भाग 2
01. जिस संख्या पर कोई चिन्ह अर्थात निशान नहीं होता। वह संख्या + या धनात्मक मानी जाती है।
01. The number without sign is known as positive number.
2. समान चिन्ह की संख्या जोड़ती हैं तथा उसी चिन्ह के साथ लिखी जाती हैं।
The same signs are added and written with their signs. As
12 + 8 = 20
+2 + 7 = 9
–12– 8 = –20
–2 –7 = –9
+12+ 5 = 17
3. विपरीत चिन्ह वाली संख्याओं को घटा आते हैं तथा बड़े चिन्ह वाली संख्या के साथ लिखते हैं।दूसरी शब्द में हम ऐसा कह सकते हैं कि जिस चिन्ह के जिस संख्या के शेष बचते हैं उसे शेष वाली संख्या के चिन्ह के साथ लिखते हैं।
The number of different signs are subtracted and written with the signs of biggest Number. As
–12 + 8 = –4
+2 – 7 = –5
+12– 8 = 4
–2 +7 = +5
+12– 3 =+9
1. Find the sum:
निम्न को जोड़ो:
(01) – 6 + 8
(02) – 13 –28
(03) – 1 + 2
(04) – 17 – 13
(05) – 7 + 25
(06) – 66 + 34
(07) 962 – 362
(08) – 966 + 566
(09) – 16 + 8
(10) – 3 – 18
(11) – 1 + 2
(12) – 27 + 13
(13) 54 – 34
(14) 262 – 362
(15) – 66 + 66
(16) 908 – 108
(17) – 111 + 11
(18) 200 – 300
(19) 373 – 245
(20) – 373 + 145
(21) – 500 + 300
(22) 54 – 85
(23) 14 – 25
(24) 524 – 534
(01) – 6 + 8
यहां पर विपरीत चिन्ह है अतः संख्या में घटेगी और जिस अंक के शेष बचेंगे उस अंक के चिन्ह के साथ लिखी जाएंगी।
यहां + 8 संख्या बड़ी है और +8 में से –6 घटाने पर 2 बचता है अतः उत्तर संख्या + 2 होगी।
–6
+8
+2
(02) – 13 –28
यहां पर समान चिन्ह है अतः संख्या में जुड़ेंगी और बड़े अंक के चिन्ह के साथ लिखी जाएंगी।
यहां – 13 और –28 को जोड़ने पर 41 आता है। जो ऋणात्मक चिन्ह अर्थात ऋण चिन्ह (–) का होगा। अतः उत्तर संख्या – 41 होगी।
– 13
– •28
– 41
2. Find the sum:
निम्न को जोड़ो:
(01) – 6 + 8 – 5
(02) – 3 – 8 + 6
(03) – 1 + 2 – 7
(04) – 17 – 13 –13
(05) – 7 + 25 – 12
(06) – 66 + 34 – 54
(07) 1262 – 362 – 400
(08) – 966 + 566 + 300
(09) – 16 + 8 + 20
(10) – 3 – 18 + 21
(11) – 1 + 2 – 8
(12) – 27 + 13 + 14
(13) – 17 – 8 + 25
(14) – 100 + 66 + 44
(15) 54 – 34 – 20
(16) 262 – 362 + 100
(17) 200 – 966 + 566
(18) 500 – 908 + 108
(19) – 111 + 11 + 99
(20) 200 – 300 + 100
(01) – 6 + 8 – 5
अगर इन सवालों को जोड़ा बनाकर करते हैं तो थोड़े आसान हो जाते हैं।
– 6 + 8
यहां पर विपरीत चिन्ह है अतः संख्या में घटेगी और जिस अंक के शेष बचेंगे उस अंक के चिन्ह के साथ लिखी जाएंगी।
यहां + 8 संख्या बड़ी है और +8 में से –6 घटाने पर 2 बचता है अतः उत्तर संख्या + 2 होगी।
–6
+8
+2
+ 2 – 5
यहां पर विपरीत चिन्ह है अतः संख्या में घटेगी और जिस अंक के शेष बचेंगे उस अंक के चिन्ह के साथ लिखी जाएंगी।
यहां –5 संख्या बड़ी है और इसमें से +2 घटाने पर –3 बचता है अतः उत्तर संख्या –3 होगी।
–5
+2
–3
3. Find the sum:
निम्न को जोड़ो:
01. – 6 + 8 – 5 – 5
02. – 3 – 8 + 6 – 5
03. – 17 – 13 –13 – 15
04. – 7 + 25 – 12 – 5
05. – 66 + 34 – 54 – 15
06. 162 – 62 – 40 – 5
07. – 96 + 56 + 30 – 15
08. – 27 + 13 + 14 – 16 + 8
09. – 100 + 66 + 44 – 8 – 25
10. 54 – 34 – 20 – 16 + 8
11. 262 – 362 + 100 – 16 + 8
12. 200 – 966 + 566 – 8 – 25
13. 500 – 908 + 108 – 16 + 8
14. – 111 + 11 + 99 – 16 + 8
15. 200 – 300 + 100 – 8 – 25
01. – 6 + 8 – 5 – 5
= – 6 + 8 – 5 – 5
= +2 –10
= –8
08. – 27 + 13 + 14 – 16 + 8
= – 27 + 13 + 14 – 16 + 8
= – 14 – 2 + 8
= –16+8
= –8
SŨNYAM SĀMYASAMUCCAYE
The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that Samuccaya is Zero.' i.e., it should be equated to zero.
