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## WHAT ARE LINEAR INEQUALITIES AND HOW TO SOLVE INEQUALITIES GRAPHICALLY

Today  we  are  going  to  discuss  linear   inequalities ,    solving inequalities , linear inequations of one variable , linear  inequalities in two variables , graphing inequalities , solving linear inequalities , system    of   linear   inequalities ,  linear     inequalities  examples , application  of  linear  inequations and its   formation with the help of some examples.

## Linear Inequation

An Inequation is a statement which involve the sign of  Inequality  < , > , ≤   ,  ≥  ,  ≠  etc

An equation which contains variables of 1st degree and does not  contains product of variables  is called Linear Inequation .
3x + 5 < 2  , 6x - 5 > 7 , 1 - 3x  10 , 2x - 3 ≤ 5

These are called Linear Inquations  Of One Variable.

3x - 5y   15 , 4x+3y  ≥ 12 ,

6x - 4y 24  , 3x + 5y   15,

2x - 5y  ≤  10,  3x + 7y   21,

These are called Linear Equations  Of two Variables .

## Solve the Inequation  1 - 2x  < - 8

1 - 2x  <  - 8

Shifting the constant term to R.H.S.

- 2x  <  - 8-1

- 2x  <  - 9
Isolating x from L.H.S by dividing  with -2 on both sides  and change the sign of Inequality.

Note : Whenever we divide or Multiply any Inequality with minus number ,then the sign of that inequality  changes , i.e.  <  to  > and  < to  >  or ≤  to   ≥  and   ≤    to   ≥ .

⇒  - 2x/-2  >  - 9/-2

⇒ x  >  9/2
It means every values greater than  9/2  will satisfy the given Inequality

## Solve   the Inequalities   -4x + 1 ≥  0  ,  3 - 4x < 0

Consider 1st Inequality
- 4x + 1 ≥ 0  Shifting 1 to RHS

⇒ - 4x  ≥ 0 -1  Dividing with -4 on both sides to isolate x

⇒ - 4x /-4   ≥  - 1/ -4  ,change the sign of Inequality

⇒ x  ≤   1/4  = 0.25  -------(1)

Consider 2nd  Inequality

⇒    3 - 4x  <  0  Shifting 3 to RHS

⇒   - 4x  <  -3 Dividing with -4 on both sides to isolate x

⇒     -4x/-4  <  -3/-4          , change the sign of Inequality

Now combine  the (1) and (2) ,we get NO Solution , Because simultaneously x can not be greater than 3/4 and less than 1/4

## Solve -12 ≤  4 -3x -5 < 2

Given   - 12  ≤  4 -3x -5 < 2 ,
To solve such types of Inequality Multiply  throughout   by -5 ( i. e. with denominator ) , if the multiplicand is negative then change the sign of all the inequalities

⇒   -12×(-5)     4 -3x -5 × (-5) > 2 × (-5) ,

⇒     60    4 - 3x  > -10 ,

Add or Subtract the -ve of constant number  appearing with x (Here -ve of 4 ) to eliminate the constant term in middle of the inequality .

⇒     60 - 4    4 - 3x - 4 > -10 - 4

⇒     56     - 3x  > -14  ,
Divide with co-eff of x to isolate  "x" i. e with -3

⇒     56/-3   ≤   - 3x/-3  < -14/-3
and change the sign of inequality if divisor is negative

⇒     -56/3   ≤  x < 14/3

⇒   All those values which are between  -56   and   14    are solution of given Inequality .

This video will help in understanding Linear Inequalities

## Solve   2x -3 4 + 8  ≥ 2+ 4x 3  and represent the Solution in number line

Given
2x - 3 4   + 8     2 +  4x 3

To find the solution of such inequality where different denominators have two different number , we just multiply with the product of these two numbers which are in the denominators (Here 4 × 3  = 12 ) to each term of the inequality

2x - 3 4 × 12 + 8 ×12    2 × 12  +  4x 3 × 12

⇒ (2x - 3) × 3 + 96    24  + ( 4x ) × 4

⇒  6x - 9 + 96    24  +  16x

⇒  6x + 87    24  +  16x

⇒  6x -  16x    24  87

⇒  - 10x    - 63   , Change the inequality sign after multiplication / division  with -ve number

