(m + n)x = - mb - na Vedic Mathematics | Paravartya-yojayet
PARĀVARTYA YOJAYET
Paravartya Yojayet means 'transpose and apply'
(i) Consider the division by divisors of more than one digit, and when the divisors are slightly greater than powers of 10.
Example 1 : Divide 1225 by 12.
Step 1 : (From left to right ) write the Divisor leaving the first digit, write the other digit or digits using negative (-) sign and place them below the divisor as shown.
12
-2
――――
Step 2 : Write down the dividend to the right. Set apart the last digit for the remainder.
i.e.,, 12 122 5
- 2
Step 3 : Write the 1st digit below the horizontal line drawn under the dividend. Multiply the digit by 2, write the product below the 2nd digit and add.
i.e.,, 12 122 5
-2 -2
――――― ――――
10
Since 1 x 2 = -2 and 2 + (-2) = 0
Step 4 : We get second digits sum as 0. Multiply the second digits sum thus obtained by 2 and writes the product under 3rd digit and add.
12 122 5
- 2 -20
―――― ――――――――――
102 5
Step 5 : Continue the process to the last digit.
i.e., 12 122 5
- 2 -20 -4
――――― ――――――――――
102 1
Step 6: The sum of the last digit is the Remainder and the result to its left is Quotient.
Thus Q = 102 and R = 1
Example 2 : Divide 1697 by 14.
14 1 6 9 7
- 4 -484
―――― ―――――――
1 2 1 3
Q = 121, R = 3.
Example 3 : Divide 2598 by 123.
Note that the divisor has 3 digits. So we have to set up the last two digits of the dividend for the remainder.
1 2 3
25 98 Step ( 1 ) & Step ( 2 )
-2-3
―――――
――――――――
Now proceed the sequence of steps write 2 and 3 as follows :
1 2 3
2 5 9 8
-2-3
-4 -6
―――――
-23
――――――――――
2 1 1 5
Since
2 X (-2, -3)= -4 , -6; 5 4 = 1
and (1 X (-2,-3); 9 6 2 = 1; 8 3 = 5.
Hence Q = 21 and R = 15.
Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last 4 digits of the dividend are to be set up for Remainder.
1 1 2 1 3 2 3
9 4 7 9
-1-2-1-3 -2
-4-2-6 with 2
――――――――
-1-2-1-3 with 1
―――――――――――――
2 1 4 0 0 6
Hence Q = 21, R = 4006.
Example 5 : Divide 13456 by 1123
1 1 2 3
1 3 4 5 6
-123
-1-2-3
―――――――
-2-4 6
―――――――――――――
1 2 02 0
Note that the remainder portion contains 20, i.e.,, a negative quantity. To over come this situation, take 1 over from the quotient column, i.e.,, 1123 over to the right side, subtract the remainder portion 20 to get the actual remainder.
Thus Q = 12 1 = 11, and R = 1123 - 20 = 1103.
Now let us consider the application of paravartya yojayet in algebra.
Example 1 : Divide 6x2 + 5x + 4 by x 1
X - 1 6x2 + 5x + 4
――――――
1
6 + 11
――――――――――――
6x + 11 + 15 Thus Q = 6x+11, R=15.
Example 2 : Divide x3
3x2 + 10x 4 by x - 5
X - 5 x3 3x2 + 10x
4
―――――
5
5 + 10 100
――――――――――――――――――
x2 + 2x + 20, + 96
Thus Q= x2 + 2x + 20, R = 96.
The procedure as a mental exercise comes as follows :
i) x3 / x gives x2 i.e.,, 1 the first coefficient in the Quotient.
ii) Multiply 1 by + 5,(obtained after reversing the sign of second term in the Quotient) and add to the next coefficient in the dividend. It gives 1 X( +5) = +5, adding to the next coefficient, i.e.,, 3 + 5 = 2. This is next coefficient in Quotient.
iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to the next coefficient 10 + 10 = 20. This is next coefficient in Quotient. Thus Quotient is x2 + 2x + 20
iv) Now multiply 20 by + 5 i.e.,, 20 x 5 = 100. Add to the next (last) term,
100 + (-4) = 96, which becomes R, i.e.,, R =9.
Example 3:
x4
3x3 + 7x2 + 5x + 7
―――――――――――――――――――――
x + 4
Now thinking the method as in example ( 1 ), we proceed as follows.
x + 4 x4 - 3x3 +
7x2 + 5x + 7
―――――
-4
- 4 + 28 - 140 + 540
――――――――――――――――――――――
x3 - 7x2 + 35x - 135 547
Thus Q = x3 7x2 + 35x 135 and R = 547.
or we proceed orally as follows:
x4 / x gives 1 as first coefficient.
i) -4 X 1 = - 4 : add to next coefficient 4 + (-3) = - 7 which gives next coefficient in Q.
ii) 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q.
iii) 35 X - 4 = - 140 : then 140 + 5 = - 135, the next coefficient in Q.
iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R.
