VEDIC MATHS-42

-


VEDIC MATHS

                           By OMKAR TENDOLKAR

Hello friends,

                      This is  post number 41 from the series of "Vedic maths" blogs. Here in this blog we will learn about "Simultaneous Linear Equations part-2".

In the previous blog we studied how to solve two types of basic equations. In this blog we will study how to solve simultaneous linear equations. 

What Are Simultaneous Linear Equations?

Simultaneous linear equations have two variables in them. Let us say x and y. Since there are two variables in the equation we cannot solve it by itself. We need another equation with the same variable values to find the answer. When these two equations are solved together we get the values of the variables x and y.

In the traditional method a new set of equations is formed in order to equalize the coefficients of any one variable. But forming new equations is a time-consuming procedure. Secondly, equalizing the coefficients is not always an easy task. If the coefficients have big numbers or decimal values it. becomes very difficult to equalize them by multiplying them with suitable numbers. the possibility of making a mistake with this method is pretty high.

Here in this blog we will study the Vedic approach to solving simultaneous linear equations.

Reference:

We had already learn about "Simultaneous Linear Equations part-1" our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-41".

Method:

In this method we will not be forming new equations but instead we will calculate the values of x and y with the given equations only. The value of the variables x and y will be expressed in the form of numerator upon denominator. 

                Numerator

x   =    ----------------------- ;  
              Denominator

                Numerator

y   =    ----------------------- ;  
              Denominator

It should be noted that although one can find the values of both x and y there is no need for doing it. If we obtain the value of either x or y then the value of the other variable can easily be obtained by substitution.

Calculating the value of 'y'

1) Find the values of x and y for the equations 6x + 4y = 50 and 5x + 5y = 50

6x + 4y = 50             .....(1)
5x + 5y = 50             .....(2)

6x + 4y = 50             .....(1)
5x + 5y = 50             .....(2)
The numerator is obtained by cross-multiplying (5 × 50) and subtracting from it the cross product of  (50 × 6).

            (5 × 50) - (50 × 6)                       -50

y   =   --------------------------------- = ---------------------
                  Denominator                Denominator

6x + 4y = 50             .....(1)
5x + 5y = 50             .....(2)
The numerator is obtained by cross-multiplying  (5 × 4) and subtracting from it the cross product of (6 × 5)

                       -50                    -50

y   =   --------------------------- = ------- =  5
            (5 × 4) - (6 × 5)           -10

In this case, we have obtained the value of y as 5. We substitute the value of y in equation (1) and
get the value of x as 5. The solution set is (5, 5).

Solution:
x=5,y=5. 


2) Find the values of x and y for the equations 5x + 4y = 3 and 2x - 3y = -8

5x + 4y =  3              .....(1)
2x  - 3y = -8              .....(2)

5x + 4y =  3              .....(1)
2x -  3y = -8              .....(2)
The numerator is obtained by cross-multiplying (3 × -2) and subtracting from it the cross product of (5 × -8)

            (3 × 2) - (5 × -8)                    46

y   =   --------------------------- = ------------------------
               Denominator              Denominator

5x + 4y =  3            .....(1)
2x  - 3y = -8            .....(2)
The numerator is obtained by cross-multiplying  (2 × 4) and subtracting from it the cross product of (-3 × 5)

                       46                      46

y   =   --------------------------- = ------- =  2
           (2 × 4) - (-3 × 5)            23

In this case, we have obtained the value of y as 2. We substitute the value of y in equation (1) and
get the value of x as -1. The solution set is (-1,2).

Solution:
x=-1, y=2


Solve the next three equations by calculating for y. Write the answer in the form of (value of x, value of y).

1. 7x + 2y = 19 and 4x + 3y = 22

2. 2x + 9y = 27 and 4x + 4y = 26

3.40x + 20y = 400 and 80x + 10y = 500


You may answer this in comment box. You may ask your any query or doubt in comment box. I will try to resolve as early as possible.

In next blog we will discuss about "Simultaneous Linear Equations part-3".     

Thanks

for giving your valuable time.

Good day😊.

Comments

Post a Comment

Popular posts from this blog

VEDIC MATHEMATICS

-
Image
  VEDIC MATHS                             By OMKAR TENDOLKAR Hello friends,                        I am Omkar Tendolkar from Kudal. I am studying in S.Y.B.Sc at SRM College, Kudal. I always enjoy sharing my knowledge. This is one of my attempts.            Some people find mathematics challenging to understand, while others struggle with calculations. Certainly, the comprehension of the subject depend on IQ (intelligent quotient), but equally crucial is one's perspective towards it. Because mastering calculation doesn't solely rely on a sharpe IQ; it also demands some basic tricks that can greatly assist in handling complex calculations.                  Starting today, I want to launch a "VEDIC MATHS" blog. Here, you will find a series of blog's explaining simple tricks that will...
Read more

VEDIC MATHS-1

-
Image
VEDIC MATHS                             By OMKAR TENDOLKAR Hello friends,                          This is our first blog in the series of 'Vedic Maths' blogs. Here, we will learn about the basic principles required in Vedic Maths. This blog will cover all the fundamental concepts necessary for understanding our next few blogs.   Base: It is the first number in the series. For a two-digit number: 10. Starting with 10, 100, 1000, 10000, etc. Complement: . Adding two numbers results in the next nearest base. 48 + 52 = 100. Then, We can say that 48 is the complement of 52 or 52 is the complement of 48. How to find: Subtract each digit from 9 and the last digit from 10. If... The last digit is 0, then take the second-to-last digit as the last one. For example: 60 is the complement of 40. All these examples, ranging from 2 digits to 6 digits, w...
Read more

VEDIC MATHS-15

-
Image
  VEDIC MATHS                             By OMKAR TENDOLKA R Hello friends,                       This is our 15th blog from the series of "Vedic maths" blogs. Here in this blog we will learn about   "Criss-Cross System of Multiplication " . Criss-Cross System of Multiplication :                In Vedic Mathematics , we have similar system as like traditional way of multiplication it helps us to get the answer much faster. This system is also universal system and can be used for any combination of numbers of any length.                      This system of Multiplication is given by the " Urdhva-Tiryak Sutra. " It means ' vertically and cross-wise '. The application of this system are manifold, but in this blog we shall confine our study only to it's utility in...
Read more