Vedic Maths

Introduction

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Introduction

Many students find maathematics as a difficult subect as they find difficulty in basic arthematical operations, to manipulate symbols , balance the equations etc. Mathematics becomes disagreeable subject to their mind as it gives an unpleasant experience to them. This is because it involves a good mental exercise . These difficulties of students if asked from an experienced teacher have a very long list .

Vedic Mathematics is the name given to the ancient system of mathematics which was rediscovered from the Vedas between 1911 and 1918 by Swami Bharti Krishna Tirthaji. In this system formulae are designed by taking into consideration the way the mind actually works . This thing naturally helps a student a lot in finding the solution to the problem as he finds a kind of entertainment in doing them instead of stress. Research is being carried out in this field for developing new , powerful and easy applications of Vedic Sutras in geometry , calculus , computing etc.

This system gives a one line approach to the problem . As it is similisr to the way the mind works hence it gives superfast calculation methods along with cross checking systems . It increases the creativity of the student as it is based on the pattern recognition . Pupils can invent their own methods . They are not limited to the one correct method . It also has with its special features has the inbuilt potential to solve the psycological problem of mathematics - anxiety.

On this site I am giving some of the teqniques which students can try in their daily routine practices and get benefits from these . Algebraic proofs of formulas are also given so as to give a hint to the student to develop his mental capability of discovering his own methods to solve difficult problenms .

Speed Division by 9

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Introduction

Method 16

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Finding the H. C .F. of two,three or more degree equation

This method is based on the sutra " Lopana Sthapana " process which means ' elemination and retention or alternate destruction of the highest and the lowest powers '

Explaining the method with the help of an example .

Example 1 : - Find the H.C.F. of
x² + 5x + 4 and x² + 7x + 6.

Answer : -

Just subtract one expression from the other and take out the common from the result .
i.e.
x² + 5x + 4
x² + 7x + 6
-2x -2

Take out common from the result
i.e. - 2

Hence the H. C . F . of the two is

(x + 1) .

Example 2 : - Find H.C.F. of 2x² – x – 3 and 2x² + x – 6 .

Answer : -

2x² – x – 3
2x² + x – 6
-2x + 3

Taske -1 out and so

2x - 3 is the H. C. F. of the two expressions .

Example 3 : - Find the H. C. F. of
x³ – 7x – 6 and x³ + 8x² + 17x + 10.

Answer : -

x³ – 7x – 6
x³ + 8x² + 17x + 10
-8x² -24x -16

Take common out from the result .
i.e.
-8

Hence the H. C. F. of the two expressions is

x² + 3x + 2

Algebraic Proof :

Let P and Q be two expressions and H is their H.C.F. Let A and B the Quotients
after their division by H.C.F.

P = A and Q = B
H H

So P = A.H & Q = B.H

(P + Q) = (A.+ B) . H & (P - Q) = (A - B) . H

Hence we see that
whatever is the H. C. F. of P & Q The same is that of (P + Q) .

Exercises : -

Find the H.C. F. of the following :
1.) x² + 2x – 8 & x² – 6x + 8
2.) x³ – 3x² – 4x + 12 & x³ – 7x² + 16x - 12
3,) x³ + 6x² + 11x + 6 & x³ – x² - 10x - 8
4.) 6x⁴ – 11x³ + 16x² – 22x + 8 &
6x⁴ – 11x³ – 8x² + 22x – 8.
5.) x³ + 6x² + 5x – 12 &
x³ + 8x² + 19x + 12.

Method 15

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Cubing of numbers near powers of 10

Explaining the method with the help of an example .

Example 1 : Find the cube of 104 .

Answer : -

Step 1 : -

In the above excess is 4 .
Find cube of 4 and retain two right most digit for answer . Remaining digits are for carry .
i.e.
4³ = 64

Obtained digit = 64 Carry = 0

Step 2 : -

Double the previous excess . Find the new excess and multiply it with the previous excess .
i.e.
2 * 4 = 8
8 + 104 = 112 (Excess is 12)
Multiply this with the previous excess 4
12 * 4 = 48

Retain two right most digit for answer . Remaining digits are for carry .

Obtained digit = 48 64 Carry = 0

Step 3 : -

Now place the digit plus the double of excess to the left .
i.e.
112 4864

One Line representation

Way of writing = 112 / ₀48 / ₀64

Hence the cube of 104 is
1124864

Example 2 : - Find the cube of 106 .

Answer : -

Step 1 : -

Find the cube of excess .
6³ = 216
Obtained digit = 16 Carry = 2

Step 2 : -

Double the excess . Add it to the original dgit . And multiply the new excess with the previous excess .
6 * 2 = 12 + 106 = 118 (New Excess = 18)
18 * 6 = 108
Add previous carry = 108 + 2 = 110

Obtained digit = 1016 Carry = 1

Step 3 : -

Add new excess(obtained in previous step) to the base , Add carry and then place it to the left of above obtained digit .

118 + 1 = 119

Obtained digit = 1191016

Hence the answer is 1191016 .

One Line representation of above question .

118 / ₁08 / ₂16
= 1191016

Example 3 : - Find the cubve of 94 .

