VILOKANAM The Sutra 'Vilokanam' means 'Observation'. Generally we come across problems which can be solved by mere observation. But we follow the same conventional procedure and obtain the solution. But the hint behind the Sutra enables us to observe the problem completely and find the pattern and finally solve the problem by just observation.Let us take the equation x + ( 1/x ) = 5/2 Without noticing the logic in the problem, the conventional process tends us to solve the problem in the following way. 1 5 x + __ = __ x 2 x2 + 1 5 _____ = __ x 2 2x2 + 2 = 5x 2x2 – 5x + 2 = 0 2x2 – 4x – x + 2 = 0 2x (x – 2) – (x – 2) = 0 (x – 2) (2x – 1) = 0 x – 2 = 0 gives x = 2 2x – 1 = 0 gives x = ½ But by Vilokanam i.e.,, observation 1 5 x + __ = __ can be viewed as x 2 1 1 x + __ = 2 + __ giving x = 2 or ½. x 2 Consider some examples. Example 1 : x x + 2 34 ____ + _____ = ___ x + 2 x 15 In the conventional process, we have to take L.C.M, cross-multiplication. simplification and factorization. But Vilokanam gives 34 9 + 25 3 5 __ = _____ = __ + __ 15 5 x 3 5 3 x x + 2 3 5 ____ + _____ = __ + __ x + 2 x 5 3 gives x 3 5 _____ = __ or __ x + 2 5 3 5x = 3x + 6 or 3x = 5x + 10 2x = 6 or -2x = 10 x = 3 or x = -5 Example 2 : x + 5 x + 6 113 ____ + _____ = ___ x + 6 x + 5 56 Now, 113 49 + 64 7 8 ___ = _______ = ___ + ___ 56 7 x 8 8 7 x + 5 7 x+5 8 ____ = __ or ____ = __ x + 6 8 x+6 7 8x + 40 = 7x+42 or 7x+35 = 8x+48 or x = 42 - 40 = 2 or -x = 48 – 35 = 13 x = 2 or x = -13. Example 3: 5x + 9 5x – 9 82 _____ + _____ = 2 ___ 5x - 9 5x + 9 319 At first sight it seems to a difficult problem. But careful observation gives 82 720 841 – 121 29 11 2 ___= ___ = ________ = ___ – __ 319 319 11 x 29 11 29 (Note: 292 = 841, 112 = 121) 5x + 9 29 –11 _____ = __ or ___ 5x – 9 11 29 (Note: 29 = 20 + 9 = 5 x 4 + 9 ; 11 = 20 – 9 = 5 x 4 – 9 ) i.e., x = 4 or 5x + 9 –11 _____ = ___ 5x – 9 29 145x + 261 = –55x + 99 145x + 55x = 99 – 261 200x = –162 –162 –81 x = ____ = ____ 200 100 Simultaneous Quadratic Equations: Example 1: x + y = 9 and xy = 14. We follow in the conventional way that (x – y)2 = (x + y)2 – 4xy = 92 – 4 (14) = 81 – 56 = 25 x – y = √ 25 = ± 5 x + y = 9 gives 7 + y = 9 y = 9 – 7 = 2. Thus the solution is x = 7, y = 2 or x = 2, y = 7. But by Vilokanam, xy = 14 gives x = 2, y = 7 or x = 7, y = 2 and these two sets satisfy x + y = 9 since 2 + 7 = 9 or 7 + 2 = 9. Hence the solution. Example 2: 5x – y = 7 and xy = 6. xy = 6 gives (1) x = 6, y = 1; (2) x = 1, y = 6; (3) x = 2, y = 3; (4) x = 3, y = 2 and of course these may be negatives of all these. (1) Observe that x = 6, y = 1: are not solutions because they do not satisfy the equation 5x – y = 7. (2) Observe that x = 1, y = 6: are not solutions because they do not satisfy the equation 5x – y = 7. (3) But for x = 2, y = 3; 5x – y = 5 (2) – 3 = 10 – 3 = 7 we have 5(3)–2≠7. Hence x = 2, y = 3 is a solution. (4) For x = 3, y = 2 we get 5 (3) – 2 = 15 – 2 ≠ 7. Hence it is not a solution. Negative values of the above are also not the solutions. Thus one set of the solutions i.e., x = 2, y = 3 can be found. Of course the other will be obtained from solving 5x – y = 7 and 5x + y = –13. i.e., x = –3 / 5, y = -10. Partial Fractions: Example 1: Resolve 2x + 7 _________ into partial fractions. (x + 3) (x + 4) 2x + 7 A B We write _______ = _____ + ______ (x + 3)(x + 4) (x + 3) (x + 4) A (x + 4) + B (x + 3) = __________________ (x + 3) (x + 4) 2x + 7 ≡ A (x + 4) + B (x + 3). We proceed by comparing coefficients on either side coefficient of x : A + B = 2 ..........