Linear Inequalities Class 11 Notes Mathematics Chapter 6 - CBSE

Chapter : 6

What Are Linear Inequalities ?

Introduction

The inequality is said to be linear, if it involves a linear function. In linear inequality variable(s) occur in first degree and there is no term involving the product of the variables.

E.g., ax + b ≤ 0, ax + by + c > 0, ax + b < c

Where a, b, c are real numbers a ≠ 0 and x is a
variable are called inequations in x.

Two algebraic expressoins, which are connected by the symbol < (less than), > (greater than), ≤ (less than or equal to) and ≥ (greater than or equal to) form the inequality in a_n.

Following are the linear inequations:

(i) 3x + 2 < 7

(ii) 5x – 10 > 2

(iii) 5x – 2 ≤ 8

(iv) 4x – 7 ≥ 8

Inequality

Two real number or two algebraic expressions related by the symbol <, >, ≤ or ≥ form an inequality.

Linear Inequality

A linear inequality is an inequality which involves a linear function. A linear inequality contains one symbol of inequlity and each variable occurs in first degree only.

e.g., ax + b ≤ 0, ax + by + c ≥ 0, ax ≤ 4.

Algebraic Soultion Of Linear Inequalities

Some rules can be applied to solve the linear
inequalities. These are as following:

On adding or subtracting the same number of expression from both side of inequality, does not change it.
On multiplying or dividing each side of the inequation by same positive number and does not change the inequality.
On multiplying or dividing each side of the inequation by the same negative number, reverses the inequality.
Any solution of an inequality in one variable is a value of the variable which makes it a true statement.

For example, consider an inequality 10x < 150
Obviously, x cannot be a negative integer or a
function. Left hand side (LHS) of this inequality is 10x and right hand side (RHS) is 150. Therefore,

we have

For x = 0, LHS = 10(0) < 150 (RHS), which is true.

For x = 1, LHS = 10(1) = 10 < 150 (RHS), which is true.

For x = 2, LHS = 10(2) = 20 < 150 (RHS), which is true.

For x = 3, LHS = 10(3) = 30 < 150 (RHS), which is true.

For x = 4, LHS = 10(4) = 40 < 150 (RHS), which is true.

For x = 5, LHS = 10(5) = 50 < 150 (RHS), which is true.

For x = 6, LHS = 10(6) = 60 < 150 (RHS), which is true.

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 {1, 2, 3, 4, 5, 6} is called its solution set.

Graphical Representation Of Inequality

$$\text{Consider an inequality}\\ 12 +1\frac{5}{6}× \leq 5 + 3x\\\text{when n ∈ N}\\\text{Then,\space} 12 + \frac{11}{6}x\leq 5 + 3x$$

On multiplying both sides by 6

$$⇒\space 6\bigg(12 + \frac{11}{6}x\bigg) = 6(5 + 3x)\\⇒\space \text{72 + 11x ≤ 30 + 18x}$$

Subtracting 18x on both sides

⇒ – 7x ≤ – 42

⇒ 7x ≥ 42

⇒ x ≥ 6

∴ Required solution set = {x∈N : x ≥ 6}

= {6, 7, 8, 9, ... }

Types Of Inequality

There are four types of inequalities they are:

Strict: The inequalities containing < or > is called strict inequality. A strict inequality can be of the following forms :

a < b or a > b

Slack: The inequalities that have ≤ or ≥ is called slack inequality.
Linear: The inequality that have a degree 1.

i.e., px + q < 0 or px + q ≤ 0

or px + q > 0 or px + q ≥ 0

Quadratic: The inequalities that have a degree 2.

i.e., px² + qx + r < 0 or px² + qx + r ≤ 0

or px² + qx + r > 0 or px² + qx + r ≥ 0

Some Useful Results

Consider 'a' be positive real number.

Then,

|x| < a ⟺ – a < x < a ∈ x ∈ (– a, a)
|x| ≤ a ⟺ – a ≤ x ≤ a ∈ x ∈ [– a, a]
|x| > a ⟺ x < – a or x > a
|x| ≥ a ⟺ x ≤ – a or x ≥ a

While solving two simultaneously inequations in one variable, we solve them simultaneously. Then, find the solution set of each of them. Then, the intersection of these solution sets is the required solution set.