Your Ultimate Guide to the Sankalana Vyavakanabhyam Sutra

The Sankalana Vyavakanabhyam is a popular Vedic Maths sutra used to solve linear equations in 2 variables. Literally meaning ‘addition and subtraction’, this 7th Vedic Maths sutra will help enthusiastic learners to breeze through linear equations. This article is dedicated to this wonderful method which will help you speed up your calculation and give accurate results. Check out other Vedic Maths lessons on the Podium Blog.

About Sankalana Vyavakanabhyam

Only specific types of problems are solved with this method. For generic lengthy multiplication, look up the Nikhilam sutra and the Anurupyena sutra on the Podium blog.

Suppose we have a linear equation as:

ax+by= c1 (1)

bx+ay=c2  (2)

The linear equation should be such that, the coefficients are interchanged i.e. where x is a coefficient of a and y is a coefficient of b in equation (1), it is vice-versa in equation (2).

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Let us try with some examples

EXAMPLE 1

5x – 4y = 34

4x – 5y =11

Addition –

(5x-4y) + (4x-5y) = 34 +11

5x – 4y + 4y- 5y = 45

9x + 9y = 45

x + y = 45…..(i)

Subtraction –

(5x-4y) – (4x-5y) = 23

5x – 4y – 4x + 5y= 23

x + y = 23…..(ii)

x + y +x – y = 28 /  x – y – (x-y) =-18

x = 14                            -2y = -18

                                       y = 18

Therefore, x = 14 and y = 18 and we have our equation solved!

EXAMPLE 2

45x + 23y = 113

23x + 45y = 91

Addition –

(45x + 23y) + (23x + 45y) = 204

68x + 68y = 204

x + y = 204/68

x + y = 3…….(i)

Subtraction –

45x + 23y – (23x + 45y) = 22

45x + 23y  – 23x – 45y = 22

22x – 22y = 22

x – y = 1……(ii)

Taking (i) and (ii) together, we will repeat the addition and subtraction procedures till we get the desired values of x and y, respectively.

x + y + x – y = 4 (ADDITION)

Which implies to –

2x = 4

Therefore, x = 2

x + y – (x – y) = 2 (SUBTRACTION)

x + y – x + y =2

2y = 2

y = 1

And here’s our answer!

More Practice

This Sutra is all about practice in solving linear equation in two variables. The more we practice it, the easier it gets and the more it allows and helps us to do quick math.

EXAMPLE 3

9x – 7y = 40

7x – 9y = 24

Addition –

9x  – 7y + 7x – 9y  = 64

16x – 16y = 64

x + y = 4…..(i)

Subtraction –

9x – 7y – (7x – 9y) = 16

9x – 7y – 7x + 9y = 16

2x – 2y = 16

(Here we will divide both sides by 2 since it is the common co-coefficient on LHS and RHS)

Which leaves us with x– y =8….(ii)

Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.

x + y + x – y = 4 (ADDITION)

Which gives us-

2x = 4

Therefore, x = 2

x + y – (x – y) = 8 (SUBTRACTION)

x + y – x + y = 8

2y = 8

y = 4

And here’s our answer!

Vedic Mathematics ppt
The meaning of Sankalana Vyavakanabhyam

Practicing Sankalana Vyavakanabhyam with More Problems

EXAMPLE 4

69y – 5z = 42

5y – 69z = 32

Addition –

69y – 5z + 5y – 69z = 74

74y – 74z = 74

x – y = 1…..(i)

Subtraction –

69y – 5z – (5y – 69z) = 74

69y – 5z – 5y + 69z = 74

y + z = 5/28……(ii)

Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.

y = 1 + z = 5/28 – z
1 + z = 5/28 – z
1 + 2z = 5/28
2z = -23/28
z = (-23/28)/2
z = -23/56 Therefore, y = 5/28 – -23/56
y = 10/56 + 23/56
y = 33/56

And here’s our answer!

EXAMPLE 5

1955x + 476y = 2482

 476x + 1955y = – 4913

Addition –

1955x + 476y + 476x + 1955y =  -2431

2431 x + 2431y = -2431

x + y = -1….(i)

Subtraction –

1955x + 476y – (476x +1955y) = -2431

1955x + 476y – 476x – 1955y = -2431

1479x + 1479y = 7395

x + y =5

Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.

x + y + x – y = 6 (ADDITION)

2x = 6

which gives us, x = 3

x + y – (x – y) = 4 (SUBTRACTION)

x + y – x + y = 4

2y = 4

y = 2

And here’s our answer!

Our Last Practice Problem!

EXAMPLE 6

3x + 2y = 18

2x  + 3y = 17

Addition –

3x + 2y + 2x + 3y = 35

5x + 5y = 35

x + y = 7…..(i)

Subtraction –

3x + 2y – (2x + 3y) = 1

3x + 2y – 2x + 3y= 1

x – y = 1……(ii)

Taking (i) and (ii) together, we will again repeat the addition and subtraction procedures till we get the desired values of x and y respectively.

x + y + x – y = 6 (ADDITION)

2x = 6

x = 3

x + y – (x – y) =  (SUBTRACTION)

2y = 4

y = 2

And here’s our answer!

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