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)

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  -563    and   143     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

Solve  - 3  ≤  4-7x2    ≤  18

Given - 3  ≤  4-7x2    ≤  18 ,

To Eliminate 2 from denominator  Multiply every term  with 2 

  ⇒   - 3 × 2  ≤  4-7x2   × 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.

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  x ≥ 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 x ≥ 0 ; y ≥  0 ; 3x + 4y   ≥  12 and 4x + 7y ≤ 28 and  shade it , after  graphing linear inequalities. The 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  x ≥ 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 x ≥ 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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WHAT IS SET, TYPES OF SETS ,UNION ,INTERSECTION AND VENN DIAGRAMS

We shall discuss introduction to sets,What is Sets , kinds of sets, representation of sets, subset , power set , universal set ,cardinal number , union, intersection of sets  , disjoint sets , complement and difference of sets .And we shall be able to solve Puzzles based on set like given below

WHAT IS SETS

Def : Any collection of well defined and distinct objects is called a set. 

        By well defined set/object we mean , for a given set and object, it must be possible to decide whether the object belonging to the set or not . i .e it is accepted by everyone as an element for that Set  or  not .The object in a set are called  elements of a set or Members.

Set are usually denoted by Capital letter A , B , C , D Etc, and their element are represented by small letters.

For any set A let 'a' , 'b'  and 'c' are members of that Set, then a ∈ A  , b ∈ A and  c ∈ A . This special character ∈ is read as belongs to symbol  . So a ∈ A  will be read as  " a belongs to A" , b ∈ A means "b belongs to A" and  c ∈ A read as " c belongs to A".

Here are some examples of sets given below

(1)   The collection of vowels in English Alphabet.
This is Set because every one will accept the letters from a , e , i , o , u  as the Vowels in English Alphabet

(2)   The collection of  Mountains in India.

(3)   The collection of all districts in India.

(4)   The collection of  even numbers .

(5)   The collection of all the M L As and M Ps in  India .

(6)   The collection of  all the football players of India.

(7)   The collection of hundred  novels of Indian writers.

(8)   The collection of Natural numbers from 100 to 55555.

(9)   The collection of prime numbers less than 100.

(10)  The collection of  Prime Ministers of India .
All those person who have accepted the post of Prime Minister are the elements of the set of Prime Ministers of India .On this point there will be no debate,The debate can be on the topic that particular Prime Minister is/was a good , popular or is/was a caring Prime minister.

These were examples of set as all the examples given above were well defined and easily acceptable to everyone.

Here are some examples which are not  Set

(1)   The collection of   beautiful  Mountains in India.

Because the word beautiful is not acceptable to all , Some peoples can says that these Mountains are beautiful But at the same time some other peoples can says that particular Mountains are not beautiful in their opinions .

  

(2)   The collection of   Popular  M L A and M P in  India .
Similar is the case with Popular Leaders, as some people likes  MLAs Or MPs But at the same time many other peoples hate them. So the word Popular is not acceptable to all .

(3)   The collection of   best football players of India.

(4)   The collection of  an Interesting novels of Indian writers.

(5)   The collection of   handsome students in any class .

(6)   The collection of  dangerous animals in this world.

(7)   The collection of  rich person in India.

(8)   The collection of  an easy subjects.

(9)   The collection of an  intelligent  students in any class.

(10) The collection of  beautiful animals .

These were not  examples of set as all the examples given above were not  well defined and were not easily acceptable to everyone. Because the some mountains are  beautiful for some person but at the same time ,same mountains are not beautiful for some persons ,  Similarly some peoples find any book interesting but at the same time same book may not be  interesting for others.

Representation of Sets

The set can be represented in two types

1  Roster Form  or Tabulation Form
2  Set builder form or Rule form

Roster Form 

To represent elements in Roster form ,we separate every element by comma and all the elements are listed in this form.

