Mathematics LINEAR INEQUALITIES
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### Topics Covered

star Introduction
star Inequalities
star Algebraic Solutions of Linear Inequalities in One Variable

### Introduction

● 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.

● 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, optimisation problems, economics, psychology, etc.

### Inequalities

\color{purple}ul(✓✓) \color{purple} \mathbf("DEFINITION ALERT")
● Two real numbers or two algebraic expressions related by the symbol ‘<’, \ \ ‘>’, \ \ ‘≤’ or ‘≥’ form an inequality.

\color{green}✍️\color{green}\mathbf("Numerical inequalities")
● 3 < 5; 7 > 5 are the examples of numerical inequalities while

\color{green}✍️\color{green}\mathbf("Literal inequalities")
● x < 5; y > 2; x ≥ 3, y ≤ 4 are the examples of literal inequalities.

\color{green}✍️\color{green}\mathbf("Double 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.

\color{green}✍️\color{green}\mathbf("Linear inequalities in one Variable")
These are linear inequalities in one Variable x when a≠ 0
● ax + b < 0
● ax + b > 0
● ax + b ≤ 0
● ax + b ≥ 0

\color{green}✍️\color{green}\mathbf("Linear inequalities in two variables")
These are linear inequalities in two variables x and y when a ≠ 0, b ≠ 0.
● ax + by < c
● ax + by > c
● ax + by ≤ c
● ax + by ≥ c

\color{green}✍️\color{green}\mathbf("Quadratic inequalities in one variable")
These are quadratic inequalities in one variable x when a ≠ 0.
● ax^2 + bx + c ≤ 0
● ax^2 + bx + c > 0

### Algebraic Solutions of Linear Inequalities in One Variable

\color{purple}ul(✓✓) \color{purple} \mathbf("DEFINITION ALERT")
● Any solution of an inequality in one variable is a value of the variable which makes it a true statement.

\color{green} ✍️ \color{green} \mathbf(KEY \ CONCEPT) we followed the following rules:
"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.

\color{red} \ox \color{red} \mathbf(COMMON \ CONFUSION) : 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.
Q 3171201126

Solve 30 x < 200 when
(i) x is a natural number,
(ii) x is an integer.

Solution:

We are given 30 x < 200
or (30x)(30) < (200)/(30) "(Rulu 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.
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}
Q 3111301220

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.
Q 3121301221

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, ∞).
Q 3141301223

Solve (5-2x)/(3)≤x/6 - 5

Solution:

We have
(5-2x)/(3)≤x/6 - 5

or 2(5-2x) ≤ x/6 - 5
or 10 - 4x ≤ x - 30
0r -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, ∞).
Q 3151301224

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

Solution:

We have 7x + 3 < 5x + 9 "or" 2x < 6 or x < 3
The graphical representation of the solutions are given in Fig 6.1.
Q 3161301225

Solve (3x-4)/(2) ≥ (x+1)/(4) - 1 Show the graph of the solutions on number line.

Solution:

We have

(3x-4)/(2) ≥ (x+1)/(4) - 1
or (3x-4)/(2) ≥ (x+3)/(4) - 1
or 2(3x-4) ≥ (x-3)
or 6x - 8 ≥ x- 3
or 5x ≥ 5 or x ≥ 1

The graphical representation of solutions is given in Fig 6.2.
Q 3111301229

The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the number of 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+x)/(3) ≥ 60
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.
Q 3121401321

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
and x + ( x + 2) < 40
Solving (2), we get
2x + 2 < 40
i.e., x < 19
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)