## Solve  - 3  ≤  4-7x2    ≤  18

Given - 3  ≤  4-7x   ≤  18 ,

To Eliminate 2 from denominator  Multiply every term  with 2

⇒   - 3 × 2  ≤  4-7x  × 2  ≤  18 × 2

⇒   -6  ≤  4 -7x    ≤  36

⇒   -6 - 4 ≤  - 4 + 4 -7x  ≤  - 4+ 36

⇒   -10   -7x  ≤   32

⇒   -10/-7      -7x/-7     32/-7 , change the sign of inequality

⇒   10/7      x   ≥  -32/7

Hence set of all those numbers lying between 10/7 and -32/7 is the solution of the given inequality

## Draw  the Diagram of the solution set of the constraints :   x  ≥ 0 ; y ≥ 0 , 4x + 7y ≤  28

In order to solve such a constraints , 1st of all draw graph of linear Equations x = 0 (which is known as y  axis) , y = 0 ( which is known as x-axis ) , 4x + 7y = 28 simultaneously .

As we know that graph of the  line x = 0 is the line y - axis and graph of the line  y = 0  is the line x - axis .

Now to draw the graph of line 4x + 7y = 28 , Put x = 0  and y = 0 in given line respectively and find the values of y and x . So if   x = 0 in the equation , we get y = 4 , and when we put y = 0 in line 4x + 7y = 28 , we get x = 7 .

Therefore       If    x  =  0  then  y = 4
and                 If    x =  7   then  y = 0
we get two points A(0,4) and B(7,0) .

Now check the feasible region of each lines , as x  ≥ 0 implies right half of the Cartesian plan ( including y - axis ) and y  ≥ 0  implies upper half of the  Cartesian plan  ( including x - axis ) . So from these two inequalities we get the 1st quadrant as the common/feasible region .

Now  plots both the points A(0,4) and B(7,0) in the plan and draw a line passing through these two points.

Now put (0,0) point in the inequality 4x + 7y  28 , If it comes out true then feasible region will be toward origin and if it comes out false then feasible region will be away from  origin .

Now mark  the common region from all the inequalities and  shade it , The shaded region will be the required/feasible region after graphing linear inequalities or system of inequalities . i.e The solution of  the given constraint.

## Solve Graphically the System of Linear constraint  :   3x + 4y   ≥  12 ; 4x + 7y ≤ 28 ; x ≥ 0 ; y ≥  0

As we know from previous problem   0 ; y   0 the common region from these two inequalities is in 1st quadrant.

Now Draw the graphs for  two  lines 3x + 4y 12 and  4x + 7y  28 .

Put x = 0 and y = 0 in 1st equation , we get  y = 3 and  x = 4 respectively . Therefore  two points will be  A(0,3) and B(4,0) .

Similarly put x = 0 and y = 0 in 2nd  equation respectively , we get  y = 4 and  x = 7 respectively .Therefore  two points will be  C(0,4) and D(7,0) .

Now  plots both the points A(0,3) and B(4,0) in the plan and draw a line passing through these two points. Also plots both the points C(0,4) and D(7,0) in the plan and draw a line passing through these two points.

Now put (0,0) point in the inequality 3x + 4y     12 and 4x + 7y  28 , If it comes out true then feasible region of that inequality will be toward origin and if it comes out false then feasible region of that inequality  will be away from  origin .
At last mark  common region from all the inequalities  0 ; y   0 ; 3x + 4y     12 and 4x + 7y  28 and  shade it , after  graphing linear inequalitiesThe shaded region will be the required/feasible region ( As shaded in the above figure ) i.e. This is the  solution of the given constraint.

## Solve Graphically the System of Linear constraint   2x+3y   ≥  6 ; x - 2y ≤ 2 ; 6x + 4y ≤  24 ; -3x +2y ≤ 3   x≥ 0 ; y ≥  0

As  from previous problems   0 ; y   0 the common region from these two inequalities is in 1st quadrant.

Now Draw the graphs for  two linear lines 2x + 3y   = 6   ; 6x + 4y = 24 and  -3x +2y = 3 .

