28x + 42 = 30x + 40
28x – 30x = 40 – 42
-2x = -2
x = -2 / -2 = 1.Vedic Mathematics | Antyayoreva
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
Solve the following problems using ‘antyayoreva’
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