Answer : -

Step 1 : -

Excess is - 6 .
Cube of -6 = - 216

Way of writing = / -₂16

Step 2 : -

Double the excess . Add it to the original digit .
- 6 * 2 = - 12
94 - 12 = 82 (New excess is - 18)
Multiply it with the original excess .
- 18 * - 6 = 108

Way of writing = / ₁08 / - ₂16

Step 3 : -

Add new excess to the base
i.e.
100 - 18 = 82
and place it to the left of obtained digit .

Way of writing = 82 / ₁08 / _-216

Step 4 : -

(82 + 1) / (08 - 3) / (100 - 16)

(08 - 3) is because -2 is for carry and 1(which converts into 100 going to right hand side is for - 16 .

Hence the answer is
830584

Exercise : -

Find the cubes of the following numbers using yavadunam sutra.
1.) 105
2.) 114
3.) 1003
4.) 10007
5.) 92
6.) 96
7.) 993
8.) 9991
9.) 1000008
10.) 999992.

Method 14

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Resolving into Partial Fractions

This method is based on the sutra " Paravartya Yojayet " which means
' transpose and apply ' .
Explaining the method with the help of an example .
Example 1 : - Resolve the following into partial fractions .
3x + 13
(x + 1) (x + 2)
Answer : -
Write the above term as
3.x + 13 = A + B
(x + 1) (x + 2) (x + 1) (x + 2)
Step 1 : -
Take one term in the denominator . Like (x = 1) & obtain the value of x by placing it to 0
i.e. x + 1 = 0 => x = -1
Place this in the remaining term
i.e.
3.x + 13
(x + 2)
= 3.(-1) + 13
(-1 + 2)
= 10
which is the value of A

Step 2 : -

Again place the other term to 0 and obtain the value of x .
x = -2
Place this in the remaining terms
i.e.
3.(-2) + 13
(-2 + 1)
= - 7
which is the value of B .

The answer to the above quuestion is

3.x + 13 = 10 - 7
(x + 1) (x + 2) (x + 1) (x + 2)

Example 2 : - Resolve the following into partial fractions
9
(x² + x - 2)

Answer : -

9 = 9
(x² + x - 2) (x - 1) (x + 2)
9 = A + B
(x -1) (x + 2) (x -1) (x + 2)

Step1 : -

x = 1
Substitute in the remaining term
9 = 9 = 3
(x + 2) 1 + 2
which is the value of A.

Step 2 : -
x = -2
Substitute in the remaining term
9 = 9 = - 3
(x - 1) -2 -1
which is the value of B .

The answer to the above question is
9 = 3 - 3
(x² + x - 2) (x - 1) (x + 2)

Algebraic Proof : -

Taking
x + 3 = x + 3
(x² + x - 2) (x - 1) (x + 2)
x + 3 = A + B
(x - 1) (x + 2) (x - 1) (x + 2)
=> x + 3 = A(x + 2) + B(x - 1)
(x - 1) (x + 2) ( x -1) (x + 2)
=> x + 3 = A(x + 2) + B(x - 1)
Now putting x - 1 = 0 (which is the term below A)we get
A = (x + 3)
(x + 2)

Similarly putting x + 2 = 0 (which is the term below B) we get
B = (x + 3)
(x - 1)
We use the same things in the various steps .

Exercises : -

Resolve the following into partial fractions

1.)

x² + 5x + 3
(x + 2)(x - 7)(x - 4)

2.)
8x² + 9x + 11

          (x - 12)(x + 13)(x + 14) 

3.)       

     2.x -5      
(x - 2) (x - 3)

4.) 

    2x + 7     
(x + 3) (x + 4)

5.)
3x + 4
3x2 + 3x + 2

Method 13

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Factorise Three or more variable Equations

This method is based on the sutra " Lopana Sthapanabyam " which means ' by alternate elemination and retention ' .

Explaining the method with the help of an example .

Example 1 : - Factorise the following : -
x² + 2y² + 3.x.y + 2.x.z + 3.y.z + z².

Answer : -

Step 1 : -

First eleminate terms containing z from the given equation and factorise the remaining equation .
i.e.
put z = 0

x² + 2y² + 3.x.y = x² + 2.x.y + x.y + 2y²
= x.(x + 2.y) + y.(x + 2.y)
= (x + 2.y) . (x + y)

Step 2 : -

Now eleminate the terms containing y and factorise the remaining equation .
i.e.
put y = 0

x² + 2.x.z + z² = x² + x.z + x.z + z²
= x.(x + z) + z.(x + z)
= (x + z) . (x + z)

Step 3 : -

Fill the gaps in the expressions .
i.e.
Fill z in (x + 2.y) &
Fill z in (x + y ) of step 1

Answer is

( x + 2.y + z) . (x + y + z)

Example 2 : - Factorise
3.x² + y² - 4.x.y - y.z - 2.z² - z.x.

Answer : -

Step 1 : -

Put z = 0

3.x² + y² - 4.x.y = 3.x² - 3.x.y - x.y + y²
= 3.x . (x - y) - y.( x - y)
= (x - y) . (3.x - y)

Step 2 : -

Put y = 0

3.x² - z.x - 2.z² = 3.x² - 3.z.x + 2.z.x - 2.z²
= 3.x(x - z) + 2.z(x - z)
= (x - z) . (3.x + 2.z)

Step 3 : -
Fill the gaps

Answer is
(x - y - z) . (3.x - y + 2.z) .

