PRACTICE AND PROPORTION
EXTENSION OF MERGER METHOD
{Multiple Merger)
We now take up and deal with equations wherein N1-(-N2+
N8(of the L.H.S.)=N of the R.H.S. and wherein the same
‘Paravartya’ (Merger) formula can be applied in exactly the same way as before. Thus:
(!) _ L | 3 I 5 - 9 x + 2 x + 3 T x + 5 x + 4
TEST: l + 3 + 5 = 9 YES.
—2 , —3 , 5 _ 0 . YES, again.
x + 2 x + 3 x + 5 i.e. 2 , 3 5__
x + 2 x + 3 x + 5 + _ = L = o
x + 2 x + 3
(i) By the Basic Formula ^ x=— na m + n __ 18+12 30
—6—6 - 1 2
or (ii) By 4Sunyam Samuccaye5 formula : (x + 2 )+ (x + 3 ) = 0 x = —
Note :—These two steps (of successive merging) can be combined into one by multiplying Nx first by (2—4) and then by (2—5) i.e. by 6 and similarly N2 first by (3—4) and then by (3—5) i.e. by 2 and proceeding as before
.-. _ J L + _ J L _ = 0 x + 2 x + 3
By either method (Basic or Sunyam), x = — The Algebraic Proof hereof is this :
m n , p _ m + n + p x + a x + b x + c x + d . m(a—d)_^n(b—d)_^p(c—d)__^
x + a x + b x + c
m(a—d) (a—c )^ n(b—d) (b—c ) ^ 0
x + a x + b
( 132 )
which is the exact shape of the formula required for the single- step merger, (vide swpra).
Similarly, the merger-formula can be extended to any number of terms as follows:
m + n 1 p + q | r + . . . x + a x + b x + c x + d x + e _ m + n + p + q + r + ...
x + w
• m(a—w) (---) (a~ e ) (a—d) (a—c) x + a
, n(b—w) (---) (b—e) (b—d) (b—c)
T ' x + b
(which is the general formula for the purpose). Thus, in the above example—
- _ ( —3)( 2)( 3) + ( — 3 ) ( — 2 )(—2) _ - 1 8 - 1 2 , - 3 0 _
( — 3 ) ( — 2 ) ( + l ) + 3 ( — 1 ) ( — 2) 6 + 6 1 2
A few more illustrations of this type are given below : (!) 3 , 4 , 48 _ 48
3 x + l 4 x + l 8 x + l 6 x + l
YES Here § + £ + ^ = 8 ; and ^ = 8 / . Y E S
24 , 24 , 144 _ 192 2 4 x + 8 2 4x+ 6 24x+ 3 2 4x+ 4
1 , 1 , 6 8 YES
24x+ 8 24x+ 6 24x+ 3 24x+ 4
4 1 2 - 6 .-. YES
24x+ 8 24x+ 6 24x+ 3 _ * _ + _ « _ = <>
24x+ 8 24x+ 6
. . 624X+168 = 0 x = — 26 (2) 2 . 18 . 75 __ _88 2x + l 3 x + l 5 x + l 4 x + l
Here i + V + V = 2 2 ; and 2» is also 22
• 60 , 360 , 900 _ 1320 . YES 6 0 X + 3 0 6 0 X + 2 0 6 0 x + 1 2 6 0 x + 1 5
• _ J __ -i. 6 .1. 15 - 22 YES 6 0 x + 3 0 6 0 x + 2 0 6 0 x + 1 2 6 0 x + 1 5
( 138 )
will cause no change in the working or the result.(3) 4 , 27 , 125 _ 144
(By Basic rule or by cross-multiplication or by (‘Sunyam, Formula), 50x—20 = 0 x = |
OR (by Multiple simultaneous merger)
6 0 x = (' -18) (—15)4~(—270) (—8) (—5) ( - 2 ) ( - 1 8 ) (—5 )+ (9 ) ( - 8 ) ( - 5 ) : 24
• •x = f
Note :—Again any change of SEQUENCE (of the terms on the L.H.S.) will cause no change in the working or the result.
Ch a p t e r X IV COMPLEX MERGERS
There is still another type—a special and complex type of equations which are usually dubbed ‘harder’ but which can be readily tackled with the aid of the Paravartya Sutra. For instance:
10 , 3 _ 2 . 15 2 x + l 3x—2 2x—3 3 x + 2 Note the TESTS : (1) 3 J > - g = § + » * ; and
(2) 1 0 x 3 = 2 x 1 5
i.e. 10 : 15 : : 2 : 3 (or 10 : 2 :: 15 : 3)
Transposing, 10 15 _ 2 3
2 x + I ~ 3 x + 2 ~ 2x—3 ~ 3 x —2 ’ and taking the L.C.M.
