Vedic Mathematics | Sunyam Samyasamuccaye
SŨNYAM SĀMYASAMUCCAYE
The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya' has several meanings under different contexts.
i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all the terms concerned and proceed as follows.
Example 1: The equation 7x + 3x = 4x + 5x has the same factor x in all its terms. Hence by the sutra it is zero, i.e., x = 0.
Otherwise we have to work like this:
7x + 3x = 4x + 5x
10x = 9x
10x 9x = 0
x = 0
This is applicable not only for x but also any such unknown quantity as follows.
Example 2: 5(x+1) = 3(x+1)
No need to proceed in the usual procedure like
5x + 5 = 3x + 3
5x 3x = 3 5
2x = -2 or x = -2 χ 2 = -1
Simply think of the contextual meaning of Samuccaya
Now Samuccaya is ( x + 1 )
x + 1 = 0 gives x = -1
ii) Now we interpret Samuccaya as product of independent terms in expressions like (x+a) (x+b)
Example 3: ( x + 3 ) ( x + 4) = ( x 2) ( x 6 )
Here Samuccaya is 3 x 4 = 12 = -2 x -6
Since it is same , we derive x = 0
This example, we have already dealt in type ( ii ) of Paravartya in solving simple equations.
iii) We interpret Samuccaya as the sum of the denominators of two fractions having the same numerical numerator.
Consider the example.
1 1
____ + ____ = 0
3x-2 2x-1
for this we proceed by taking
L.C.M.
(2x-1)+(3x2)
____________ = 0
(3x2)(2x1)
5x3
__________ = 0
(3x2)(2x1)
5x 3 = 0 5x = 3
3
x = __
5
Instead of this, we can directly put the Samuccaya i.e., sum of the denominators
i.e., 3x 2 + 2x - 1 = 5x - 3 = 0
giving 5x = 3 x = 3 / 5
It is true and applicable for all problems of the type
m m
____ + _____ = 0
ax+b cx+d
Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 )
- ( b + d )
x = _________
( a + c )
iii) We now interpret Samuccaya as combination or total.
If the sum of the numerators and the sum of the denominators be the same, then that sum = 0.
Consider examples of type
ax
+ b ax + c
_____
= ______
ax
+ c ax + b
In this case, (ax+b) (ax+b) =
(ax+c) (ax+c)
a2x2 + 2abx + b2 = a2x2 + 2acx +
c2
2abx 2acx = c2 b2
x ( 2ab 2ac ) = c2 b2
c2b2 (c+b)(c-b)
-(c+b)
x
= ______ = _________ = _____
2a(b-c)
2a(b-c) 2a
As per Samuccaya (ax+b) + (ax+c) = 0
2ax+b+c = 0
2ax = -b-c
-(c+b)
x = ______
2a Hence the statement.
Example 4:
3x
+ 4 3x + 5
______ = ______
3x
+ 5 3x + 4
Since N1 + N2 = 3x + 4 + 3x + 5 = 6x + 9 ,
And
D1 + D2 = 3x + 4 + 3x + 5 = 6x + 9
We have
N1 + N2 = D1 + D2 = 6x + 9
Hence from Sunya Samuccaya we get 6x + 9 = 0
6x = -9
-9 -3
x = __ = __
6 2
Example 5:
5x +
7 5x + 12
_____
= _______
5x
+12 5x + 7
Hence N1 + N2 = 5x + 7 + 5x + 12 = 10x + 19
And
D1 + D2 = 5x + 12 + 5x + 7 = 10x + 19
N1 + N2 = D1 + D2 gives 10x + 19 = 0
10x = -19
-19
x = ____
10
Consider the examples of the type, where N1 + N2 = K ( D1 + D2 ), where K is a numerical constant, then also by removing the numerical constant K, we can proceed as above.
Example 6:
2x + 3
x + 1
_____
= ______
4x + 5 2x + 3
Here N1 + N2 = 2x + 3 +
x + 1 = 3x + 4
D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8
= 2 ( 3x + 4 )
Removing the numerical factor 2, we get 3x + 4 on both sides.
3x + 4 = 0
3x = -4 x = - 4 / 3.
v) Samuccaya with the same meaning as above, i.e., case (iv), we solve the problems leading to quadratic equations. In this context, we take the problems as follows;
If N1 + N2 = D1 +
D2 and also the differences
N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the solution gives the two values for x.
