VEDIC MATHS-93

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VEDIC MATHS

                           By OMKAR TENDOLKAR

Hello friends,

                      This is  post number 93 from the series of "Vedic maths" blogs. Here in this blog we will learn about "Algebraic Calculations-III"

Reference:

We had already learn about "Algebraic Calculations-I" in our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-91".

We had already learn about "Algebraic Calculations-II" in our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-92".

Algebraic Calculations:

Here in this post we will solve algebraic calculations by maths sutra.

Examples :
let’s solve our next example: (2x + 3y + 4)(x – 3y + 5)
Step 1: 
Now these kind of sums can be done easily using the vertically and crosswise method of multiplication.

We had already learn about "Criss-Cross system of multiplication for 3-digits Number" in our previous blog. If you have missed my last blog then please visit "VEDIC MATHS-15".

There you have it. Now with the pattern shown, let’s straightaway go to our example and apply it.

Step 1:
Here we’ll solve the sum right to left, so we first multiply vertically 5 and 4, and that gives us 20.

      2x + 3y +  4
         x - 3y +  5
-----------------------
                  + 20

Step 2:
Next, we’ll solve it crosswise: (5 × 3y) + (4 × 3y).
So we get 15y - 12y = 3y.

      2x + 3y +  4
         x - 3y +  5
----------------------
          + 3y + 20

Step 3:
Now comes the third step—the star multiplication step. So we multiply (5 ×
2x) + (4 × x) + (3y × -3y).
And this becomes 10x + 4x - 9(y)^2 = 14x - 9(y)^2.

                2x + 3y +  4
                   x - 3y +  5
------------------------------
14x - 9(y)^2.+ 3y + 20

Step 4:
Now we’ll solve it crosswise again as per the pattern.
So we have (2x × -3y) + (x × 3y).
So this becomes -6xy + 3xy = -3xy.

                           2x + 3y +  4
                              x - 3y +  5
----------------------------------------
-3xy + 14x - 9(y)^2.+ 3y + 20

Step 5:
In our final step, we will follow the vertical pattern. So when we multiply 2x
and x, it gives us 2(x)^2.

                                      2x + 3y +  4
                                         x - 3y +  5
-------------------------------------------------
2(x)^2 -3xy + 14x - 9(y)^2.+ 3y + 20

Answer:
(2x + 3y + 4)(x – 3y + 5) =2(x)^2 -3xy + 14x - 9(y)^2.+ 3y + 20.

The simplicity of this method can be vouched from examples given above.


In next blog we will discuss about "Quadratic Equations-I".     

Are you excited for this?...
Then, please wait for it.
I will post my new blog in next week.

We will meet very soon through our next  blog. Till that stay connected, stay healthy and stay safe.

Thanks

for giving your valuable time.

Good day😊

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