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ANTYAYOREVA
'Atyayoreva' means 'only the last terms'. This is useful in solving simple
equations of the following type. The type of equations are those whose numerator and denominator on the L.H.S. bearing the independent terms stand in the same ratio to each other as the entire numerator and the entire denominator of the R.H.S. stand to each other. Let us have a look at the following example. Example 1: x2 + 2x + 7 x + 2 __________ = _____ x2 + 3x + 5 x + 3 In the conventional method we proceed as x2 + 2x + 7 x + 2 __________ = _____ x2 + 3x + 5 x + 3 (x + 3) (x2 + 2x + 7) = (x + 2) (x2 + 3x + 5) x3 + 2x2 + 7x + 3x2 + 6x + 21 = x3 + 3x2 + 5x + 2x2 + 6x + 10 x3 + 5x2 + 13x + 21 = x3 + 5x2 + 11x + 10 Canceling like terms on both sides 13x + 21 = 11x + 10 13x – 11x = 10 – 21 2x = -11 x = -11 / 2 Now we solve the problem using anatyayoreva. x2 + 2x + 7 x + 2 __________ = _____ x2 + 3x + 5 x + 3 Consider x2 + 2x + 7 x + 2 __________ = _____ x2 + 3x + 5 x + 3 Observe that x2 + 2x x (x + 2) x + 2 ______ = ________ = _____ x2 + 3x x (x + 3) x + 3 This is according to the condition in the sutra. Hence from the sutra x + 2 7 _____ = __ x + 3 5 5x + 10 = 7x + 21 7x – 5x = -21 + 10 2x = -11 x = -11 / 2 Algebraic Proof: Consider the equation AC + D A ______ = ___ ------------- (i) BC + E B This satisfies the condition in the sutra since AC A ___ = ___ BC B Now cross–multiply the equation (i) B (AC + D) = A (BC + E) BAC + BD = ABC + AE BD = AE which gives A D __ = __ --------(ii) B E i.e., the result obtained in solving equation (i) is same as the result obtained in solving equation (ii). Example 2: solve 2x2 + 3x + 10 2x + 3 ___________ = _____ 3x2 + 4x + 14 3x + 4 Since 2x2 + 3x x (2x + 3) 2x+3 _______ = ________ = ____ 3x2 + 4x x (3x + 4) 3x+4 We can apply the sutra. 2x + 3 10 _____ = __ 3x+4 14 Cross–multiplying 28x + 42 = 30x + 40 28x – 30x = 40 – 42 -2x = -2  x = -2 / -2 = 1.Let us see the application of the sutra in another type of problem. Example 3: (x + 1) (x + 2) (x + 9) = (x + 3) (x + 4) (x + 5) Re–arranging the equation, we have (x + 1) (x + 2) x + 3 ____________ = _____ (x + 4) (x + 5) x + 9 i.e., x2 + 3x + 2x + 3 = ______________ x2 + 9x + 20x + 9 Now x2 + 3x x (x + 3) x + 3 ______ = _______ = _____ gives the solution by antyayoreva x2 + 9x x (x + 9) x + 9 Solution is obtained from x + 3 2 ____ = __ x + 9 20 20x + 60 = 2x + 18 20x – 2x = 18 – 60 18x = -42 x = -42 / 18 = -7 / 3.Once again look into the problem (x + 1) (x + 2) (x + 9) = (x + 3) (x + 4) (x + 5) Sum of the binomials on each side x + 1 + x + 2 + x + 9 = 3x + 12 x + 3 + x + 4 + x + 5 = 3x + 12 It is same. In such a case the equation can be adjusted into the form suitable for application of antyayoreva. Example 4: (x + 2) (x + 3) (x + 11) = (x + 4) (x + 5) (x + 7) Sum of the binomials on L.H.S. = 3x + 16 Sum of the binomials on R.H.S. = 3x + 16 They are same. Hence antyayoreva can be applied. Adjusting we get (x + 2) (x + 3) x + 5 2 x 3 6 ____________ = _____ = _____ = ___ (x + 4) (x + 7) x + 11 4 x 7 28 28x + 140 = 6x + 66 28x – 6x = 66 – 140 22x = -74 -74 -37 x = ___ = ___ 22 11 1. 3x2 + 5x + 8 3x + 5 __________ = ______ 5x2 + 6x +12 5x + 6 2. 4x2 + 5x + 3 4x + 5 __________ = ______ 3x2 + 2x + 4 3x + 2 3. (x + 3) (x + 4) (x + 6) = (x + 5) (x + 1) (x + 7) 4. (x + 1) (x + 6) (x + 9) = (x + 4) (x + 5) (x + 7) 5. 2x2 + 3x + 9 2x + 3 __________ = ______ 4x2 +5x+17 4x + 5 |