Thus Q = x3 7x2 + 35x 135 , R = 547.
Note :
1. We can follow the same procedure even the number of terms is more.
2. If any term is missing, we have to take the coefficient of the term as zero and proceed.
Now consider the divisors of second degree or more as in the following example.
Example :4 2x4 3x3 3x + 2 by x2 + 1.
Here x2 term is missing in the dividend. Hence treat it as 0 .
x2 or 0 . And the x term in divisor is also absent we treat it as 0 . x. Now
x2
+ 1 2x4 -
3x3 + 0 . x2 - 3x + 2
x2 + 0 . x + 1 0
- 2
――――――――――――
0 - 1
0 + 3
0 + 2
―――――――――――――――――――――――
2 - 3 - 2
0 4
Thus Q = 2x2 - 3x - 2 and R = 0 . x + 4 = 4.
Example 5 : 2x5 5x4 + 3x2 4x + 7
by
x3 2x2 + 3.
We treat the dividend as 2x5
5x4 + 0. x3 + 3x2 4x + 7 and divisor as x3
2x2 + 0 . x + 3 and proceed as follows :
x3 2x2 + 0 . x + 3
2x5 5x4 + 0.x3 + 3x2 4x + 7
―――――――――――――――――
2 0 - 3
4 0
- 6
-2 0 + 3
- 4
0 + 6
―――――――――――――――――――――――――――――
2 - 1 -
2 - 7
- 1 +13
Thus Q = 2x2 x 2, R = - 7 x2 x + 13.
You may observe a very close relation of the method paravartya in this aspect with regard to REMAINDER THEOREM and HORNER PROCESS of Synthetic division. And yet paravartya goes much farther and is capable of numerous applications in other directions also.
Paravartya in solving simple equations :
Recall that 'paravartya yojayet' means 'transpose and apply'. The rule relating to transposition enjoins invariable change of sign with every change of side. i.e., + becomes - and conversely ; and X becomes χ and conversely. Further it can be extended to the transposition of terms from left to right and conversely and from numerator to denominator and conversely in the concerned problems.
Type ( i ) :
Consider the problem 7x 5 = 5x + 1
7x 5x = 1 + 5
i.e.,, 2x = 6 x = 6 χ 2 = 3.
Observe that the problem is of the type ax + b = cx + d from which we get by transpose (d b), (a c) and
d - b.
x = ――――――――
a - c
In this example a = 7, b = - 5, c = 5, d = 1
Hence
1 (- 5) 1+5
6
x = _______ = ____
= __ = 3
7 5
7-5 2
Example 2: Solve for x, 3x + 4 = 2x + 6
d - b 6 - 4
2
x = _____ =
_____ = __
= 2
a - c 3 - 2
1
Type ( ii ) : Consider problems of the type (x + a) (x+b) = (x+c) (x+d). By paravartya, we get
cd - ab
x =
______________
(a + b) (c + d)
It is trivial form the following steps
(x + a) (x + b) = (x + c) (x + d)
x2 + bx + ax + ab = x2 + dx + cx + cd
bx + ax dx cx = cd ab
x( a + b c d) = cd ab
cd ab
cd - ab
x =
____________ x = _________________
a + b c d
( a + b ) (c + d.)
Example 1 : (x 3) (x 2 ) = (x + 1 ) (x + 2 ).
By paravartya
cd ab
1 (2) (-3) (-2)
x = __________
= ______________
a + b c d
- 3 2 1 2
2 - 6
- 4 1
= _______
= ___
= __
- 8
- 8 2
Example 2 : (x + 7) (x 6) = (x +3) (x 4).
Now
cd - ab
(3) (-4) (7) (-6)
x = ___________
= ________________
a + b c d
7 + (-6) 3 - (-4)
- 12 + 42
30
= ____________ =
___ = 15
7 6 3 + 4
2
Note that if cd - ab = 0 i.e.,, cd = ab, i.e.,, if the product of the absolute terms be the same on both sides, the numerator becomes zero giving x = 0.
For the problem (x + 4) (x + 3) = (x 2 ) ( x 6 )
Solution is x = 0 since 4 X 3 = - 2 X - 6. = 12
Type ( iii) :
Consider the problems of the type ax + b
m
______ = __
cx + d n
By cross multiplication,
n ( ax + b) = m (cx + d)
nax + nb = mcx + md
nax - mcx = md nb
x( na mc ) = md nb
md - nb
x = ________
na - mc.