 For example

Let A is the collection of  even Natural Numbers . Then

A = {2, 4, 6, 8, 10, ..............}

Let P is the collection of prime numbers less than 100

P = {2, 3, 5 , 7, 11, 13, 17, 23, 29, ........, 97}

Let V is the collection vowels in English Alphabet.Then

V = {a, e, i, o, u }

Let S is set of square of the natural number between 1 to 10 Then 

S= {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

Set Builder form

In this form of set , we write the elements of set by special rule, and that rule must be satisfied by every element of the set in this form all the elements of the set are not written. Given below are Set builder notation examples.

Let A is the collection of  even Natural numbers . then

A = {x : x = 2n , n ∈ N } 

Let V is the collection of vowels in English Alphabet.Then

V = { x : x is a vowel in English Alphabet }

Let S is set of square of the natural number between 1 to 10 Then 

S= { x : x = n²  :  n  ∈ N and   1 ≤ n  ≤  10 }

Also  Read  HOW TO UNDERSTAND RELATIONS AND FUNCTIONS ,INVERSE OF A FUNCTION

Empty (Null Set or Void Set)

When a set having No element in it then that Set is Called an 

Empty Set and Empty or null set symbol is denoted by ф or { }.  

 Empty set examples are given below

(1)  ф = {x : x ∈ N ,  5 < x < 6}  

(2)   ф = {x : x ∈ R ,  x² +3 = 0 }

(3)   ф = { x ; 99 >  x and  x > 101}

Non  Empty Set

Any set which contain at least one element is called Non empty set .

Some examples of Non Empty set are given below

(1)  A = { 0 } 

(2)   B = { x : x ∈ N ,  5 < x < 8}   

(3)  C = { x ; 99 <  x and  x < 105}

(4)  D = { x : x ∈ R ,  x² - 3 = 0 }

Singleton Set

Any set which contain only one element is called Singleton Set .

e. g

(1)  A = { 0 } 

(2)   B = { x : x ∈ N ,  6< x < 8}   

(3)   C = { x ; x ∈ R , 99 <  x and  x < 101}

(4)   D = { x : x ∈ R , x - 3 = 0 }

Finite Set 

The set in which total numbers of elements can be counted, is called Finite set.

Some of the Finite set examples are given below

(1) The set of  all the countries in the world.

(2) The set containing all the prime number less than 1000.

(3) C  = { x ; x ∈ N , 99 <  x and  x < 109}.

(4) D  = { x : x ∈ R ,  x² - 3 = 0 }.

(5) E  = The set of all the even Natural Numbers less than 100.

Infinite Set

The set in which total numbers of elements can 
Not be counted, i. e. their counting of elements 
do not come to an end ,is called Finite set.

Here are Infinite set examples given below

(1) The set of all points in a plane.
(2) The set containing all the prime number .
(3) C = { x ; x ∈ R , 99 < x and x < 109}.
(4) D = Set of lines parallel to given line

(5) E  = The set of all the even numbers greater than 100.

(6)  S = Set of stars in the sky,

as the stars can nit be counted , they are numberless.

Cardinal Number of a Set

The numbers of distinct elements present in any finite set is called cardinal number of finite set. And it is denoted by n(A) .

If A =  { 1 , 3 , 5 , 8 , 9 } then n(A) = 5 and 

if  

B= {a , b , c , d , e , f , g , h , i , j } then  n(B) = 10 and if

C = {1 , 2 , 3 , 4 , 5 , 6 , ...... , 100 }then  n(C) = 100

Because The set A , B and C have 5 ,10 and 100 elements respectively .

What is a subset

The set B is said to be the subset of a set A if every element of set B is also  an element of set A . Subset symbol  is denoted by ⊆   , if we write A ⊆ B that means A is subset of B and if we write B ⊆ A means B is subset of A .

If A =  { 1 , 3 , 5 , 8 , 9 } and  B = { 1 , 3 , 5 } then B is called subset of A as all the elements of set B are in Set A.

Subset and proper subset

Any set B is called proper subset of set A if every element of set B is an element of set A whereas every element of set A is not an element of Set B . It is denoted by ⊂ .