Example 3 : - Factorise
3.x² + 4.y² + 7.x.y - 2.x.z - 3.y.z - z² + 17x + 21y – z + 20.

Answer : -

Step 1 : -

In this eleminate two variables at a time taking independent term as a separate variable .
i.e.
First eleminate y and z put y = 0 & z = 0
3.x² + 17.x + 20 = 3.x² + 12.x + 5.x + 20
= 3.x.(x + 4) + 5.(x + 4)
= (x + 4) . (3.x + 5)

Step 2 : -

Now eleminate x and z put x = 0 & z = 0
4.y² + 21.y + 20 = 4.y² + 16.y + 5.y + 20
= 4.y.(y + 4) + 5.(y + 4)
= (y + 4) . (4.y + 5)

Step 3 : -

Now eleminate x & y put x = 0 & y = 0
-z² - z + 20 = -z² - 5.z + 4.z + 20
= -z.(z + 5) + 4 (z + 5)
= (z + 5) . (-z + 4)

Step 4 : -

Now fill the gaps of various brackets
(x + y - z + 4) . (3.x + 4.y + z + 5)
which is the answer .

Note : - If you want to find the remainder then you have to compare the terms containing x.y & x.z & y.z of gin\ven expression and obtained quatient .

Exercises : -

Factorise the following : -
1.)3x² + 7xy + 2y² + 11xz + 7yz + 6z²
2.) 12x² + 11xy + 2y² - 13xz - 7yz + 3z²
3.) 2p² + 2q² + 5pq + 2p – 5q - 12.
4.) u² + v² – 4u + 6v – 12.
5.) 3x² + 4y² + 7xy - 2xz - 3yz - z² + 17x + 21y – z + 20

Method 12

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Solution of some simultaneous equations

This method is based on the sutra " Anurupye Sunyamanyat " which means ' If one is in the ratio the other one is zero ' .

Explaining the method by an example

Example 1 : - Find the solution of the following simultaneous equation .
3x + 7y = 2
4x + 21y = 6

Answer : -

In above we notice that the cofficients of y are in the same ratio as the constants .
i.e.

7 : 21 = 2 : 6 = 1 : 3

Hence Sutra says that if one is in the ratio the other one is zero .
So x =0
and
7.y = 2 => y = 2 / 7

Example 2 : - Find the solution of the following simultaneous equation .
12x + 78y = 12
16x + 96y = 16

Answer : -

In the above we see that
12 : 16 = 12 : 16

Hence y = 0 and
12 .x = 12
=> x =1

Example 3 : - Find the solution of the following simultaneous equation .
a.x + b.y = 2.b.m
c.x + d.y = 2.d.m

Answer : -

In the above we see that cofficients of y are in the same ratio as constants

b : d = 2.b.m : 2.d.m = b : d

Hence according to the sutra
x = 0 and
b.y = 2.b.m
=> y = 2.m

Method 11

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Method 10
Division By 9

This method is based on the sutra " Nikhilam Navatas' Caramam Dashath "which means ' all from 9 and the last from 10 ' .
Explaining the method with the help of an example .
Example 1 : - Divide 341 by 9 .

Answer : -

Step 1 : -

Write the first digit from left as it is .
i.e.
Obtained digit = 3
Step 2 : -
Now add First digit to the next digit .
i.e.
                 3 + 4 = 7
and write it to the right of above obtained digit .
Obtained digit =   37

Step 3 : -

Now add the digit obtained in previous step to the next digit .
i.e.
              7 + 1 =   8
The quotient is the digit leaving the last obtained digit .
Take the remainder as the last digit thus obtained .
If last digit is 9 then increase the quotient by 1 and make the remainder as zero .
Hence
Quotient = 37        Remainder = 8.

Example 2 : - Divide 40214 by 9 .
Answer : -
Step 1 : -
Obtained digit = 4
Step 2 : -
Add next digit
i.e.   4 +   0 =   4
Obtained digit = 44

Step 3 : -

Add next digit
i.e.
            4 + 2 = 6
Place it to the right
Obtained digit = 446
Step 4 : -
Add the next digit tot the previously obtained digit .
i.e.
              6 + 1 = 7
Obtained digit = 4467
Step 5 : -
Add the next digit tot the previously obtained digit .
i.e.
               7 + 4 = 11
Subtract as many 9 from this as posssible and add the number of subtracted nine
to the quotient .
11 - 9.a (value of a is 1)
11 - 9 = 2
Hence
Qotient = 4467 + a = 4468
Remainder = 2

Method 10

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Multiplication of two digit numbers whose first digit are same and sum of last digits is 10 .

This method is based on the sutra " Antyayor Dasakepi " which means 'sum of last digit is 10 ' .

Explaining the method with the help of an example .

Example 1 : - Find the product of 47 * 43.

Answer : -

Just write the product of two end digits to the write and product of first digit and the digit next to it to the left .
Like This

4 * (4 + 1) / ( 7 * 3 )
4 * 5 / 21
20 / 21

Hence the answer to the question is

47 * 43 = 2021

Example 2 : -Find the product of 84 * 86 .