Put x = 0 and y = 0 in 1st equation 2x + 3y  =  6 , we get  y = 2 and  x = 3 respectively . Therefore  two points will be  A(0,2) and B(3,0) .

Now  put x = 0 and y = 0 in 2nd  equation x - 2y = 2, we get  y = -1 and  x = 2 respectively .Therefore  two points will be  C(0,-1) and D(2,0) .

Put x = 0 and y = 0 in 3rd equation 6x + 4y =  24 , we get  y = 6 and  x = 4 respectively . Therefore  two points will be  E(0,6) and F(4,0) .

Also Put x = 0 and y = 0 in 1st equation -3x + 2y = 3, we get  y = 3/2 and  x = -1 respectively . Therefore  two points will be  G(0,3/2) and H(-1,0) .

Now  plots both the points A(0,2) and B(3,0) in the plan and draw a line passing through these two points. Also plots both the points C(0,-1) and  D(2,0)  in the plan and draw a line passing through these two points.

Also plots both the points E(0,6) ,F(4,0)  and G(0,3/2) and H(-1,0)
in the plan and draw a line passing through these two points. therefore there will be four lines in 1st quadrant.

Now put (0,0) point in the inequality 2x + 3y     6 ;  x - 2y    2 ; 6x + 4y   24 ; -3x + 2y ≤ 3 , If it comes out true then feasible region of that inequality will be toward origin and if it comes out false then feasible region of that inequality  will be away from  origin .

At last mark  common region from all the inequalities  0 ; y   0 ; 3x + 4y     12 and 4x + 7y  28 and  shade it , The shaded region will be the required/feasible region. ( As shown in the figure above  ) i.e. This is the  solution of the given constraint.

Also Read  : Sets , Types of Set , Union ,Intersection , Complement of  set  and venn Diagrams

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## HOW TO LEARN INTEGRATION FORMULAE/FORMULAS USING TRICKS

Let us learn and remember most Important formulas of Integration , tips and tricks to learn algebraic ,most important differentiation questions for plus 2 maths, indefinite integration tricks and shortcuts trigonometric and by parts formulas in an easy and short cut manners.

## 1  ∫ sin x dx           =  - cos x +c

where "c" is called constant of Integration.

The integration of sin x is  - cos x ,then divide it with the derivative of its angle.

If we have to find the integration of  sin 2x , then we shall find it as

Step1    1st find the integration of sin x which is - cos x .

Step2    Divide it with the derivative of 2x ,which is 2, so

∫ sin 2x dx     =  - ( cos 2x) 2 + c ,
∫ sin 8x dx      =  - ( cos 8x) 8 + c ,

∫ sin  3x4  dx  =   - ( cos  3x4 )   3 4  + c ,
Therefore   ∫ sin nx dx =  - ( cos nx) n +c ,

## 6 ∫ cosec x dx = - log |cosec x - cot x | +c

If we want to integrate cosec√x .Then 1st of all we apply the formula of integration of cosec (any angle) then formula of integration of √x,So we have

∫  cosec√x dx = - (log | cosec√x -co√x | )(2√x) +c

## 7  ∫ sec2 x dx = tan x + c

Because the derivative of tan x is sec 2 x , So the Antiderivative or Integration of sec 2 x  is tan x .

∫ sec 2 √x dx     =  (2√x ) tan √x  + c

∫ cosec 2 dx = - cot x +c

Because the derivative of  cot x is  - cosec 2 x , So the  Anti derivative or Integration of  cosec 2 x is - cot x .

## 8 ∫ sec x tan x dx = sec x +c

Because the derivative of sec x is sec x tan x ,Therefore the integration of tan x sec x is sec
x .

If we want to integrate sec√x .tan √x .Then its  integration  will be sec √x,

sec √x tan √x   dx      =  √x   sec √x + c

## 9 ∫ cosec x cot x dx = - cosec x +c

Because the derivative of cosec x is    - cosec x cot x , Therefore the integration of tan x sec x is sec x .

∫ cosec √x cot √x dx = - ( 2 √x  cosec √x ) +c

## Conclusion

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