Example 7:
3x +
2 2x + 5
_____
= ______
2x +
5 3x + 2
In the conventional text book method, we work as follows :
3x + 2 2x + 5
_____ = ______
2x + 5 3x + 2
( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 )
9x2 + 12x + 4 = 4x2 + 20x + 25
9x2 + 12x + 4 - 4x2 - 20x 25 = 0
5x2 8x 21 = 0
5x2 15x + 7x 21 = 0
5x ( x 3 ) + 7 ( x 3 ) = 0
(x 3 ) ( 5x + 7 ) = 0
x 3 = 0 or 5x + 7 = 0
x = 3 or - 7 / 5
Now Samuccaya sutra comes to help us in a beautiful way as follows :
Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7
D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7
Further N1 ~ D1 = ( 3x + 2 ) ( 2x + 5 ) = x 3
N2 ~ D2 = ( 2x + 5) ( 3x + 2 ) = - x + 3 = - ( x 3 )
Hence 5x + 7 = 0 , x 3 = 0
5x = -7 , x = 3
i.e., x = -7 / 5 , x = 3
Note that all these can be easily calculated by mere observation.
Example 8:
3x + 4 5x + 6
______ = _____
6x + 7 2x + 3
Observe that
N1 + N2 = 3x + 4 + 5x + 6 = 8x + 10
and
D1 + D2 = 6x + 7 + 2x + 3 = 8x + 10
Further N1 ~
D1 = (3x + 4) (6x + 7)
= 3x + 4 6x 7
= -3x 3 = -3 ( x + 1 )
N2 ~ D2 = (5x + 6) (2x + 3) = 3x + 3 = 3( x + 1)
By Sunyam Samuccaye we have
8x + 10 = 0
3( x + 1 ) = 0
8x = -10
x + 1 = 0
x = - 10 / 8
x = -1
= - 5 / 4
vi)Samuccaya with the same sense but with a different context and application .
Example 9:
1
1 1 1
____ + _____ = ____ + ____
x - 4 x 6
x - 2 x - 8
Usually we proceed as follows.
x6+x-4
x8+x-2
___________ = ___________
(x4) (x6)
(x2) (x-8)
2x-10
2x-10
_________ = _________
x210x+24
x210x+16
( 2x 10 ) ( x2 10x + 16 ) = ( 2x 10 ) ( x2 10x + 24)
2x320x2+32x10x2+100x160 = 2x320x2+48x10x2+100x-240
2x3 30x2 + 132x 160 = 2x3 30x2 + 148x 240
132x 160 = 148x 240
132x 148x = 160 240
16x = - 80
x = - 80 / - 16 = 5
Now Samuccaya sutra, tell us that, if other elements being equal, the sum-total of the denominators on the L.H.S. and their total on the R.H.S. be the same, that total is zero.
Now D1 + D2 = x 4 + x 6 = 2x 10, and
D3 + D4 = x 2 + x 8 = 2x 10
By Samuccaya, 2x 10 gives 2x = 10
10
x = __ = 5
2
Example 10:
1
1
1 1
____ + ____ = ____ + _____
x -
8 x 9 x -
5 x 12
D1 +
D2 = x 8 + x 9 = 2x 17, and
D3 +
D4 = x 5 + x 12 = 2x 17
Now 2x 17 = 0 gives 2x = 17
17
x = __ = 8½
2
Example 11:
1 1
1 1
____ - _____ = ____ - _____
x +
7 x + 10 x +
6 x + 9
This is not in the expected form. But a little work regarding transposition makes the above as follows.
1 1
1 1
____ + ____ = ____ + _____
x +
7 x + 9 x +
6 x + 10
Now Samuccaya sutra applies
D1 +
D2 = x + 7 + x + 9 = 2x + 16, and
D3 +
D4 = x + 6 + x + 10 = 2x + 16
Solution is given by 2x + 16 = 0 i.e., 2 x = - 16.
x = - 16 / 2 = - 8.
Solve the following problems using Sunyam Samya-Samuccaye process.
1. 7 ( x + 2 ) + 3 ( x + 2 ) = 6 ( x + 2 ) + 5 ( x + 2 )
2. ( x + 6 ) ( x + 3 ) = ( x 9 ) ( x 2
)
3. ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x 7
)
1
1
4.
______ + ____ = 0
4 x - 3 x 2
4
4
5.
_____ + _____ = 0
3x + 1 5x + 7
2x + 11 2x+5
6.