Answer : -

8 * ( 8 + 1 ) / ( 4 * 6 )
72 / 24

84 * 86 = 7224

Example 3 : - Find the product of 72 * 78 .

Answer : -
7 * ( 7+ 1 ) / ( 2 * 8 )
56 / 16

72 * 78 = 5616

It is further interesting to note that the same rule works when the sum of the last 2, last 3, last 4 - - - digits added respectively equal to 100, 1000, 10000 -- - - . The simple point to remember is to multiply each product by 10, 100, 1000, - - as the case may be . Your can observe that this is more convenient while working with the product of 3 digit numbers.

Example 4 : - Find the product of 292 * 208 .

Answer : -

Since 92 + 8 = 100 .
Hence the sutra is applicable .

2 * (2 + 1 ) * 10 / 92 * 8
60 / 736
(Retain the three digits as sum is 100)

292 * 208 = 60736

Method 9

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Square of four digit numbers
(from the same process of duplex of method 7 )

Example 3 : - Find the square of 3242 .

Answer : -

Step 1 : -

Write according to the format
number of digits = 4 so number of zeros = 4 - 1 = 3

0003242

Step 2 : -

Now take the duplex of 2 = 2² = 4

Obtained digit = 4

Step 3 : -

Now take the duplex of 42 = 2.(4 * 2) = 16

Retain 6 and 1 is for carry .

Obtained digit = 64

Step 3 : -

Now take the duplex of 242 = 2.(2 * 2) + 4² = 24

Add carry from the previous step
i.e.
24 + 1 = 25

Retain 5 and 2 is for carry .

Obtained digit = 564

Step 4 : -

Now take the duplex of 3242 = 2.(3 * 2) + 2.(2 * 4) = 12 + 16
= 28

Add carry from previous step
i.e.
28 + 2 = 30

Retain 0 and 3 is for carry to the next step

Obtained digit = 0564

Step 5 : -

Now take the duplex of 03242 = 2.(0 * 2) + 2.(3 * 4) + 2² = 28

Add carry from the previous step
i.e.
28 + 3 = 31

Retain 1 and 3 is for carry .

Obtained digit = 10564

Step 6 : -

Now take the duplex of 003242 = 2.(0 * 2) + 2.(0 * 4) + 2.(3 * 2) = 12

Add carry
i.e.
12 + 3 = 15

Retain 5 and 1 is for carry

Obtained digit = 510564

Step 7 : -

Now take the duplex of 0003242
= 2.(0 * 2) + 2.(0 * 4) + 2.(0 * 2) + 3² = 9

Add carry
i.e.
9 + 1 = 10

Place it to the left

10510564

Hence the square of the 3242 is 10510564 .

Algebraic Proof

Try the proof yourself by seeing the proofs of method 7 and 8 .

Exercises :

Find the squares of the following numbers :

1.) 5532
2.) 2286
3.) 8732
4.) 5467

Method 8

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Square of three digit number
(from the same process of duplex of method 7)

Note : - This method only looks long but after a small practise it will become so easy that you will complete the square in one line .

Example 2 : - Find the square of 215 .

Answer : -

Step 1 : -

Write according to the way
number of digits = 3 number of zeros = 3 -1 = 2

00215

Step 2 : -

Write the duplex of 5 = 5² = 25

Retain 5 and 2 is for carry .

Obtained digit = 5

Step 3 : -

Write the duplex of 15 = 2.( 1 * 5 ) = 10

Add carry of previous step
i.e.
10 + 2 = 12

Retain 2 and 1 is for carry .

Obtained digit = 25

Step 4 : -

Write the duplex of 215 = 2.(2*5) + 1² = 21

Add carry
i.e.
21 + 1 = 22

Retain 2 and 2 is for carry .

Obtained digit = 225

Step 5 : -

Write the duplex of 0215 = 2.(0 * 5) + 2.(2 * 1) = 4

Add carry
i.e.
4 + 2 = 6

Obtained digit = 6225

Step 6 : -

Write the duplex of 00215 = 2.(0 * 5) + 2.(0 * 1) + 2² = 4

Place it to the left

= 46225

Hence the square of 215 is 46225 .

Algebraic Proof : -

Any three digit nuumber ' abc ' can be represented as

(100.a + 10.b + c) .

The square of above number is

(100.a + 10.b + c)²

= (100.a)² + (10.b)² + c² + 2.(100.a).(10b) + 2.(10.b).c + 2.(100.a).c

= 10000.a² + 100.b² + c² + 2.a.b.1000 + 2.b.c.10 + 2.a.c.100
(arranging in increasing powers of 10)

= 10⁴.a² + 10³.2.a.b +10².(2.a.c + b²) + 10.2.b.c + c²

In the above

c² is represented by Step 2 .

2.b.c is represented by Step 3.

(2.a.c. + b²) is represented by Step 4 .

2.a.b. is represented by Step 5 . &

a² is represented by Step 6 .

Exercises : -

Find the square of the following numbers :

1.) 332
2.) 442
3.) 112
4.) 887
5.) 264

Method 7

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Finding the square of a number

This method is based on the sutra "Dwadanda Yoga" which means a 'Duplex combination process ' . For learning this method we have to learn one small extra thing which is known as Duplex .