• 30 _ 30 __ 6 _ 6 _ 6x-(-3 6x-j-4 6x—9 6x—4
Simple CROSS-MULTIPLICATION leads us to the main TEST:
30 _ 30
' '(6 x + 3 ) (6x+ 4) (6x—9) (6x—4)
Here comes the third TEST i.e. that the numerator (of the final derived equation) is the same on both sides—
(6x-j-3) (t a + 4 ) = (6 x - 9 ) (6x—4) , , 6 r = - 3? - 12 = ^ 6 .
3 + 4 + 9 + 4 20 5 1
CLUE— This gives us the necessary clue, namely, that, after putting up the L.C.M. coefficient for x in all the denominators. (Dj) (D2) = ( D 3) (D4). As the trans
position, the L.C.M. etc., can be done mentally, this clue amounts to a solution of the equation at sight.
In these examples, we should transpose the 4 fractions in such a manner that, after the cross-multiplication etc., are over, all the four denominators (of the final derived equation)
( 138 )
have the same (L.C.M.) coefficient for x and the numerator is the same on the L.H.S. and the R.H.S (of the same equation).
A few more illustrations will be found helpful:
(1) 2 , 2 _ 9 , 1 0X+1 2 x - l 9 x - 5 3 x + l
(i) Transposing etc., we have :
6 6 18 18
18x+3 18x+ 6 18x—10 18x—9
Here the N on both sides (of the final derived equation) is 18 .\ The Sutra applies.
,\(18x+3) (18x+6) = (1 8 x -1 0 ) (1 8 x -9 ) , . 1 8 x = _ ^ = 1 8 _ = 7_2 , ' X==± = 1 3 + 6 + 1 0 + 9 28 28 7
N ote:—In some cases (details of which we need not now enter into but which will be dealt with later), the original fractions themselves (after the transposition) fulfil the conditions of the Test. In such cases, we need not bother about the L.C.M. etc., but may straightaway transpose the terms and apply the ‘Paravartya formula.
In fact, the case just now dealt with is of this type, as will be evident from the following :
(ii) 2 _ 1 ^ 9 _ 2 6 x + l 3 x + l 9 x — 5 2 x — 1
Here | = | ; | = f ; and the numerator (on both sides of the final derived equation) is 1.
/.T h e Sutra applies and can be applied immediately (without bothering about the L.C.M. etc.).
.\ (6 x + l) (3 x + l) = (9x—5) (2x—1)
18x2+ 9 x + l = 1 8 x 2—19x+5 .\ 28x = 4 .*.x = ^ (2) _ 2 , __3 _ 1 . __6
2 x + 3 3 x + 2 x + l ^ x - f - V 2 1 __ 6 3
” 2 x + 3 x + 1 6 x + 7 3 x + 2
(i) By L.C.M. method, (6x+ 9) (6 x + 6 )= (6 x + 7 ) (6x+4)
( 130 )
(ii) In this case, there is another peculiarity i.e. that the transposition may be done in the other way too and yet the conditions are satisfied. So, we have :
(6x+ 9) (6x+ 7) = (6x+ 6) (6x+4) - « - = %
-(iii) And even, by CROSS-multiplication at the very outset, we get 12x+13 = 0 (by Samya Samuccaye). x = -^-§
In such cases, SEQUENCE (in transposition) does not matter ! (This will be explained later).
(3) 51 _ 68 = 52 _ 39 3 x + 5 4 x + l l 4x—15 3x—7
TESTS : ~ and are both 17 ; and ~ and ^ are both 13.
This equation can be solved in several ways (all of them very simple and easy):
(i) By the L.C.M. process:
204 204 156 156
12x+20 12x+33 12x—45 12x—28 In the derived equation (in its final form),
1^ = 204X13 = 12X13X17 ; and N2 = 156x17 = 1 2 x 1 3 x 1 7
The Sutra applies.
