Author : PRADEEP KOSHY

Category : Research Papers

Sub-Category:

Mathematics and Applied Mathematics

Language: English

Title: MIRACLE EQUATION-CAN BE USED TO SOLVE 3 VARIABLES IN A SINGLE EQUATION

Date Published: 27-9-2013

Keyords:

NumberTheory, Theorem.

Abstract:

Before 2013, you required 3 equations to solve 3 variables. Now it isn’t necessary.

PROOF

MIRACLE EQUATION-CAN BE USED TO SOLVE 3 VARIABLES IN A SINGLE EQUATION.

Seems Impossible.

But here is the proof.

MIRACLE EQUATION:

[(NX )² ― {(N―2) X }² ] =[N―(1―X²)]²―[N―(1+X²)]²

The above equation which is true for all real values of N and X,is actually analogous to the equation

[A²―B²] = C ²―D ² where A = NX ,B = (N―2) X , C = [N―(1―X²)] & D=[N―(1+X²)] where .A,B.C & D are four variables. One way of analyzing the same is, if anyone chooses one of these variables either A,B,C or D,the remaining 3 variables can be found out,by applying suitable values to N and X,in the considered variable and the other variables turn out correspondingly to the same.

The second case or application is given below.

Now ,there is an interesting application wherein ,we can utilize this equation to solve 3 unknown

variables in a single equation.

Assuming the 3 variabled equation is of the form

ax + by + dz = k where a ,b ,d are coefficients x , y , z are variables and k is the constant . Solution is given

by x = A²/a, y = B²/(-b) and z= D ² /d , since equaton is of the form A²―B²+D ² = C ²

Hence the solution to the equation 2x + 3y + 4z = 16

HERE C = 4 . Arbitrarily selected values of N = 1 , X = 2 to satisfy C=[ N―(1―X²)]

Ergo , x = 2, y = - 4/3 and z = 4. Alternatively. let us substitute X as any rational number .X can assume infinite values.Albeit, if X is real we get only approx. solutions. We could generate different values of N = C + 1 - X² ,corresponding to X equal any rational number)we can threreby get infinite solutions to this equation.We could resort to algorithm and programming at this stage,since a general equation is involved and a trinitarian aspect could be proved. Please note that C = √k

PN: When k is a perfect square , calculations are simple. Otherwise,multiply k by itself.For the equation to remain unchanged multiply each term of LHS by k and then resort to the steps like below

Suppose one need to solve

2x+3y+4z=13 Taking the necessary steps,the equation becomes ie multiplying each term in the given equation by k = 13,it transforms into

26x+39y+52 z= 169,therefore x= A²/a,y= B²/(-b) and z= D ² /d

HERE C= 13, If selected value of X=2, N = k+1 - X²=13+1- 4=10

Therefore x= 400/26=200/13, y=256/-39= -256/39 and z= 25/52

Take another value of X = 25,then N= k+1 - X²= 13+1-225= -211.

A= NX= 5275

B= (N―2) X= 5325

D=[N―(1+X²)] = 837

x= A²/a= 1070216.345

,y= B²/(-b)= - 727067.3077 and

z= D ² /d =13472.48077

Verification 26x+39y+52 z= 169

26(1070216.345)- 39(727067.3077+52(13472.48077)=169 (hence verified)