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 ÷ 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)+(3x–2)
                   ____________   =   0
                    (3x–2)(2x–1)

                          5x–3
                    __________   =   0
                    (3x–2)(2x–1)

                        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

                        c²–b²        (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 pro