 litres but less than 1280 litres.
13. How many litres of water will have to be added to 1125 litres of the $45 %$ solution of acid so that the resulting mixture will contain more than $25 %$ but less than $30 %$ acid content?
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Answer :
Let $x$ litres of water is required to be added.
Then,total mixture $=(x+1125)$ litres
It is evident that the amount of acid contained in the resulting mixture is $45 %$ of 1125 litres.
This resulting mixture will contain more than $25 %$ but less than $30 %$ acid content.
$\therefore 30 %$ of $(1125+x)>45 %$ of 1125
And, $25 %$ of $(1125+x)<45 %$ of 1125
$30 %$ of $(1125+x)>45 %$ of 1125 $\Rightarrow \dfrac{30}{100}(1125+x)>\dfrac{45}{100} \times 1125$
$\Rightarrow 30(1125+x)>45 \times 1125$
$\Rightarrow 30 \times 1125+30 x>45 \times 1125$
$\Rightarrow 30 x>45 \times 112530 \times 1125$
$\Rightarrow 30 x>(4530) \times 1125$
$\Rightarrow x>\dfrac{15 \times 1125}{30}=562.5$
$25 %$ of $(1125+x)<45 %$ of 1125
$\Rightarrow \dfrac{25}{100}(1125+x)<\dfrac{45}{100} \times 1125$
$\Rightarrow 25(1125+x)>45 \times 1125$
$\Rightarrow 25 \times 1125+25 x>45 \times 1125$
$\Rightarrow 25 x>45 \times 112525 \times 1125$
$\Rightarrow 25 x>(4525) \times 1125$
$\Rightarrow x>\dfrac{20 \times 1125}{25}=900$
$\therefore 562.5<x<900$
Thus, the required number of litres of water that is to be added will have to be more than 562.5 but less than 900 .
14. IQ of a person is given by the formula
$$ IQ=\dfrac{MA}{CA} \times 100 $$
where MA is mental age and CA is chronological age. If $80 \leq IQ \leq 140$ for a group of 12 years old children, find the range of their mental age.
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Answer :
It is given that for a group of 12 years old children, $80 \leq IQ \leq 140 \ldots$ (i)
For a group of 12 years old children, $CA=12$ years
$ IQ=\dfrac{MA}{12} \times 100 $
Putting this value of IQ in (i), we obtain
$ \begin{aligned} & 80 \leq \dfrac{MA}{12} \times 100 \leq 140 \\ & \Rightarrow 80 \times \dfrac{12}{100} \leq MA \leq 140 \times \dfrac{12}{100} \\ & \Rightarrow 9.6 \leq MA \leq 16.8 \end{aligned} $
Thus, the range of mental age of the group of 12 years old children is $9.6 \leq MA \leq 16.8$.
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 $x$.