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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 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. x–6+x-4 x–8+x-2 ___________ = ___________ (x–4) (x–6) (x–2) (x-8) 2x-10 2x-10 _________ = _________ x2–10x+24 x2–10x+16 ( 2x – 10 ) ( x2 – 10x + 16 ) = ( 2x – 10 ) ( x2 – 10x + 24) 2x3–20x2+32x–10x2+100x–160 = 2x3–20x2+48x–10x2+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. 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 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 (x–249)3 + (x+247)3 = 2(x–1)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(x3–3x2a–3x2b+3xa2+3xb2+6axb–a3–3a2b–3ab2–b3) = 2x3–6x2a–6x2b+6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3 cancel the common terms on both sides 12xa2+12xb2–8a3–8b3 = 6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3 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 since x = a + b 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 (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. 1. (x + 3)3 x+1 ______ = ____ (x + 5)3 x+7 2. (x - 5)3 x - 3 ______ = ____ (x - 7)3 x - 9 |