Let A = { 1 , 2 , 3}  and B = { 1 , 2 } and C = { 1 , 3 } then B ⊂ A  and also  C  ⊂ A . It is to be noted that A ⊆ A i.e. every set is a subset of itself but it is not a proper subset. As we know that  ⊂ is notation for Proper subset symbol .

Power Set

If we form a set which consist of all the subset of a given set A , then that set is called Power Set and is denoted by P(A) .

Let us consider Power set example    if A = { a , b , c } then 

P(A) = { ф , { a } , { b } , {c} , {a , b}, { a , c} , { b , c } , { a , b , c } }

Equal Sets

Two sets are said to be equal to each others if they have same and same numbers of elements ,

Given below is   an example of  equal sets examples  
  If  A =  {  1 , 2} and B = {x :  x ∈ N , 1 ≤  x ≤ 2 } , Then these sets A and B are equal to each other because both the sets have same numbers of elements and also same elements  1 and 2 .

Equivalent sets

Two sets are said to be equivalent if both the sets have same numbers of elements , And Equivalent Sets need not be equal to each other but have same cardinal numbers .

e.g  if A = { a , b , c } and B = { 1 , 2 , 3 } are Equivalent Sets

Comparable Sets

Two sets are said to be comparable if one of them is a subset of other  i.e. either A ⊆ B or B ⊆ A .

A = { a , b , c } and B = { b , c } are comparable sets.

Universal Set

A set that contains all the sets under consideration , and is denoted by U . 

When we are using set containing  real numbers , then Set R is called Universal set .

If we are considering set of equilateral triangles , then Set of Triangles is called Universal set .

Let us understand it with Universal set example

If A = { 2 , 4 , 6 },  B = { 1 , 2 , 3 , 4 } and C { 5 , 6 , 8 , 9 , 10 }

then Set U = { 1 , 2 , 3 , 4 ,  5 , 6 , 8 , 9 ,10 } is called Universal set.

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Sets and venn diagrams

Let us discuss below algebra of sets with the help of some examples.

What is union set

The union of two sets A and B is the set which contains all those elements which are either in set A or in from set B or in both A and B. It is denoted by A U B .

Thus A U B = { x : x ∈ A or x ∈ B }If A = { 1 , 2 , 5 , 6  } and B = { 8 , 9 , 10 }    then  

A U B = {1 , 2 , 5 , 6 , 8 , 9 , 10 } , as this set contains  the elements either  from Set A or from Set B.

Again if A = { a , b , c , d, e , f , g } and B = { a , b , c , d  }

A U B = { a , b , c , d, e , f , g } = A

Thus  the union of a given set and its subset is always the given set 

Intersection of sets

Intersection of two sets A and B is the set which contains all those elements which are in both the set A and Set B . It is denoted by A ∩ B . So ∩  is called  intersection symbol .

Thus A ∩ B = { x : x ∈ A and x ∈ B }

If A = { 1 , 2 , 5 , 6  } and B = { 5 , 6 , 10 }

then    A∩ B = { 5 , 6  }

Again if A = { a , b , c , d, e , f , g } and B = { a , b , c , d  }
then A∩ B = { a , b , c , d  } = B 
Thus  the intersection  of a given set and its subset is always the subset of  a given set .
So these were some of examples of union and intersection examples.

Disjoint Sets

Two sets are said to be disjoint sets if they have no common element. i.e. if their intersection is a null set A⋂ B = {  } = ф
if  A = { a , b , c , d, e , f , g } and B = { h , i , j , k , l , m } then these sets A and B are disjoint as they have no common element

i. e A∩ B = ф .