Now look at the problem once again
ax + b m
_____ = __
cx + d n
paravartya gives md - nb, na - mc
and
md - nb
x = _______
na - mc
Example 1: 3x + 1
13
_______ = ___
4x + 3 19
md - nb 13 (3) - 19(1)
39 - 19 20
x = ______
= ____________
= _______ = __
na - mc 19 (3) - 13(4)
57 - 52 5
= 4
Example 2: 4x + 5
7
________ = __
3x + 13/2 8
(7) (13/2) - (8)(5)
x = _______________
(8) (4) - (7)(3)
(91/2) - 40 (91 - 80)/2
11
1
= __________
= _________
= ______ = __
32 21
32 21
2 X 11 2
Type (iv) :
Consider the problems of the type m n
_____ + ____ = 0
x + a x + b
Take L.C.M and proceed.
m(x+b) + n (x+a)
______________ = 0
(x + a) (x +b)
mx + mb + nx + na
________________ = 0
(x + a)(x + b)
(m + n)x + mb + na = 0
(m + n)x = - mb - na
-mb - na
x = ________
(m + n)
Thus the problem m
n
____ + ____ = 0,
by paravartya process
x + a x + b
gives directly
-mb - na
x = ________
(m + n)
Example 1 : 3
4
____ + ____ = 0
x + 4 x 6
gives
-mb - na
x = ________ Note that m = 3, n = 4, a = 4, b = - 6
(m + n)
-(3)(-6) (4) (4) 18 - 16
2
= _______________ =
______ = __
( 3 + 4)
7 7
Example 2 :
5
6
____ + _____ = 0
x + 1 x 21
gives
-(5) (-21) - (6) (1)
105 - 6 99
x = ________________ = ______ = __ =
9
5 + 6
11 11
II)
1. Show that for the type of equations
m
n
p
____ + ____ +
____ = 0, the solution is
x + a x + b
x + c
- mbc nca pab
x = ________________________ , if m + n + p =0.
m(b + c) + n(c+a) + p(a + b)
2. Apply the above formula to set the solution for the problem
Problem 3
2
5
____ + ____ -
____ = 0
x + 4 x + 6
x + 5
some more simple solutions :
m n
m + n
____ + ____ =
_____
x + a
x + b x + c
Now this can be written as,
m n
m
n
____ + ____ =
_____ + _____
x + a x + b x + c
x + c
m
m
n n
____ - ____
= _____ - _____
x + a x + c
x + c x + b
m(x +c) m(x + a)
n(x + b) n(x + c)
________________ = ________________
(x + a) (x + c)
(x + c) (x + b)
mx + mc mx ma
nx + nb nx nc
________________ = _______________
(x + a) (x + c)
(x +c ) (x + b)
m (c a)
n (b c)
____________ =
___________
x +a
x + b
m (c - a).x + m (c - a).b
= n (b - c). x + n(b - c).a
x [ m(c - a)
- n(b - c) ]
= na(b - c) mb (c - a)
or x [ m(c - a) + n(c - b) ]
= na(b - c) + mb (a - c)
Thus
mb(a - c) + na (b - c)
x = ___________________
m(c-a) + n(c-b).
By paravartya rule we can easily remember the formula.
Example 1 : solve
3 4
7
____ + _____ = ____
x + 1 x + 2
x + 3
In the usual procedure, we proceed as follows.
3 4
7
____ + ____ = ____
x + 1 x + 2
x + 3
3(x + 2) + 4(x + 1)
7
________________
= _____
(x + 1) (x + 2)
x + 3
3x + 6 + 4x + 4
7
_____________ = ____
x2 + 2x + x + 2
x + 3
7x + 10
7
_________ = ____
x2 + 3x + 2
x + 3
(7x + 10) (x + 3) = 7(x2 + 3x + 2)
7x2 + 21x + 10x + 30 = 7x2 + 21x + 14.
