- Obtain the linear inequation
- Pull all the terms having variable on one side and all the constant term on another side of the inequation
- Simplify the equation in the form given above

$ ax> b $

or

$ax \geq b$

or

$ax< b $

or $ax \leq b$

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

$ x- 2 > 2x+15 $

$x-2 > 2x+15$

Subtract x from both the side $x-2 -x > 2x+15 -x $

$-2 >x +15$

Subtract 15 from both the sides $-17 > x $

Solution set

$(-\infty,-17)$

- $2x > 9 $
- $x + 5 > 111 $
- $3x < 4 $
- $2(x + 3) < x+ 1$

- Take k on the LHS
- Simplify LHS to obtain the inequation in the form

$ \frac {px+q}{ex+f} > 0 $

Make the coefficent positive if not

- Find out the end points solving the equatoon $px+q=0 $ and $ex+f=0$
- Plot these numbers on the Number line. This divide the number into three segment
- 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
- Write down the solution set in interval form

- Take k on the LHS
- Simplify LHS to obtain the inequation in the form $ \frac {px+q}{ex+f} > 0 $
- For the equation to satisfy both the numerator and denominator must have the same sign
- So taking both the part +, find out the variable x interval
- So taking both the part -, find out the variable x interval
- Write down the solution set in interval form

Make the coefficent positive if not

$ \frac{x - 3}{x + 5} > 0$

- Lets find the end points of the equation

Here it is clearly

x=3 and x=-5

- Now plots them on the Number line
- Now lets start from left part of the most left number

i.e

__Case 1__

$x < -5 $ ,Let takes x=-6 then $\frac {-6-3}{-6+5} > 0$

$3 > 0$

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 then $\frac {1-3}{1+5} > 0$

$\frac {-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

$\frac {4-3}{4+5} > 0 $

$\frac {1}{9} > 0 $

So this is good - So the solution is

x < -5 or x > 3

or

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

- 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

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

Class 11 Maths Class 11 Physics