## Do you struggle with solving simultaneous equations?

By simultaneous I mean equations with multiple unknown variables. Generally the number of equations given will be equal to the number of equations.

Let’s take an example,

3x + 4y = 18

5x + 7y = 31

### Methods Taught at Schools

In our schools, we are taught to solve for x by equating the co-efficient of y by multiplying both the equations by some constants in such a manner that you get the same resultant value and then subtracting one equation from the other.

For instance in this case, to find the value of x, we will multiply first equation by 7 and second equation by 4 and get 28 in both the cases. This is done so that we get a zero on subtracting.

3x + 4y = 18 …………………(i) x 7

5x + 7y = 31 …………………(ii) x 4

We get 2 equations, where co-efficient of y is same

21x + 28y = 126

20x + 28y = 124

Subtracting second equation obtained above from the first one we get

(21x – 20x) + (28y – 28y) = 2

Hence, x = 2

Plugging the value of x in equation (i) we get y = 3

### Problems with the Above Methods

- This method can become quite laborious, especially when the co-efficients of the unknowns are such that they have to be multiplied by large numbers to make them equal to eliminate one of the unknown by adding or subtracting as the case may be.
- The above calculations become cumbersome, when the co-efficient(s) are large prime numbers.
- This method involves multiple steps where we need to do multiplication and addition/subtraction.
- Also, there is no chance of using this method to solve the problem mentally as one has to keep track of the equations and various computations.

### Solving Simultaneous Equations the Smarter Way

Our new method will give us the final answer in fractions, i.e. in Numerator and Denominator for both the variables: x and y.

First, we need to find the numerator of the value of x in the above case, take the simple following steps:

Step #1: Cross-multilply the coefficient of y in the first equation by the constant term (RHS) of the second equation

Step #2: Subtract from it the cross-product of the y coefficient in the second equation and the constant term (RHS) of the first equation.

So the numerator is 4×31 – 18×7 = 124-126 = -2.

Second, we need to find the denominator of the value of x:

Step #1: Cross-multiply the coefficient of y in the first equation by the coefficient of x in the second equation

Step #2: Subtract from it the cross-product of the y coefficient in the second equation and x coefficient in the first equation.

Hence the denominator is 4×5 – 7×3 = -1

Hence, the value of x = -2/-1 = 2

Now, let’s try with a simpler example,

x+2y = 8

3x + y = 9

Using the above method, in a single line calculation you can say,

x = (2×9 – 1×8)/(2×3 – 1×1)

x = 10/5 = 2

Therefore, y = 3

Isn’t this amazingly simpler? With some practice you can comfortably apply this technique to solve simultaneous equation mentally.

If you liked this method, you must explore another Vedic Mathematics trick of solving a special class of simultaneous equations in seconds.

Do you find simultaneous linear equations difficult to solve? Do you think you can start using above method in solving equations?

(1). From the example what is the value of y if the value of x equal to 2

(2). How do we get the value of y in the second example

I am glad a Chartered Accountant is so good at maths , i am also a ca student . loved all your post sir.

Thanks a lot sir