Duplex

For a single digit ' a ' number the duplex is ' a² '

For a two digit number ' ab ' the duplex is 2.(a * b)

For a three digit number ' abc ' the duplex is 2.(a * c) + b²

For a four digit number ' abcd ' the duplex is 2.(a * d) + 2.(b * c)

For a five digit number ' abcde ' the duplex is 2.(a * e) + 2.(b * d) + c²

And so on .........................

The concept is that take the two numbers equidistant from the beginning and from the end , then multiply these two numbers and then multiply the result by 2 And then add the result in the previously result obtained by the same process . Carry on in the same way until middle digit is reached in case of odd digit number only . Then add the square of this middle digit to the previous result to obtain the duplex .

Now explaining the method with the help of an example

Example 1 : - Find the square of 64 .

Answer : -

Step 1 : -

First write the number and to the left of it write as many zeros equal to one less than the number of digits in the number .
In number of digits in the number = 2
So number of zeros equal to = 1

Write in this way
064

Step 2 : -

First find the duplex of 4 = 4² = 16

Take 6 as the left most digit of the equation and 1 is for carry

Step 3 : -

Then find the duplex of 64 = 2.(6 * 4) = 48

Add carry to it
i.e.
48 + 1 = 49

Retain 9 as next digit from left and 4 is for carry

So obtained digit is 96

Step 4 : -

Find the duplex of 064 = 2.(0 * 4) + 6² = 36

Add carry to it
i.e.
36 + 4 = 40

Place it to the left of digits obtained till now

4096

Hence the square of 64 is 4096 .

Note : - This method only looks long . After small practise you will become so perfect that you will even calculate the square without even writing anything . In the beginning you can also practise one line implementation with paper

Algebraic Proof

Any number of the form ' ab ' can be represented as

( 10.a + b )
Square of the number is

( 10.a + b )² = (10.a)² + 2.(10.a).b + b²
= a².100 + 2.a.b.10 + b²

which shows that the square goes in the same format as above .

This is because in step 1 we do b² . Then we do 2.(a * b )
and then we do a² and all are placed on their respective places .i.e. first step
gives unit's place ,second step gives ten's place and third step gives hundred's
place .

Exercises : -

Find the square of the folowing numbers : -

1.) 54
2.) 22
3.) 34
4.) 92
5.) 45

Method 6

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Equation of a straight line

This method is based on the sutra " Paravartaya Yojayet "which means 'transpose and apply'.

Explaining with the help of an example .

Example 1 : Find the equation of a line passing through the points
( 3 , 4 ) and ( 4 , 2 )

Answer : -

Step 1 : -

The difference of the y coordinates acts as the coefficient of x and
the difference of the x coordinates acts as the coefficient of y .
Like this

diff. of x coordinates = 3 - 4 = -1 acts as y coeficient

diff. of y coordinates = 4 - 2 = 2 acts as x coefficient

Write x first and put a - negative sign in between .

So our equation is

2.x - (-1).y
2.x + y which is the right hand part of the equation .

Step 2 : -

Substitute any given point from ( 3 , 4 ) and ( 4 , 2 ) into the above obtained equation to find the right hand side of the straight line equation .

2.(3) + 4 = 10

Hence the equation of the straight line is

2.x - y = 10

Example 2 : - Find the equation of a straight line passing through the points
( 5, 2 ) and ( 9 , 1 ) .

Answer : -

Step 1 : -

diff. of x coordinates = 5 - 9 = -4 acts as coefficient of y

diff. of y coordinates = 2 - 1 = 1 acts as coefficient of x

x - (-4).y

Hence L.H.S. of the equation is

x + 4.y

Step 2 : -

Substitute ( 5 , 2 ) or ( 9 , 1 ) in the above obtained equation

5 + 4.(2) = 13

which is the R.H.S. of the straight line equation .

Hence the equation of the straight line is

x + 4.y = 13

Example 3 : - Find the equation of the straight line passing through the points
(5 , 5) and (3, 4) .

Answer : -

Step 1 : -

diff. of x coordinates = 5 - 3 = 2 = y coefficient
diff. of y coordinates = 5 - 4 = 1 = x coefficient

Hence the L.H.S of the equation is
x - 2.y

Step 2 : -

Substitute (5 , 5) or (3 , 4) in the equation

5 - 2.5 = 5 - 10 = -5

Hence the equation of line is

x - 2.y = -5

Algebraic Proof : -

For a straight line passing through two points ( x1 , y1 ) and ( x2 , y2 )

the equation is

y - y1 = (y2 - y1 ) ( x - x1 )
(x2 - x1 )

Cross multiplying

(x2 - x1 ) (y - y1 ) = (y2 - y1 ) (x - x1 )
Solving

(x2 - x1 ) (- y1) + ( y2 - y1) (x1) = (y2 - y1) x - (x2 - x1 ) y

In the above we see that coefficient of x is difference of y coordinates and coefficient of y is difference of x coordinates .

For obtaining the constant term we know that on substituting any one point we can easily know the term .