31x + 30 = 21x + 14
31 x 21 x = 14 30
i.e.,, 10x = - 16
x = - 16 / 10 = - 8 / 5
Now by Paravartya process
3 4
7
____ + ____ =
____ ( ... N1 + N2 = 3+4 = 7 = N3)
x + 1 x + 2
x + 3
mb( a c ) + na ( b c )
x
= _____________________ here N1 = m = 3 ,
N2 = n = 4 ;
m ( c a ) + n ( c b ) a = 1, b = 2, c = 3
3 . 2 ( 1 3 ) + 4 . 1 . ( 2 3)
= __________________________
3 ( 3 1 ) + 4 ( 3 2 )
6 ( -2)+ 4 (-1) - 12 4
- 16 - 8
= _____________ =
_______ = ____ = ___
3 (2) + 4(1)
6 + 4 10
5
Example 2 :
3 5
8
____ + ____ = _____ Here N1 + N2 = 3 + 5 = 8.
x - 2 x 6
x + 3
mb ( a c ) + na ( b c)
x = _____________________
m ( c a ) + n ( c b )
3 . ( -6 ) ( - 2 - 3 ) + 5 .( -2 ) ( -6 3 )
=
__________________________________
3 ( 3 ( -2 ) ) + 5 ( 3 ( - 6 ) )
3 ( - 6 ) ( - 5 ) + 5 ( - 2 ) ( - 9 )
=
____________________________
3( 3 + 2 ) + 5 ( 3 + 6 )
90 + 90
= _______ = 180 / 60
= 3.
15 + 45
Solve the problems using the methods explained above.
1)
2
3
5
____ + ____ = ____
x + 2 x + 3
x + 5
2) 4
6
10
____ + ____ =
____
x + 1 x + 3
x + 4
3) 5
2 3
____ + ___ = ____
x - 2 3 - x
x 4
4) 4
9 15
_____ + _____ = _____
2x + 1
3x + 2 3x + 1
Note : The problem ( 4 ) appears to be not in the model said above.
But 3 (4)
2 (9)
2(15)
________ + ________
= _______ gives
3(2x + 1) 2( 3x + 2)
2(3x + 1)
12 18
30
_____ + _____ =
_____ Now proceed.
6x + 3 6x + 4
6x + 2
Simultaneous simple equations:
By applying Paravartya sutra we can derive the values of x and y which are given by two simultaneous equations. The values of x and y are given by ration form. The method to find out the numerator and denominator of the ratio is given below.
Example 1: 2x + 3y = 13, 4x + 5y = 23.
i) To get x, start with y coefficients and the independent terms and cross-multiply forward, i.e.,, right ward. Start from the upper row and multiply across by the lower one, and conversely, the connecting link between the two cross-products being a minus. This becomes numerator.
i.e.,, 2x + 3y = 13
4x + 5y = 23
Numerator of the x value is 3 x 23 5 x 13 = 69 65 = 4
ii) Go from the upper row across to the lower one, i.e.,, the x- coefficient but backward, i.e.,, leftward.
Denominator of the x value is 3 x 4 2 x 5 = 12 10 = 2
Hence value of x = 4 χ 2 = 2.
iii) To get y, follow the cyclic system, i.e.,, start with the independent term on the upper row towards the xcoefficient on the lower row. So numerator of the yvalue is
13 x 4 23 x 2 = 52 46 = 6.
iv) The denominator is the same as obtained in Step(ii) i.e.,, 2. Hence value of y is 6χ2=3.
Thus the solution to the given equation is x = 2 and y = 3.
Example 2: 5x 3y = 11
6x 5y = 09
Now Nr. of x is (-3) (9) (5) (11) = - 27 + 55 = 28
Dr. of x is (-3) (6) (5) (-5) = - 18 + 25 = 07
x = Nr χ Dr = 28 χ 7 = 4
and for y,
Nr is (11) (6) (9)(5) = 66 45 = 21
Dr is 7
Hence y = 21 χ 7 = 3.
Example 3: solve 3x + y = 5
4x y = 9
Now we can straight away write the values as follows:
(1)(9) (-1)(5) 9 + 5
14
x
= _____________ = _____ = ___
= 2
(1)(4) (3)(-1) 4 + 3
7
(5)(4) (9)(3) 20 27
-7
y =
____________ = _______ =
___ = -1
(1)(4) (3)(-1) 4 + 3
7
Hence x = 2 and y = -1 is the solution.
Algebraic Proof:
ax + by = m
( i )
cx + dy = n
. ( ii )
Multiply ( i ) by d and ( ii ) by b, then subtract
adx + bdy = m.d
cbx + dby = n.b
____________________
( ad cb ) .x = md nb
md - nb bn - md
x = ______ = ______
ad - cb bc - ad
Multiply ( i ) by c and ( ii ) by a, then subtract
acx + bcy = m.c
cax + day = n.a
_____________________
( bc ad ) . y = mc - na
mc - na
y = ______
bc - ad
You feel comfort in the Paravartya process because it avoids the confusion in multiplication, change of sign and such other processes.
Find the values of x and y in each of the following problems using Paravartya process.
1.
2x + y = 5
2. 3x 4y = 7
3x 4y = 2
5x + y = 4
3.
4x + 3y = 8
4. x + 3y = 7
6x - y = 1
2x + 5y = 11