Difference of Sets

The difference of two sets A and B is the set of all those elements of  A which which are not in B . It is denoted by A - B.
Thus A - B = {x : x ∈ A but x ∉ B  }
If A = { 2 , 3 , 4 , 5 , 6 } and B = { 5 , 6 , 7 , 8 } then
A - B = { 2 , 3  , 4 }

Similarly  The difference of two sets B and A is the set of all those elements of  B which which are not in A. It is denoted by B - A.
Thus A - B = {x : x ∈ B but x ∉ A  }

If A = { 2 , 3 , 4 , 5 , 6 } and B = { 5 , 6 , 7 , 8 } then

B - A = { 7 , 8 }

what is complement of a set

If we have a universal set U and  a given set A ⊂ U then  complement of  a set A is the set which contains all those elements of universal set U which are in not in A . It is denoted by A^c 

Thus A^c = { x : x ∉ A  and x ∈ U }

Some Important Results 

U ^c  = ф 

ф ^c  =U  

U ^c   = ф 

( A ^c )  ^{c   =}    A 

A U  A ^{c   =}  U

A ∩  A ^{c   =}  ф

(A U B )  ^{c   =}  A ^c ∩ B ^{c    De Morgan's Law}

^{(A ∩ B )  c   =  A c U B c}    De Morgan's Law

Practical Application of Sets

If A and  B are disjoint sets

n (A U B ) = n(A) + n(B) 

If A and B are not disjoint sets

n (A U B ) = n(A) + n(B) - n (A ∩ B )

Question 

In a class of 50 students , 35 opted Mathematics 25 opted Biology then how many opted both Mathematics and Biology ?

Solution

Here   n ( M U B ) represents the total ( sum ) of students in a class and n ( M ∩ B ) represents the common students who have opted both the subjects Biology and Mathematics in that class.

Here we can use this formula 

n ( M U B ) = n ( M ) + n ( B ) - n ( M ∩ B ) --- -- -- (1)

where n (M U B ) = 50 , n ( M ) = 35 , n( B ) = 25 ,

n ( M ∩ B ) = ? 

so putting the above values in (1)

50 = 35 + 25 - n ( M ∩ B )n ( M ∩ B ) = ( 35 + 25 ) - 50

n ( M ∩ B ) = 60- 50

n ( M ∩ B ) = 10 

So There are 10 students in that class who opted both the subjects Biology and Mathematics .

Question 

In a village 350 peoples read Hindi News Paper , 235 peoples read English News Paper and 150 People read both the News Papers , How many peoples read News Papers ?

Solution

Here n ( E U H ) represents the total ( sum ) of people in a village and n ( E ∩ H ) represents the common Reader who read both the News Papers Hindi and English in that village .

Here we can use this formula 

n ( H U E ) = n ( H ) + n ( E ) - n ( H ∩ E ) --- -- -- (1)


where n (H ∩  B ) = 150  , n ( H ) = 350 , n( E ) = 235 ,
 n ( H U B ) = ? 

so putting the above values in (1)

n ( H U E ) = 350 + 235 - 150n ( M U B ) = ( 350 + 235 ) - 150

n ( M U B ) = 585 - 150

n ( M U B ) = 435 

So there are 435 Peoples who  read News Papers .

Question 

In a group of 50 teachers , 30 teachers drinks coffee and 25 drinks both coffee and tea, How many teachers drinks tea and how many drinks tea only

Answer

Here  C  and T  represents Coffee and Tea Respectively . Therefore Total numbers of teachers  in that group are 50  i.e.  n (T U C ) , and numbers of teachers who take both drinks are 30 . i . e. 

n ( C ∩ T ) = 25 And  n( C ) = 30 so n (T) = ? and Teachers drinks tea only will be represented  with n ( T - C ) ,which we have to find

n ( T U C ) = n ( T ) + n ( C ) - n ( T ∩ C ) --- -- -- (1)

              50  = n (T) + 30 - 25

              n ( T ) = 45

It means that 45 Teachers drinks tea , this implies that in these 45

Teachers there are some Teachers who  drinks both coffee as well . so by the statement  how many drinks tea only , we have to drop those Teachers who drinks both ,

 n( T - C ) = n ( T ) - n ( T ∩ C )

 n( T - C ) = 45  - 25

 n( T - C ) =  20

Also Read What is Quadratic Equation , discriminant and What are different method of  its  Solutions  

Conclusion

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