(i) × 3 Independent of x : 4A + 3B = 7 .............(ii) Solving (ii) – (i) x 3 4A + 3B = 7 + 3A + 3B = 6 – – – _______________ A = 1 A = 1 in (i) gives, 1 + B = 2 i.e., B = 1 Or we proceed as 2x + 7 ≡ A (x + 4) + B (x + 3). Put x = –3, 2 (–3) + 7 ≡ A (–3 + 4) + B (-3 + 3) 1 = A (1) ... A = 1. x = –4, 2 (– 4) +7 = A (–4 + 4) + B (–4 + 3) –1 = B(–1) ... B = 1. 2x + 7 1 1 Thus _________ = _____ + _____ (x + 3) (x + 4) (x + 3) (x + 4) But by Vilokanam can be resolved as 2x + 7 ____________ (x + 3) (x + 4) (x + 3) + (x + 4) =2x + 7, directly we write the answer. Example 2: 3x + 13 ____________ (x + 1) (x + 2) from (x + 1),(x + 2) we can observe that 10 (x + 2) – 7(x + 1) = 10x + 20 – 7x – 7 = 3x + 13 3x + 13 10 7 Thus __________ = _____ – _____ (x + 1) (x + 2) x + 1 x + 2 Example 3: 9 ________ x2 + x - 2 As x2 + x – 2 = (x – 1) (x + 2) and 9 = 3 (x + 2) – 3 (x – 1) (3x + 6 – 3x + 3 = 9) 9 3 3 We get by Vilokanam, ____________ = ____ - ____ x2 + x – 2 x - 1 x + 2 1. 1 25 x + __ = __ x 12 2. 1 5 x – __ = __ x 6 3. x x + 1 1 _____ + _____ = 9 __ x + 1 x 9 4. x + 7 x + 9 32 ____ – ____ = ___ x + 9 x + 7 63 II. Solve the following simultaneous equations by vilokanam. 1. x – y = 1, xy = 6 2. x + y = 7, xy = 10 3. 2x + 3y = 19, xy = 15 4. x + y = 4, x2 + xy + 4x = 24. III. Resolve the following into partial fractions. 1. 2x - 5 ____________ (x – 2) (x – 3) 2. 9 ____________ (x + 1) (x – 2) 3. x – 13 ____________ x2 - 2x - 15 4. 3x + 4 __________ 3x2 + 3x + 2 |
एकयुनेन पुर्वेन्
English Translation:
By one less than the previous.
Condition:
Ekanyunena Purvena of Vedic Mathematics is applicable whenever the multiplier completely consists of nines.
Either Multiplicand or multiplier should be entirely consisting of 9’s. But for simplicity we will call number with all 9’s as multiplier.
Ekanyunena Purvena Sutra Multiplication has below 3 cases:
I: Case 1:
Digits of multiplicand and multiplier (number with 9’s) are same.
Columns of answer:
- 1st column: 1 less of multiplicand
- 2nd column: complement of above column or (Multiplier – previous column).
II: Case 2:
Multiplier (number with 9’s) has digits less than multiplicand.
Columns of answer:
- 1st Column: If multiplicand starts with 1, subtract 2 from it.If multiplicand starts with 2, subtract 3 from it and so on.
- 2nd Column: Take compliment of last digit of multiplicand.
III: Case 3:
Multiplier (number with 9’s) has digits more than multiplicand.
Columns of answer:
- 1st Column: 1 less than Multiplicand
- 2nd column: all 9’s
- 3rd Column: 10’s compliment of multiplicand or (9’s complement of 1st columns). column: all 9’
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