This article,i am explaining Inequalities Rules and how to solve inequalities

Contents

## What is inequalities

In mathematics, an inequality is a relation that holds between two values when they are different

Solving linear inequalities is very similar to solving linear equations, except for one small but important detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative

## Symbols used in inequalities or Inequalities Signs

The symbol **<** means less than. The symbol > means greater than.

The symbol **<** with a bar underneath means less than or equal to. Usually this is written as $\leq$

The symbol > with a bar underneath means greater than or equal to. Usually this is written as $\geq $

The symbol $\ne $ means the quantities on left and right side are not equal**Examples**

a < b means a is less then b or b is greater a

a $\leq$ b means a is less then or equal to b

a > b means a is greater than b

a $\geq $ b means a is greater or equal to b

## Inequalities Rules

It can be divided into part

a. Things which can be done in inequality and it does change the sign

b. Things which change the sign of the inequality

Things are which are safe to do in inequality which does not change in direction

1. addition of same number on both sides

$a > b$

$a+c > b +c$

2. Subtraction of same number on both sides

$a > b$

$a-c > b-c$

3. Multiplication/Division by same positive number on both sides

$a > b$

if c is positive number then

$ac > bc$

or

$ \frac {a}{c} > \frac {b}{c}$

Things which changes the direction of the inequality

1. swapping the left and right sides

$a > b$

$b < a$

2. Multiplication/Division by negative number on both sides

$a > b $

$ -a < -b $

3. Don’t multiple by variable whose values you dont know as you dont know the nature of the variable

Another Important Concept which we must keep in mind while work on inequalities is Number line

## Concept Of Number line

A number line is a horizontal line that has points which correspond to numbers. The points are spaced according to the value of the number they correspond to; in a number line containing only whole numbers or integers, the points are equally spaced.

It is very useful in solving problem related to inequalities and also representing it

Suppose x >2(1/ 3), this can represent this on number line like that

## How to solve inequalities questions in one variable

We will check out how to solve inequalities for both Linear, Quadratic and Rational Fraction

(1)**Simplify the Linear inequality **

$x- 2 > 2x+15$

$-15 -2 > 2x -x $

$ -17 > x$

$ x < -17$

(**2) Simplify the quadratic inequalities**

x^{2}-5x+6 > 0

Which can be simplified as

(x-2)(x-3) > 0

or this can be done using equal sign in inequalities and solve it to find end points i,e 2 and 3 in above case.

Now plot those points on Number line clearly

Now start from left of most left point on the Number line and look out the if inequalities looks good or not. Check for greater ,less than and inequalities at all the end points

So in above case of

x^{2}-5x+6 > 0

We have two ends points 2 ,3

*Case 1*

So for x < 2 ,Let take x=1,then (1-2)(1-3) > 0

2 > 0

So it is good

So This inequalities is good for x < 2

*Case 2*

Now for x =2,it makes it zero,so not true. Now takes the case of x > 2 but less 3.Lets takes 2.5

(2.5-2)(2.5-3) > 0

-.25 > 0

Which is not true so this solution is not good

*Case 3*

Now lets take the right most part i.e x > 3

Lets take x=4

(4-2)(4-3) > 0

2> 0

So it is good.

Now the solution can either be represented on number line or we can say like this

$(-\infty,2)\cup (3,\infty)$

(**3) Simplify the rational or fractional inequality**

$ \frac {(x – 3)}{(x + 5)} > 0$

**Solution**

*Method A*

(1) Lets find the end points of the equation

Here it is clearly

x=3 and x=-5

(2) Now plots them on the Number line

(3) Now lets start from left part of the most left number

i.e

case 1

x < -5 ,Let takes x=-6 (-6-3)/(-6+5) > 0

3 > 0

So it is good

Case 2

Now take x=-5

as x+5 becomes zero and we cannot have zero in denominator,it is not the solution

Case 3

Now x > -5 and x < 3, lets take x=1 (1-3)/(1+5) > 0

-1/6 > 0

Which is not true

Case 4

Now take x =3,then

0> 0 ,So this is also not true

case 5

x> 3 ,Lets x=4

(4-3)(4+5) > 0

1/9 > 0

So this is good

So the solution is

x < -5 or x > 3

or

$(-\infty,-5)\cup (3,\infty)$

Method B

(1) the numerator and denominator must have the same sign. Therefore, either

x – 3 > 0 and x + 5 > 0,

or

(2) x – 3 < 0 and x + 5 < 0.

Now, (1) implies x > 3 and x > -5.

Which numbers are these that are both greater than 3 and greater than -5?

Clearly, any number greater than 3 will also be greater than -5. Therefore, 1) has the solution

x > 3.

Next, (2) implies

x < 3 and x < -5.

Which numbers are these that are both less than 3 and less than -5?

Clearly, any number less than -5 will also be less than 3. Therefore, (2) has the solution

x < -5.

The solution, therefore, is

x < -5 or x > 3

## Some Important points to note

1) We cannot have zero in denominator

2) We should be checking for inequalities at all the end points

## Some Problems to practice

1. $2x > 9$

2. $x + 5 > 111$

3. $3x < 4$

4. $2(x + 3) < x+ 1$

5. $x^2 -1 < 0$

6. $\frac {x-1}{x+5} < 0$

7. $x^2 -5x + 6 > 0$

8. $\frac {x-5}{x+1} < 0$

**Related Articles**

iit jee questions on relations and functions

vedic maths tricks

how to study maths

how to solve assertion reason questions

sin cos tan table

https://en.wikipedia.org/wiki/Inequality_(mathematics)

Please send it again to me