Exercises : -

Find the equation of the straight linepassing through the points

1.) ( 9 , 0 ) and ( 2 , 1 )
2.) ( 2 , 4 ) and ( 1 , 9 )
3.) ( 5 , 3 ) and ( 4 , 2 )
4.) ( 8 , 4 ) and ( 5 , 9 )
5.) ( 6 , 3 ) and ( 4 , 9 )

Method 5

2010-01-02T06:02:00.000-08:00

Solution of Some General Form Equations

This method os based on the Sutra "Paravartya Yojayet " which means 'transpose and apply ' . In this rule + is converted into - and vice versa and
* is converted into / and vice versa .

These formula although very simple and in the eye , even then not used by many so try this to speed up your calculations .

Type 1 :

Equations of the form ax +b = cx + d . (very simple one )

Direct answer can be obtained by the formula

x = d - b
a - c

Example 1 : - Find the value of x in the equation
3x + 5 = 2x + 7

Answer :
x = 7 - 5
3 - 2
=> x = 2.
Exercises : -
Find the value of x in the following equqtions : -
1.) 3x + 2 = 7x + 5
2.) 6x + 2 = 2x + 5
3.) 9x + 2 = 2x + 4
4.) 23x + 12 = 12x + 4
5.) 7x - 21 = 9x - 31

.

Type 2 :

Equations of the form (x +a )(x + b) = (x + c)(x + d)

The above is solved by the formula

x = c.d - a.b
(a + b) - (c + d)
Derivation is simple . Try it yourself .
Use it directly to solve .

Example 1 : - Find the value of x in the Equation
(x – 3) (x – 2 ) = (x + 1 ) (x + 2 ).

Answer : -

x = 1.2 - (-3).(-2)
(-3 + (-2)) - (1 + 2)

=> x = 2 - 6 = -4
-5 - 3 -8

=> x = 1 / 2

Exercises : -

Find the value of x in the following equqtions : -

1.) (x - 4)(x-2) = (x + 3)(x + 1)
2.) (x + 1)(x - 4) = (x + 9)(x - 1)
3.) (a - 7)(a + 1) = (a + 3)(a + 8)
4.) (y + 5)(y - 5) = (y +1)(y + 2)
5.) (x - 6)(x - 1) = ( x + 5)(x - 9)

Type 3 :

Equations of type m + n = 0
(x + a ) (x + b)

Take L.C.M and proceed.
m(x+b) + n (x+a) = 0
(x + a) (x +b)
mx + mb + nx + na = 0
(x + a)(x + b)
(m + n)x + mb + na = 0
Therefore

x = - (m.b + n.a)
(m + n)

Example : - Find the value of x in the following equation .
5 + 2 = 0
(x + 3) (x - 4)

Answer : -

Use the formula directly : -

x = - (5.(-4) + 2.3)
(5 + 2)
=> x = - (-20 + 6)
7
x = 14 / 7
x = 2

Exercises : -
Find the value of x in the following equqtions : -
1.) 4 + 9 = 0
(x + 3) (x + 2)
2.) 6 + 5 = 0
(x - 5) (x + 2)
3.) 7 + 1 = 0
(x+1) (x + 9)

Type 4 :

Equations of the form (ax + b) = m
(cx + d) n

By cross – multiplication,
n ( ax + b) = m (cx + d)
nax + nb = mcx + md
nax - mcx = md – nb
x( na – mc ) = md – nb
x = (md - nb)
(na - mc)
Example : - Find the solution of x in equation
7.x + 2 = 5
3.x - 5 8

Answer : -

x = (5.(-5) - 8.2)
(8.7 - 5.3)
x = (-25 -16 )
(56 - 15 )
x = - 41
41
x = -1
Exercises : -

Find the value of x in the following equqtions : -

1.) 3x + 6 = 8
4x - 2 3
2.) 5x + 3 = 6
6x +8 4
3.) 9x + 7 = 2
7x - 3 5

Method 4

2010-01-02T06:01:00.000-08:00

Solution of Simultaneous Equations

This method is based on the Sutra "Patavartya Yogayet" which means ' transpose and apply ' .

Explaining the method by an esample .

Example 1 : - Find the values of x and y .
2x + 3y = 9
3x + y = 4

Answer : -

Step 1 : -

First finding the value of x .
Cross multipling the cofficients of y and constants and subtracting rightward .

2.x + 3 y 9

3.x + 1 y 4

( 3 * 4 ) - ( 9 * 1 ) (Subtract from left to right )

= 3 which is numerator for x .

Step 2 : -

Now cross multipling the cofficients of x and y and subtracting leftward .

2.x + 3.y 9

3.x + 1.y 4

( 3 * 3 ) - ( 2 * 1 ) (Subtract from right to left )
= 9 - 2
= 7 which is the denominator .
x = 3 / 7

Step 3 : -

Finding the value of y .
Now cross multipling the cofficients of x and constants and subtracting leftward .

3.y + 2.x 9

1.y + 3.x 4

( 9 * 3 ) - ( 2 * 4 )
(Subtract from right to left Starting from constants)
= 27 - 8
= 19 which is numerator .

Denominator is same as step 2 .