______ = _____
2x+ 5 2x+11
3x + 4 x + 1
7.
______
= _____
6x + 7 2x + 3
4x - 3 x
+ 4
8.
______
= _____
2x+ 3 3x - 2
1 1
1
1
9. ____
+ ____ = ____ + _____
x - 2
x - 5 x - 3
x - 4
1
1
1 1
10.
____ - ____ = _____ - _____
x - 7
x - 6 x - 10
x - 9
Sunyam Samya Samuccaye in Certain Cubes:
Consider the problem ( x 4 )3 + ( x 6 )3 = 2 ( x 5
)3. For the solution by the traditional method we follow the steps as given below:
( x 4 )3 + ( x 6 )3 = 2 ( x 5 )3
x3 12x2 + 48x 64 + x3 18x2 + 108x 216
= 2 ( x3 15x2 + 75x 125 )
2x3 30x2 + 156x 280 = 2x3 30x2 + 150x 250
156x 280 = 150x 250
156x 150x = 280 250
6x = 30
x = 30 / 6 = 5
But once again observe the problem in the vedic sense
We have ( x 4 ) + ( x 6 ) = 2x 10. Taking out the numerical factor 2, we have ( x 5 ) = 0, which is the factor under the cube on R.H.S. In such a case Sunyam samya Samuccaye formula gives that x 5 = 0. Hence x = 5
Think of solving the problem (x249)3 + (x+247)3 = 2(x1)3
The traditional method will be horrible even to think of.
But ( x 249 ) + ( x + 247 ) = 2x 2 = 2 ( x 1 ). And x 1. on R.H.S. cube, it is enough to state that x 1 = 0 by the sutra.
x = 1 is the solution. No cubing or any other mathematical operations.
Algebraic Proof :
Consider ( x 2a
)3 + ( x 2b )3 = 2 ( x a b )3 it is clear that
x 2a + x 2b = 2x 2a 2b
= 2 ( x a b )
Now the expression,
x3 -
6x2a + 12xa2 8a3 + x3 6x2b +
12xb2 8b3 =
2(x33x2a3x2b+3xa2+3xb2+6axba33a2b3ab2b3)
= 2x36x2a6x2b+6xa2+6xb2+12xab2a36a2b6ab22b3
cancel the common terms on both sides
12xa2+12xb28a38b3 = 6xa2+6xb2+12xab2a36a2b6ab22b3
6xa2 + 6xb2 12xab = 6a3 + 6b3
6a2b 6ab2
6x ( a2 + b2 2ab ) = 6 [ a3 + b3 ab ( a + b )]
x ( a b )2 = [ ( a + b ) ( a2 + b2 ab ) ( a + b )ab]
= ( a + b ) ( a2 + b2 2ab )
= ( a + b ) ( a b )2
x = a + b
Solve the following using Sunyam Samuccaye process :
1.
( x 3 )3 + ( x 9 )3 = 2 ( x 6 )3
2.
( x + 4 )3 + ( x 10 )3 = 2 ( x 3 )3
3.
( x + a + b c )3 + ( x + b + c a )3 = 2 ( x + b )3
Example :
(x + 2)3 x + 1
______ = _____
(x + 3)3 x + 4
with the text book procedures we proceed as follows
x3 + 6x2 + 12x +8
x + 1
_______________ = _____
x3 + 9x2 + 27x +27
x + 4
Now by cross multiplication,
( x + 4 ) ( x3 +
6x2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9x2 + 27x + 27 )
x4 + 6x3 + 12x2
+ 8x + 4x3 + 24x2 + 48x + 32 =
x4 + 9x3 + 27x2 + 27x + x3 + 9x2 + 27x + 27
x4 +
10x3 + 36x2 + 56x + 32 = x4 + 10x3 +
36x2 + 54x + 27
56x + 32 = 54x + 27
56x 54x = 27 32
2x = - 5
x = - 5 / 2
Observe that ( N1 + D1 ) with in the cubes on
L.H.S. is x + 2 + x + 3 = 2x + 5 and
N2 + D2 on the right hand side
is x + 1 + x + 4 = 2x + 5.
By vedic formula we have 2x + 5 = 0
x = - 5 / 2.
Solve the following by using vedic method :
1.
(x + 3)3 x+1
______ = ____
(x + 5)3 x+7
2.
(x - 5)3 x - 3
______ = ____
(x - 7)3 x - 9