.-. (12X+20) (1 2 x + 3 3 )= (1 2 x -4 5 ) (1 2 x -2 8 )
• ! oT — 28 X 45 ~ 20 X 33 - 600 • x = ~ 2 0 + 3 3 + 4 5 + 2 8 126 ' ’ 63 (ii) or, removing the common factor (12) :
17 17 13 13
1 2 X + 2 0 1 2 x + 3 3 1 2 x — 4 5 1 2 x — 2 8
In the (final) derived Equation,
Nx = 1 7X l3 ; and N2==1 3 x l7 The Sutra applies--
. ■ • D j X D j ^ D j X D j .-. 1 2 x = ™ > , . x = | |
(iii) or, at the very outset (i.e. without L.C.M. etc.) :
5 1 _ 6 8 _ 5 2 _ 3 9 3 x + 5 4 x + l l 4 x — 1 5 3 x — 7
.-.L.H.S. N = 5 6 1 — 3 4 0 = 2 2 1 ; and R.H.S. N = — 3 6 4 + 5 8 5 = 2 2 1
( 137 )
The Sutra applies straightaway.
.-. (3x+ 5) (4x4-11) = (4x—15) (3x—7) .-. 12x24-53x4-55 = 12x2—73x4-105
126x = 50 .’ . x = g f N ote:— In the second method, note that
N1 = N 2 = D1- D 3 an dN 3= N 4 = D a- D 1 TESTS The General Formula applicable in such cases i s :
m —n^ p —q ^ m —n_^ p —q x4-p x4-n x + q x -fm
••• (m" n) ( i T p - i T q ) := (p _ q ) U i - V - U
• (in—n) (q —p) = (p—q) (n—m) (x4-p) (x4-q) (x4-m) (x4-n) As the numerators are the same,
••• The Sutra applies
••• (x4-p) (x4-q)=(x+m ) (x4-n)
• m n -p q p - f q —m—n ( 4 ) __ 1 . 8 _ 6 , 3
2x—1 4x—1 3x—1 6x—1
(i) . __ 8 _ 6 _ 24 _ 24 U ' ' 12x—6 12x—2 12x—4 12x—3 .-.In the final derived equation,
L.H.S. N = 2 4 ; and R.H.S. N is also 24 The Sutra applies.
I 2 x = 1 2 ^ ? = 0 x = 0
(ii) ‘ Vilokand (i.e. mere observation) too will suffice in this case.
(«) 3 2 - 3 2
3x4-1 2x—1 3x—2 2x4-1
(i) Here the resultant N is the same (1) (on both sides) .-.Y E S
6x24-5x4-1 = 6x2—7x4-2 ••• 12x = l .-. x = T1, (ii) or, by cross-multiplication at the very outset and
Sunyam Samuccaye, 12x—1 = 0 x = T1s is
( 138 ) (6) 5 . 3 __ 5 , 15 6 x + 2 3 x + l 5 x + 3 15x+2
15 15 15 15
(i) 15x+6 15x+9 15x+2 15X+5
The resultant Numerator on both sides is 45 The Sutra applies.
8 8 30
(ii) Or, by cross-multiplication at the very outset and Sunyatfi etc., formula, we get 3 0 x + ll and 150x+55 on the L.H.S. and the R.H.S. respectively; and the numerical factor (5) being removed, both give us 3 0 x + ll = 0 .\ x = ~JJ
(7) 2 x + l l . 6 x + l l __4 x + 4 , 3x+19 x + 5 2 x + 3 2 x + l x + 0 (i) .• (By Paravartya division):
1 . 2 _ 2 , 1 x + 5 2 x + 3 2 x + l x + 6
2 2 2 2
2x+ 1 0 2 x+ 1 2 2 x + l 2 x + 3
.*.4 is the N on both sides (of the derived equation) .•. The Sutra applies.
(2x+10) (2x+12) = (2 x + l) (2x+ 3)
18 18 4
(ii) or by cross-multiplication at the very outset and Sunyam Sutra, we h a ve:
4x+13 = 0 .-. x = ~i|
(8) 2 x + l l . 15x—4 7 _9x—9 , 4x+ 13 x + 5 3x—10 3x—4 x + 3
3 , 3 3 , 3
3x+ 15 3x—10 3x—4 3 x + 9
3 3 3 3
’ 3x+ 15 3 x + 9 3x—4 3x—10 In the resultant equation,
—18 is the numerator on both sides The Sutra applies.
( 139 )
3x = ^9.— x = = i
38 2 6
(ii) or by cross-multiplication at the very outset and Sunyam formula,
1 8 x + 1 5 = 0 x =
(9) 12x2+ 9 x + 7 , 12x2+ x + 3 = 24x2+ 1 4 x + 3 , 5x2+ 6 x + 2
3 x + 4 4x—1 1 2 x + l x + 1
.•.By (Pardvartya) devisioD twice over.