So y = -19 / 7

Hence
x = 3 / 7
y = -19 / 7

Algebraic Proof : -

Let there be two equations
a.x + b.y = m ...........( i )
c.x + d.y = n ...........( ii )
Multiply ( i ) by d and ( ii ) by b .

a.d.x + b.d.y = md
c.b.x + d.b.y = nb (Subtracting)
(a.d - c.b).x = m.d - n.b

Hence x = (md - nb)
(a.d - b.c)
x = ( b.n - m.d)
( b.c - a.d)

which is same as step ( i ).

Now multipling ( i ) by c and ( ii ) by a And subtracting ( ii ) from ( i )

We get
y = ( m.c - n.a )
( b.c - a.d )
which is same as step 2 .

Exercises : -

Find the value of x and y in the following equations : -
1.) 2x + y = 5
3x – 4y = 2
2.) 3x – 4y = 7
5x + y = 4
3.) 4x + 3y = 8
6x - y = 1
4.) x + 3y = 7
2x + 5y = 11
5.) 3x + y = 5
4x – y = 9

Method 3

2010-01-02T05:54:00.000-08:00

Squares of numbers ending with 5

This method of finding the square of a number ending with five is based on the Sutra of " Ekadhikena Purvena " of respectable Vedas . It means "one more than the one before it " or " one more than the previous one .

Explaining the method by an example : -

Example 1 : - Find the square of 95 .

Answer : -

Step 1 : -

In this we simply make the square by simply multipling the digit(s) previous to 5 by one more than themselves .
i.e.
9 * (9 + 1) = 9 * 10 = 90

Step 2 : -

Now we place 25 next to the digit obtained above .
i.e.
9025
which gives us the answer.

=>Answer is 9025.

Example 2 : - Find the square of 55.

Answer : -

Proceeding as above :

Step 1 : -

5 * (5 + 1) = 5 * 6 = 30

Step 2 : -

Place 25 next to it .

Answer is 3025.

Note : - We can extend the rule to any number of digits .

Example 3 : - Find the square of 125 .

Answer : -

Step 1 : -
12 * ( 12 + 1) = 12 * 13 = 156

Step 2 : -

Place 25 next to it .

Answer is 15625 .

Algebraic proof : -

Consider the expantion

(a.x + b )² = a².x²+ 2a.x.b + b²Substituting x = 10 and b = 5

(10.a + 5)² = 10².a² + 2.10.a.5 + 5²
= 100.a² + 100.a + 25
= (a² + a)100 + 25
= a.(a + 1)10² + 25

In the above a.(a + 1).100 gives a.(a + 1)/00 and + 25 gives

a.(a + 1)/25 .

which is our answer

Exercises : -

Find the square of following numbers.

1.) 35
2.) 75
3.) 215
4.) 85
5.) 165

Method 2

2010-01-02T05:49:00.001-08:00

Product of three digit numbers

The method of ' Urdhva Tiryagbhyam ' can also be used for findiing the product of three digit numbers . This although a bit difficult from the previous one can work effectively after a small amount of practice.Explaining the method by an example . This methodonly looks long but is very small if compared to the normal method of multipling three digtit numbers . It can be completed in one single line only instead of four lines of normal method .

Example 1 : - Find the product of 234 * 342 .

Answer : -

Step 1 : -

Multiply the last two digit of the multiplier and the multiplicand.
i.e.
4 * 2 = 8

Obtained digit = 8 Carry = 0
This is the right most digit of the answer. If any carry is produced then it is taken to the next step . Like if 18 is produced then 1 is taken to the next step .

Step 2 : -

Coss Multiply the last two - two digits of the multiplier and the multiplicand .

Multiply last digit of multiplicand(23 4) with last second digit of multiplicand(342)
i.e.
4 * 4 = 16
and
Multiply last second digit of multiplier(234) with the ladt digit of multiplicand(342)
i.e.
3 * 2 = 6

Add the two digits and add carry from the fiirst step.
i.e.
16 + 6 + 0 =22
place the right most digit to left of digit obtained in first step and use rest for carry.
Obtained digit = _______28 Carry = 2

Step 3 : -

Now cross multiply the first and the last digit of the multiplicand and multiplier
(Multiplicand) 234 (Multiplier)342
2 * 2 = 4
and
234 342
4 * 3 = 12
also multiply the middle digits
234 342
3 * 4 = 12
Add the above three ovtained digits and carry from the last step
i.e.
4 + 12 + 12 + (Carry)2 = 30
Place right most digit in obtained digit and rest for carry to next step.
Obtained digit = 028 Carry = 3

Step 4 : -

Now cross multiply the first two - two digits of the multiplier and multiplicand.
(Multiplicand) 234 (Multiplier)342
2 * 4 = 8
(Multiplicand) 234 (Multiplier)342
3 * 3 = 9
Add the two digits + the carry from above step
i.e.
8 + 9 + (Carry)3 = 20
Place right most digit in obtained digit and rest for carry to next step.
Obtained Digit = ________0028 Carry = 2

Step 5 : -

Now multiply the first two digits of multipler and multiplicand
234 342
i.e. 2 * 3 = 6
Add the carry from the last step
6 + 2 = 8
Place it to the left and here cames the answer.........

