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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

⇒    x  >  3/4  = 0.75 ---------(2)
WHAT ARE LINEAR INEQUALITIES  AND HOW TO SOLVE INEQUALITIES GRAPHICALLY

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

WHAT ARE LINEAR INEQUALITIES  AND HOW TO SOLVE INEQUALITIES GRAPHICALLY

⇒   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

⇒  x  ≤   63/10
WHAT ARE LINEAR INEQUALITIES  AND HOW TO SOLVE INEQUALITIES GRAPHICALLY

Solve  - 3  ≤  4-7x   ≤  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
WHAT ARE LINEAR INEQUALITIES  AND HOW TO SOLVE INEQUALITIES GRAPHICALLY


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 .
WHAT ARE LINEAR INEQUALITIES  AND HOW TO SOLVE INEQUALITIES GRAPHICALLY

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.


Also  Read  HOW TO SOLVE HARD AND IMPOSSIBLE PUZZLES PART 3


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 .


    WHAT ARE LINEAR INEQUALITIES  AND HOW TO SOLVE INEQUALITIES GRAPHICALLY

    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


    Application of Linear Inequalities


    Problem :   In the  first four  papers each of 100 marks , Ravi got 99 , 88 , 77 , 96 marks . If he wants an average of  ≥ 80 marks and   85 marks , Find  Find the range of Marks he should score in the fifth paper . 



    Let Marks secured by Ravi in Fifth Paper  = x
    Then as we know that average is the sum of  the marks secured in 
     five papers and divided by 5. 

    Then according to given conditions the average must be grater than 80  and less than 85. Then Mathematically it can  be translated in linear equation  as follows  

    80  ≤  99+88+77+96 + x 5  ≤  85

    80  ≤  360 + x 5  ≤  85 , 
    Now Multiplying with 5 to eliminate denominator 

    80 × 5 ≤  360 + x 5 × 5 ≤  85 × 5


    400 ≤ 360 + x ≤  425 

    Subtracting 360 throughout the inequation

    400 - 360 ≤ 360 + x -360 ≤  425 - 360 

    40 ≤ x ≤  65


    • It means in order to get average of  greater than equal to  75  and  less than equal to   85  marks , he should score in the range of  40 to 65 marks in the Fifth Paper.



    Problem : The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side . If the perimeter of the triangle is more than  166 cm then find the minimum length of the shortest side. 



    Let the shortest side of the triangle  = x cm
    It is given that longest side is twice the shortest side 
    ∴ Longest side of triangle  = 2x cm,

    given that third  side is 2 cm longer than shortest side.
    ∴ Third   side of triangle =  x + 2  cm 

    Then translating these line to mathematical form ( perimeter of triangle is more than 166 ) , we get

    ( x  ) + ( 2x ) + ( x + 2) > 166

     ⇒   4x + 2  > 166 

    ⇒   4x   > 166 - 2 

    ⇒   4x  > 164

    ⇒   x  >  166 /4

    ⇒   x  >  41

    This implies the shortest side of the triangle must be 41 cm in order to satisfies  all the conditions .


    Perimeter : Perimeter of a triangle  is the sum of all the sides of  given triangle .



    Conclusion


    In this post I have discussed Linear Inequations , linear inequations of one variable , solving linear inequalities, graphing linear inequalities,system of inequalities, linear inequalities in two variables, system of linear inequalities, linear inequalities examples,  and Application of linear inequations . If this post helped you little bit, then please share it with your friends to benefit them, comment your views on it and also like this post to boost me and to do better, and also follow me on my Blog .We shell meet in next post till then Bye .




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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.


    Trigonometric Formulae



      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 ,






    2  ∫ cos x dx          =    sin x +c  

    The integration of cos x is  sin x ,then divide it with the derivative of its angle.
    If we have to find the integration of  cos 2x , then we shall find it as

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

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


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

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

    ∫ cos  3x4  dx  = ( cos  3x4 )   3 4  + c ,

    Therefore   ∫ cos nx dx =   ( sin nx) n +c ,



    3  ∫ tan x dx = log |sec x| +c or - log |cos x| + c


    The integration of tan x is   log |sec x| +c   or -log |cos x| +c  , then divide it with the derivative of its angle.


    If we want to find the integration of tan  x2  . The integration of  tan  x2 is  log |sec x2 | +c , then divide it with the derivative of its angle.