The term 'Samuccaya' has several meanings under different contexts.
i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all the terms concerned and proceed as follows.
Example 1: Find the value of x by the equation 7x + 3x = 4x + 5x.
From General Method
7x + 3x = 4x + 5x
10x = 9x
10x – 9x = 0
x = 0
From the Vedic Method
The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all its terms. So that by the Sunyam Samyasamuccaye sutra it is zero, i.e., x = 0.
This Sutra is applicable not only for ‘x’ but also any such unknown quantity as follows.
Example 2: Solved for x,
5(x+1) = 3(x+1)
From General Method
5x + 5 = 3x + 3
5x – 3x = 3 – 5
2x = –2
x = –2/ 2
x = –1
From the Vedic Method
Simply think of the contextual meaning of ‘ Samuccaya ‘
Now Samuccaya is ( x + 1 )
x + 1 = 0 gives x = -1
ii) Now we interpret ‘Samuccaya ‘as product of independent terms in expressions like (x+a) (x+b)
Example 3:
( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 )
Here Samuccaya is
3 x 4 =–2 x –6
12 = 12
Since it is same , we derive x = 0
This example, we have already dealt in type ( ii ) of Paravartya in solving simple equations.
iii) We interpret ‘ Samuccaya ‘ as the sum of the denominators of two fractions having the same numerical numerator.
Consider the example.
1 1
____ + ____ = 0
3x-2 2x-1
for this we proceed by taking L.C.M.
(2x-1)+(3x–2)
____________ = 0
(3x–2)(2x–1)
5x–3
__________ = 0
(3x–2)(2x–1)
5x – 3 = 0 5x = 3
3
x = __
5
Instead of this, we can directly put the Samuccaya i.e., sum of the denominators
i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0
giving 5x = 3 x = 3 / 5
It is true and applicable for all problems of the type
m m
____ + _____ = 0
ax+b cx+d
Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 )
- ( b + d )
x = _________
( a + c )
iii) We now interpret ‘Samuccaya’ as combination or total.
If the sum of the numerators and the sum of the denominators be the same, then that sum = 0.
Consider examples of type
ax + b ax + c
_____ = ______
ax + c ax + b
In this case,
(ax+b) (ax+b) = (ax+c) (ax+c)
a2x2 + 2abx + b2 = a2x2 + 2acx + c2
2abx – 2acx = c2 – b2
x ( 2ab – 2ac ) = c2 – b2
c2–b2 (c+b)(c-b) -(c+b)
x = ______ = _________ = _____
2a(b-c) 2a(b-c) 2a
As per Samuccaya
(ax+b) + (ax+c) = 0
2ax+b+c = 0
2ax = -b-c
-(c+b)
x = ______
2a
Hence the statement.
Example 4:
3x + 4 3x + 5
______ = ______
3x + 5 3x + 4
Since N1 + N2 = 3x + 4 + 3x + 5 = 6x + 9 ,
And D1 + D2 = 3x + 4 + 3x + 5 = 6x + 9
We have N1 + N2 = D1 + D2 = 6x + 9
Hence from Sunya Samuccaya we get 6x + 9 = 0
6x = -9
-9 -3
x = __ = __
6 2
Example 5:
5x + 7 5x + 12
_____ = _______
5x +12 5x + 7
Hence N1 + N2 = 5x + 7 + 5x + 12 = 10x + 19
And D1 + D2 = 5x + 12 + 5x + 7 = 10x + 19
N1 + N2 = D1 + D2 gives 10x + 19 = 0
10x = -19
-19
x = ____
10
Consider the examples of the type, where N1 + N2 = K ( D1 + D2 ), where K is a numerical constant, then also by removing the numerical constant K, we can proceed as above.
