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उपसूत्र ९. अन्त्ययोरेव

(Upasūtra 9. antyayoreva) - Only the last terms.
 

The Upasūtra: antyayoreva (Only the last terms) is used to solve certain equations where the numerator and the dinominator, without the independent terms, of one side are in ratio of the entire numerator and dinominator of the other side.
 
As an illustration, let us use this Upasūtra to solve:
x2 + 2x + 7   x + 2
----------- = -----
x2 + 3x + 5   x + 3
  Note, that the LHS without the independent terms:
x2 + 2x   x(x + 2)
------- = --------
x2 + 3x   x(x + 3)
And so, is in the same ratio as the entire RHS.
 
By the Upasūtra,
x + 2   7
----- = -
x + 3   5
or, 5x + 10 = 7x + 21
or, -2x = 11
or, x = -11/2
 
Thus, the solution is:
x = -11/2
So, for a practitioner of Vedic Mathematics, for something like::
2x2 + 3x + 10   2x + 3
------------- = ------
3x2 + 4x + 14   3x + 4
  Note, that the LHS without the independent terms:
2x2 + 3x   x(2x + 3)
-------- = ---------
3x2 + 4x   x(3x + 4)
And so, is in the same ratio as the entire RHS.
 
By the Upasūtra,
2x + 3   10
------ = --
3x + 4   14
or, 28x + 42 = 30x + 40
or, -2x = -2
or, x = 1
 
Thus, the solution is:
x = 1
Again, for something like:
(x+1)(x+2)(x+9) = (x+3)(x+4)(x+5)
  Before re-arranging, let us find the sum of binomials to check if the condition of this Upasūtra is met:
For LHS, (x+1)+(x+2)+(x+9) = 3x + 12
For RHS, (x+3)+(x+4)+(x+5) = 3x + 12
Hence, this Upasūtra is applicable.
 
Re-arranging, we get:
(x+1)(x+2)   (x+3)
---------- = -----
(x+4)(x+5)   (x+9)
or,
x2 + 3x + 2    x + 3
------------ = -----
x2 + 9x + 20   x + 9
Note, that the LHS without the independent terms:
x2 + 3x   x(x + 3)
------- = --------
x2 + 9x   x(x + 9)
And so, is in the same ratio as the entire RHS.
 
By the Upasūtra,
x + 3   2
----- = --
x + 9   20
or, 20x + 60 = 2x + 18
or, 18x = -42
or, x = -7/3
 
Thus, the solution is:
x = -7/3
Also, for something like:
(x+2)(x+3)(x+11) = (x+4)(x+5)(x+7)
  Before re-arranging, let us find the sum of binomials to check if the condition of this Upasūtra is met:
For LHS, (x+2)+(x+3)+(x+11) = 3x + 16
For RHS, (x+4)+(x+5)+(x+7) = 3x + 16
Hence, this Upasūtra is applicable.
 
Re-arranging, we get:
(x+2)(x+3)   (x+5)
---------- = ------
(x+4)(x+7)   (x+11)
or,
x2 + 5x + 6     x + 5
------------- = ------
x2 + 11x + 28   x + 11
Note, that the LHS without the independent terms:
x2 + 5x    x(x + 5)
-------- = ---------
x2 + 11x   x(x + 11)
And so, is in the same ratio as the entire RHS.
 
By the Upasūtra,
x + 5    6
------ = --
x + 11   28
or, 28x + 140 = 6x + 66
or, 22x = -74
or, x = -37/11
 
Thus, the solution is:
x = -37/11
But, why does it work? For this Upasūtra (antyayoreva), let us consider the following:
  Consider an equation of the form,
Ac + d   A
------ = -
Bc + e   B
Then,
B(Ac + d) = A(Bc + e)
or, ABc + Bd = ABc + Ae
or, Bd = Ae
or, A/B = d/e
 
Hence, solving this equation would provide the solution for the original equation.
 
Lastly, please remember that, as in any other form of mathematics, the mastery of Vedic Mathematics require practice and the judgement of applying the optimal method for a given scenario - a guideline of which, is presented in Applications »
 
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