234 * 342 = 80028

One line approach of method 2

One Line Approach
234
342
8

2 3 4
3 42
4*4=16 + 3*2=6 + (carry) 0 =22

2 3 4

3 4 2
12 +12 + 4 + (carry) 2 = 30

2 3 4

3 4 2
9 + 8 + (carry) 3 = 20

2 34
3 42 Answer = 60028 (Starting from down)
6

Algebraic Proof : -

Let there be two numbers (ax ² + bx + c) and (dx² + ex + f) . If we have to multiply them then
(a.x² + b.x + c)(d.x² + e.x + f) =
ad.x4 + (bd + ae).x3 + (cd + be + af).x2 + (ce + bf)x + cf

From above we can see that in the expression the cofficients of various terms are the same as described in the various steps .

Note : - We can also start the same procedure from the left most digit also.

Exersises : -

Find the product of the following numbers : -

1.) 230 * 453
2.) 321 * 546
3.) 342 * 786
4.) 467 * 598
5.) 104 * 599

Method 1

2010-01-02T05:36:00.000-08:00

Product of two digit numbers

This method is based on the sutra 'Urdhva Triyagbhyam' of the respectable Vedas . It means vertically and cross wise . It is applicable to all cases of multiplication of a large number by another large number . It is applicable in division also .
Explaining the method by example :-

Example 1 : - Find the product of 41 * 23 ?

Answer : -

Step 1 :-

First multiply the right most digit of the multiplicand and the multiplier.
i.e.
1 * 3 = 3
This givess the right most digit of the answer .
If a two digit number is generated then retain the right mast digit for the answer and rest are used as a carry to the next step .

Obtained digit : - 3
Step 2 : -

Now cross multiply .
Multiply first digit of the multiplicand (41) with the second digit of multiplier(23)
i.e.
4 * 3 = 12
and
Multiply second digit of the multiplicand (41) with the first digit of the multiplier (23)
i.e.
1 * 2 = 2
Add the above two digts + carry from the the first step( if any )
i.e.
12 + 2 + 0 = 14
4 is placed to the left of digit of step 1 and 2 is retained for carry.

Obtained digit : - 43

Step 3 : -

Now multiply the left most digit of multiplier and multiplicand
i.e.
4 * 2 = 8
Add carry from step 2
i.e.
8 + 1 = 9
Place it to the left of obtained digit .
i.e.
943
which is the answer.

Hence 41 * 23 = 943

The above proceure is very simple and can be completed in one single line only as shown .

   4 1
2 3
8 : 12 : 1 * 3 = 3
   : 2 :
   1: 0 :
9 4 3

Example 2 : - Find the product of 59 * 45 .

Answer : -

   5 9
4 5
20 : 36 : 9 * 5 = 45
   : 25 :
  6 : 4 :
26 5 5

Answer 2655.

Example 3 : - Find the product of 33 * 24 .

Answer : -

   3 3
2 4
6 : 6 : 3 * 4 = 12
   :12 :
1 : 1 :
7 9 2
Answer is 792.

Algebraic Proof : -

Let the two 2 digit numbers be (ax+b) and (cx+d). Note that x = 10. Now
consider the product
(ax + b) (cx + d) = ac.x2 + adx + bcx + b.d
= ac.x2 + (ad + bc)x + b.d

You can observe that if x = 10 then cofficient of 10 ⁰ i.e. unit's place is according to step 1 and cofficient of 10¹ is step 2 and cofficient of 10² is step 3.

Practice Exercises : -

Find the product of following numbers : -
1.) 29 * 13
2.) 34 * 98
3.) 43 * 56
4.) 55 * 87
5.) 23 * 79

Vedic Maths

Introduction

Introduction

Speed Division by 9

Home

Introduction

Method 16

Finding the H. C .F. of two,three or more degree equation

Algebraic Proof :

Exercises : -

Method 15

Cubing of numbers near powers of 10

Exercise : -

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Method 14

Resolving into Partial Fractions

Algebraic Proof : -

Exercises : -

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Method 13

Factorise Three or more variable Equations

Exercises : -

Method 12

Solution of some simultaneous equations

Method 11

Method 10

Method 10

Multiplication of two digit numbers whose first digit are same and sum of last digits is 10 .

Method 9

Square of four digit numbers (from the same process of duplex of method 7 )

Algebraic Proof

Exercises :

Method 8

Square of three digit number (from the same process of duplex of method 7)

Algebraic Proof : -

Exercises : -

Method 7

Finding the square of a number

Algebraic Proof

Exercises : -

Method 6

Equation of a straight line

Algebraic Proof : -

Exercises : -

Method 5

Solution of Some General Form Equations

Type 1 :

Type 2 :

Type 3 :

Type 4 :

Method 4

Solution of Simultaneous Equations

Algebraic Proof : -

Exercises : -

Method 3

Squares of numbers ending with 5

Algebraic proof : -

Exercises : -

Method 2

Product of three digit numbers

One line approach of method 2

Algebraic Proof : -

Exersises : -

Method 1

Product of two digit numbers

Algebraic Proof : -

Practice Exercises : -

Square of four digit numbers
(from the same process of duplex of method 7 )

Square of three digit number
(from the same process of duplex of method 7)