Example 6:
2x + 3 x + 1
_____ = ______
4x + 5 2x + 3
Here N1 + N2 = 2x + 3 + x + 1 = 3x + 4
D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8
= 2 ( 3x + 4 )
Removing the numerical factor 2, we get 3x + 4 on both sides.
3x + 4 = 0 3x = -4 x = - 4 / 3.
v) ‘Samuccaya‘ with the same meaning as above, i.e., case (iv), we solve the problems leading to quadratic equations. In this context, we take the problems as follows;
If N1 + N2 = D1 + D2 and also the differences
N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the solution gives the two values for x.
Example 7:
3x + 2 2x + 5
_____ = ______
2x + 5 3x + 2
In the conventional text book method, we work as follows :
3x + 2 2x + 5
_____ = ______
2x + 5 3x + 2
( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 )
9x2 + 12x + 4 = 4x2 + 20x + 25
9x2 + 12x + 4 - 4x2 - 20x – 25 = 0
5x2 – 8x – 21 = 0
5x2 – 15x + 7x – 21 = 0
5x ( x – 3 ) + 7 ( x – 3 ) = 0
(x – 3 ) ( 5x + 7 ) = 0
x – 3 = 0 or 5x + 7 = 0
x = 3 or - 7 / 5
Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows :
Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7
D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7
Further N1 ~ D1 = ( 3x + 2 ) – ( 2x + 5 ) = x – 3
N2 ~ D2 = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 )
Hence 5x + 7 = 0 , x – 3 = 0
5x = -7 , x = 3
i.e., x = -7 / 5 , x = 3
Note that all these can be easily calculated by mere observation.
Example 8:
3x + 4 5x + 6
______ = _____
6x + 7 2x + 3
Observe that
N1 + N2 = 3x + 4 + 5x + 6 = 8x + 10
and D1 + D2 = 6x + 7 + 2x + 3 = 8x + 10
Further N1 ~ D1 = (3x + 4) – (6x + 7)
= 3x + 4 – 6x – 7
= -3x – 3 = -3 ( x + 1 )
N2 ~ D2 = (5x + 6) – (2x + 3) = 3x + 3 = 3( x + 1)
By ‘Sunyam Samuccaye’ we have
8x + 10 = 0 3( x + 1 ) = 0
8x = -10 x + 1 = 0
x = - 10 / 8 x = -1
= - 5 / 4
vi)‘Samuccaya’ with the same sense but with a different context and application .
Example 9:
1 1 1 1
____ + _____ = ____ + ____
x - 4 x – 6 x - 2 x - 8
Usually we proceed as follows.
x–6+x-4 x–8+x-2
___________ = ___________
(x–4) (x–6) (x–2) (x-8)
2x-10 2x-10
_________ = _________
x2–10x+24 x2–10x+16
( 2x – 10 ) ( x2 – 10x + 16 ) = ( 2x – 10 ) ( x2 – 10x + 24)
2x3–20x2+32x–10x2+100x–160 = 2x3–20x2+48x–10x2+100x-240
2x3 – 30x2 + 132x – 160 = 2x3 – 30x2 + 148x – 240
132x – 160 = 148x – 240
132x – 148x = 160 – 240
– 16x = - 80
x = - 80 / - 16 = 5
Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-total of the denominators on the L.H.S. and their total on the R.H.S. be the same, that total is zero.
Now D1 + D2 = x – 4 + x – 6 = 2x – 10, and
D3 + D4 = x – 2 + x – 8 = 2x – 10
By Samuccaya, 2x – 10 gives 2x = 10
10
x = __ = 5
2
Example 10:
1 1 1 1
____ + ____ = ____ + _____
x - 8 x – 9 x - 5 x – 12
D1 + D2 = x – 8 + x – 9 = 2x – 17, and
D3 + D4 = x – 5 + x –12 = 2x – 17
Now 2x – 17 = 0 gives 2x = 17
17
x = __ = 8½
2
Example 11:
1 1 1 1
____ - _____ = ____ - _____
x + 7 x + 10 x + 6 x + 9
This is not in the expected form. But a little work regarding transposition makes the above as follows.
1 1 1 1
____ + ____ = ____ + _____
x + 7 x + 9 x + 6 x + 10
Now ‘Samuccaya’ sutra applies
D1 + D2 = x + 7 + x + 9 = 2x + 16, and
D3 + D4 = x + 6 + x + 10 = 2x + 16
Solution is given by 2x + 16 = 0 i.e., 2 x = - 16.
x = - 16 / 2 = - 8.
