- What are inequalities
- |
- Things which changes the direction of the inequality
- |
- Linear Inequation in One Variable
- |
- Linear Inequation in Two Variable
- |
- Quadratic Inequation
- |
- Steps to solve the inequalities in one variable
- |
- Steps to solve the inequality of the another form
- |
- Steps to solve Quadratic or polynomial inequalities
- |
- Absolute value equation
- |
- Absolute value inequation
- |
- Graphical Solution of Linear inequalities in Two Variable

We have learned by Quadratic equation is previous chapter

so for a $\neq$ 0

ax^{2}+bx+c > 0

or

ax^{2}+bx+c < 0

or

ax^{2}+bx+c $\geq$ 0

or

ax^{2}+bx+c $\leq$ 0

are called Quadratic Equation in one Variable

1) Obtain the linear inequation

2) Pull all the terms having variable on one side and all the constant term on another side of the inequation

3) Simplify the equation in the form given above

ax> b

or

$ax \geq b$

or

ax< b

or $ax \leq b$

4) Divide the coefficent of the variable on the both side.If the coefficent is positive,direction of the inequality does not changes,but if it is negative, direction of the inequation changed

5) Put the result of this equation on number line and get the solution set in interval form

**Example:**

x- 2 > 2x+15

**Solution**

x-2 > 2x+15

-17 > x

Solution set

$(-\infty,-17)$

**Some Problems to practice**

1. 2x > 9

2) x + 5 > 111

3) 3x < 4,

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

(ax+b/cx+d) > k or similar type

1) take k on the LHS

2) Simplify LHS to obtain the inequation in the form

(px+q/ex+f) > 0

Make the coefficent positive if not

3) Find out the end points solving the equatoon px+q=0 and ex+f=0

4) Plot these numbers on the Number line. This divide the number into three segment

5) Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation

6) Write down the solution set in interval form

Or there is more method to solves these

1) take k on the LHS

2) Simplify LHS to obtain the inequation in the form

(px+q/ex+f) > 0

Make the coefficent positive if not

3) For the equation to satisfy both the numerator and denominator must have the same sign

4) So taking both the part +, find out the variable x interval

5) So taking both the part -, find out the variable x interval

6) Write down the solution set in interval form

Lets take one example to clarify the points

(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

1) 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

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