**Simultaneous Linear Equations**

Simultaneous linear equation is nothing but two linear equation where **two variables **are taken together. Let’s say x & y are the two variables of simultaneous equations.

As we have two variable means we need two equations to solve the sum with same variable value.

We already knew the general method of solving such sums here we will apply **Vedic sutra Paravartya** to solve Simultaneous linear equations.

In Vedic mathematics the value of variable x and y will be expressed in the form of Numerator upon Denominator as shown below.

Where the formula for Numerator of both the variables is same but the **formula for finding denominator of x is different than denominator of y.**

Once we get the value of x no need to find y by formula, we will get it easily by putting the value of x in any given linear equation or vice versa.

First, we will solve the sum by applying formula of x later we will solve by second method means find y by formula.

**Method I: By calculating x**

**Ex. i) 6x + y = 4 & 5x + 2y = 1**

**Step 1: Apply formula,**

We know the formula to find variable x,

**Step 2: Find the Numerator**

Here the **Numerator** is obtained by cross multiplying (1 x 1) and from it subtracting the cross product of (2 x 4) as shown by the arrow of the diagram.

Numerator = (1 x 1) –(2 x 4) = 1-8 = -7

**Step 3: Find the denominator**

Now **Denominator** is obtained by cross multiplying (1 x 5) & subtract from it the cross product (6 x 2) as shown**–**

Denominator = (5×1) – (6×2) = 5 – 12 = -7

**Step 4: Put the values in formula**

Put the value of Nr. & Dr. in the formula of step 1

X = -7/-7 = 1

**Step 5: Find y by using value of x**

Now put the value of x in given equ. to find y,

So, 6(1) + y = 4

Y = 4 – 6 = -2

**Ans, x = 1 & y = -2**

**Ex. ii) 4x + 3y = 25 & 2x + 6y = 26**

**As we know the formula,**

** **

**To find numerator,**

= (3 x 26) – (6 x 25) = 78 – 150 = -72

** ****To find denominator**

Dr. = (3 x 2 ) – (4 x 6) = 6 – 24 = – 18

x = Nr./ Dr. = -72/-18 = 4

Now put x = 4 in given equ.

4 x 4 + 3y = 25

3y = 25-16 = 9

So, y = 3

**Ans, x = 4 & y = 3**

**Ex. iii) 2x – 3y = 1 & 3x – 4y = 1**

** ****As we know , **

** Numerator,**

Nr. =(-3 x 1) – (-4 x 1) = -3 + 4 = 1

** Denominator,**

Dr. = (-3 x3)-(2x-4) = -9 +8 = -1

x = 1/-1 = -1

put the value of x in given equ.

2(-1) -3y = 1

-3y = 1+ 2 =3

so, y = -1

**Ans, x = -1 & y = -1**

** **

**Method II: By calculating y:**

The value of y will also be calculated by the same formula of finding value x i.e. numerator upon denominator. **Here the formula for calculating the denominator is same as x but for numerator we will study new formula.**

**Ex. iv) 7x + 2y =19 & 4x + 3y = 22**

**As we know the formula,**

** **

**Numerator,**

The numerator will obtain by cross multiplying (7 x 22) & subtracting from it the cross product (4 x 19) as shown in below diagram:

**Nr. = (19 x 4) – ( 7 x22) = 76 – 154 = -78**

**Denominator,**

We will get denominator by the same technique of denominator in the above technique as shown below,

Dr. = (2 x 4) – (7 x 3) = 8 – 21 = -13

Y = Nr./ Dr. = -78/-13

= 6

Now put y = 6 in the given equ. to find x,

7x + 2 x ( 6) = 19

7x = 19 – 12 = 7

So, x = 7/7 = 1

**Ans, x=1 & y = 6 **

**Ex. v) 5x + 4y = 3 & 2x – 3y = -8**

**Numerator:**

**= (3 x 2) – (-8 x 5)**

**= 6 + 40 = 46**

**Denominator:**

**= (4 x 2 ) – ( 5 x -3)**

**= 8 + 15 = 23**

**Y = Nr./ Dr.**

** = 46 / 23 = 2**

**So, y = 2**

**Substitute the value of y = 2 in given equ.**

**5x + 4 x (2) = 3**

**5x = 3- 8 = -5**

**x = -1**

**Ans, x = -1 & y = 2**

**Specific case:**

Here we find the specific case if any one coefficient of the variable is constant.

There is special sutra in vedic maths for such cases called **Sunyam Anyat **which means **‘If one is in ratio, the other is zero’**. This sutra is very useful when the coefficients of either x or y are in certain ratio.

**Ex. vi) 67x + 302y = 1510 & 466x + 906y = 4530**

**Here the y coefficient are in ratio of 1:3 (302:906) and the constants are also in the same ratio 1:3(1510:4530).**

Now here our sutra says ‘**If one (variable) is in ratio, the other(variable) is zero’.**

In this case we see that the variable y is in ratio with the constant term so, other variable x is zero.

So, we put x = 0 in our given equ. 1

0 + 302y = 1510

x = 1510/302 =5.

Ans, x = 5 & y = 5.

**Ex. vii) 507x + 922y =1014 & 2028x + 1634 y = 4056**

**Here the x coefficient are in ratio of 1:4 (507:2028) and the constants are also in the same ratio 1:4(1000:4000).**

Now here our sutra says ‘**If one (variable) is in ratio, the other(variable) is zero’.**

In this case we see that the variable x is in ratio with the constant term so, other variable y is zero.

So, we put y = 0 in our given equ. 1

**507 x + 0 = 1014**

**x = 1014 / 507**

**x = 2**

**Ans, x = 2 & y = 0**