Solve the following problems using Sunyam Samya-Samuccaye process.
1. 7 ( x + 2 ) + 3 ( x + 2 ) = 6 ( x + 2 ) + 5 ( x + 2 )
2. ( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2 )
3. ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x – 7 )
1 1
4. ______ + ____ = 0
4 x - 3 x – 2
4 4
5. _____ + _____ = 0
3x + 1 5x + 7
2x + 11 2x+5
6. ______ = _____
2x+ 5 2x+11
3x + 4 x + 1
7. ______ = _____
6x + 7 2x + 3
4x - 3 x + 4
8. ______ = _____
2x+ 3 3x - 2
1 1 1 1
9. ____ + ____ = ____ + _____
x - 2 x - 5 x - 3 x - 4
1 1 1 1
10. ____ - ____ = _____ - _____
x - 7 x - 6 x - 10 x - 9
Sunyam Samya Samuccaye in Certain Cubes:
Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3. For the solution by the traditional method we follow the steps as given below:
( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3
x3 – 12x2 + 48x – 64 + x3 – 18x2 + 108x – 216
= 2 ( x3 – 15x2 + 75x – 125 )
2x3 – 30x2 + 156x – 280 = 2x3 – 30x2 + 150x – 250
156x – 280 = 150x – 250
156x – 150x = 280 – 250
6x = 30
x = 30 / 6 = 5
But once again observe the problem in the vedic sense
We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2, we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5
Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3
The traditional method will be horrible even to think of.
But ( x – 249 ) + ( x + 247 ) = 2x – 2 = 2 ( x – 1 ). And x – 1. on R.H.S. cube, it is enough to state that x – 1 = 0 by the ‘sutra’.
x = 1 is the solution. No cubing or any other mathematical operations.
Algebraic Proof :
Consider ( x – 2a )3 + ( x – 2b )3 = 2 ( x – a – b )3 it is clear that
x – 2a + x – 2b = 2x – 2a – 2b
= 2 ( x – a – b )
Now the expression,
x3 - 6x2a + 12xa2 – 8a3 + x3 – 6x2b + 12xb2 – 8b3 =
2(x3–3x2a–3x2b+3xa2+3xb2+6axb–a3–3a2b–3ab2–b3)
= 2x3–6x2a–6x2b+6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3
cancel the common terms on both sides
12xa2+12xb2–8a3–8b3 = 6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3
6xa2 + 6xb2 – 12xab = 6a3 + 6b3 – 6a2b – 6ab2
6x ( a2 + b2 – 2ab ) = 6 [ a3 + b3 – ab ( a + b )]
x ( a – b )2 = [ ( a + b ) ( a2 + b2 –ab ) – ( a + b )ab]
= ( a + b ) ( a2 + b2 – 2ab )
= ( a + b ) ( a – b )2
since x = a + b
Solve the following using “Sunyam Samuccaye” process :
1. ( x – 3 )3 + ( x – 9 )3 = 2 ( x – 6 )3
2. ( x + 4 )3 + ( x – 10 )3 = 2 ( x – 3 )3
3. ( x + a + b – c )3 + ( x + b + c – a )3 = 2 ( x + b )3
Example :
(x + 2)3 x + 1
______ = _____
(x + 3)3 x + 4
with the text book procedures we proceed as follows
x3 + 6x2 + 12x +8 x + 1
_______________ = _____
x3 + 9x2 + 27x +27 x + 4
Now by cross multiplication,
( x + 4 ) ( x3 + 6x2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9x2 + 27x + 27 )
x4 + 6x3 + 12x2 + 8x + 4x3 + 24x2 + 48x + 32 =
x4 + 9x3 + 27x2 + 27x + x3 + 9x2 + 27x + 27
x4 + 10x3 + 36x2 + 56x + 32 = x4 + 10x3 + 36x2 + 54x + 27
56x + 32 = 54x + 27
56x – 54x = 27 – 32
2x = - 5
x = - 5 / 2
Observe that ( N1 + D1 ) with in the cubes on
L.H.S. is x + 2 + x + 3 = 2x + 5 and
N2 + D2 on the right hand side
is x + 1 + x + 4 = 2x + 5.
By vedic formula we have 2x + 5 = 0 x = - 5 / 2.
Solve the following by using vedic method :
1.
(x + 3)3 x+1
______ = ____
(x + 5)3 x+7
2.
(x - 5)3 x - 3
______ = ____
(x - 7)3 x - 9
Post a Comment
Post a Comment