3 ,___4 ^ 12 , 1 3 x + 4 4x—1 1 2 x + l x + 1
12 , 12 12 , 12
(3x+15) (3x+ 9) = ( 3 x - 4 ) (3 x -1 0 )
1 2 x + 1 6 1 2x — 3 1 2 x + l 1 2 x + 1 2
By ‘Sunyam’ Sutra, we immediately obtain : 24x+13 = 0 x = i f _IQ
24
Note :—The Cross-multiplication and 4Sunyam5 method is so simple, easy and straight before us here that there is no need to try any other process at all. The student may, however, for the sake of practice try the other methods also and get further verification therefrom for the correctness of the answer just hereinabove arrived at.
C h a p t e r X V
SIMULTANEOUS SIMPLE EQUATIONS
Here too, we have the GENERAL FORMULA applicable to all cases (under the ‘Pardvartya5 Sutra) and also the special Sutras applicable only to special types of cases.
THE GENERAL FORMULA
The current system may congratulate and felicitate itself on having a fairly satisfactory method—known as the Cross
multiplication method—for the solving of simultaneous simple equations, which is somewhat akin to th.6 Vedic ‘Pardvartya method and comes very near thereto.
But even here, the unfortunate drawback still remains that, in spite of all the arrow-directions etc., intended to facilitate its use, the students (and sometimes even the teachers) of Mathematics often get confused as regards the plus and the minus signs ( + and —) and how exactly they should be used ; and, consequently, we find most of them preferring—in actual daily practice—the substitution method or the elimination method (by which they frame new equations involving only x or only y). And this, of course, does not permit a one-line mental-method answer ; and it entails the expenditure of more time and more toil.
The Vedic method (by the Pardvartya Rule) enables us to give the answer immediately (by mere mental Arithmetic).
Thus— 2 x + 3 y = 8?
4 x + 5 y = 14 )
The rule followed is the “ Cyclic” o n e :
(i) For the value of x, we start with the y-coefficients and the independent terms and cross-multiply forward (i.e.
rightward) (i.e. we start from the upper row and multiply
( 141 )
across by the lower one; and conversely; and the connecting link between the two cross-prodcts is always a minus). And this gives us our Numerator;
(ii) For finding the Denominator, we go from the upper row across to the lower one (i.e. the x coefficient) but backward (i.e. leftward). Thus,
2 x + 3 y = 8 / for the value of x, the numerator is 3 x 1 4 — 4x-f-5y= 14 ) 5 x 8 = 2 ; and the Denominator is 3 x 4 —
In other words x = | = l .
And, as for the value of y, we follow the cyclic system (i.e.
start with the independent term on the upper row towards the x coefficient on the lower row). So, our Numerator is :
And NOTE that the Denominator is invariably the SAME as before (for x) and thus we avoid the confusion caused in the current system by another set of multiplications, a change of sign etc. In other words, V ;= f = 2
2 X 5 = 2
8 x 4 - 1 4 X 2 = 3 2 - 2 8 = 4
(2) x — y = 7 5 x + 2 y = 42
- 4 2 - 1 4 _ - 5 6
— 5 — 2 — '7 3 5 — 4 2 = — 7 _ 1 and y =
- 7 - 7 (3) 2 x + y = 5
3 x — 4y = 2
(4) 5 x — 3y = 11 6 x — 5y = 9
(5) ll x -f -6 y = 28 7 x — 4y = 10
( 142 ) A SPECIAL TYPE
There is a special type of simultaneous simple equations which may involve big numbers and may therefore seem “ hard”
but which, owing to a certain ratio between the coefficients, can be readily i.e. mentally solved with the aid of the Sutra
SRr spirt (Sunyam Anyat) (which cryptically says : If one is in ratio, the other one is Zero).
An example will make the meaning and the application clear :
6 x + 7y = 81 1 9 x + 1 4 v = 16 J
Here we note that the v-coefficients are in the same ratio to each other as the independent terms are to each other.
And the Sutra says that, in such a case, the other one, namely, x = 0. This gives us two simple equations in y, which give us the same value f for y. Thus x = 0 ; y = f
N.B. :—Look for the ratio of the coefficients of one of the un
known quantities being the same as that of the inde
pendent terms (on the R.H.S.) ; and if the four are in proportion, put the other unknown quantity down as zero ; and equate the first unknown quantity to the absolute term on the right.
The Algebraical Proof is this : a x + b y = bm ?
c x + d y = dm >
adx-fbdy = bdm \ bcx-J-bdy = bdm J
x(ad—bc) = 0 x = 0 1 and y - m )
A few more illustrations may be taken :
(1) 12x+ 8 y = 7 ? Here, v 8 : 16 :: 7 : 14 (mentally) 16x+16y = 14> x = 0 l
and y = | )