    Step 1

    Find the integration of  tan  x2 ,which is log |sec  x2 | or  - log |cos  x2 |,


    Step 2
     Divide it with the derivative of angle x2 ,which is 2x.

    Therefore
    ∫ tan  x2  dx = -(1/2x) log |cos x2 | +c or (1/2x)log |sec x2 | + c

    4  ∫ cot x dx = log |sin x| +c or - log |cosec x| + c


     The integration of cot x is   -log |cosec x| +c   , then divide it with the derivative of its angle.



    If we want to find the integration of  cot x2  . Then integration of  cot  x2 is  -log |cosec x2 | +c  or  log |sin x2 | +c  , then divide it with the derivative of its angle.

    Step 1 

     Find the integration of  cot  x2 ,which is -log |coec  x2 | or  log |sin x2 |,

    Step 2

      Divide it with the derivative of angle x2 , which is 2x.   

    Therefore      ∫ cot  x2  dx = -(1/2x) log |cosec x2 | +c or (1/2x ) log |sin x2 | + c



     5  ∫ sec x dx         =  log |sec x - tan x | +c 


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


     ∫  sec√x dx = ( log |sec√x - tan √x | )(2√x) +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 


     Also Read   WHAT IS SET, TYPES OF SETS, UNION INTERSECTION AND VENN DIAGRAMS


    Algebraic Formulae


    1 ∫ (constant) dx = (constant ) x +c

    Integration of constant function is the constant function itself multiplied by the variable .

    ∫ 5 dx   = 5x  +c

    2  ∫  xn  dx  = xn+1  n+1dx  + c ,

    ∫ x3   dx  =  x4  4  + c ,


    HOW TO LEARN INTEGRATION  FORMULAE/FORMULAS USING TRICKS

       
    To find the integration of function where variable "x" or f(x) has power 'n' , where "n" is any real number, we shall increase the power of "x"  by 1 and divide it with increased power.
    e.g 

     ∫  x2  dx   =  {1/(2+1)} x2+1  + c

    ∫  (x ) 2/3   dx     =   (x ) (2/3)+1  (2/3)+1 + c ,

                      =   3(x ) 5/3  5 + c ,

    ∫  (ax+b ) n   dx     =   (ax+b ) n+1  a(n+1) + c ,

    ∫  (3x + 7 ) 2   dx     =   (3x+7 )2+1  3(2+1) + c ,

                                  =   (3x+7 )3  9 + c ,

    If we have to integrate sum of two functions ,then we shall integrate it separately as follows

    4  ∫  [ f(x) + g(x)] dx = ∫f(x)dx +  g(x) dx + c


    ∫  [{ x2}3  + (2x) ]dx = ∫ { x2}3 dx +  ∫ (2x) dx

      =∫   x6dx +  ∫ 2x dx =  
     x  6+1  6+1 +(2/2)x2  + c

     =   x  7  7 x2  + c

    ∫  {4x2  + 3x }dx = 4x2   + ∫ 3x dx
                             =  4×(1/3)x3  (3/2)x2 +c


     ∫ [ 6x / 3x2] dx = log |  3x2 | + c

    HOW TO LEARN INTEGRATION  FORMULAE/FORMULAS USING TRICKS




    Memorize  these integration formulas along with differentiation in Hindi




    Integration By Parts 

    ∫ [ f(x) g(x)] dx = f(x) ∫ g(x) dx -  {f '(x) ∫ g(x) dx}dx + c


        
    ∫ x sin x dx = x ∫ sin x dx - x' { ∫ sin x dx}dx + c
    = x(-cos x) -  (-cos x)dx + c
     = -x cos x - sin x +c

    ∫ log x dx =  ∫ log x.1 dx 
    = log x  - f '(log x) ( x )dx + c
    = log x .1  - ∫(1/x)  x dx + c  
    = log x  - ∫ 1 dx + c 
    = log x  - x  + c


     Integration   Exponential Function

    1  ∫  ex  dx   =  ex  + c

    2 ∫  ax  dx = ax / log a    + c  

    3 ∫ log x dx = x log x - x + c

    4 ∫ (1/x ) dx = ln |x | + c




    Exponential and Derivative Mixed Formula

       ex  [ f(x) + f '(x)] dx =ex  f(x) + c


      ex  [ sin x + cos x] dx = ex  sin x + c

    Conclusion


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