Math, asked by Anonymous, 3 months ago


 \bf \: find \:  \alpha  \: and \:  \beta \: satisfying \:   \mapsto3 \alpha  + 2 \beta  =  \alpha  \beta  + 1.
●need detailed solution!


Anonymous: sorry i forgot to put some information in it. lemme clear you.
Arceus02: if it's a Diophantine equation, then I found (although not me, with some help from a calculator xD) I found 4 pairs of solution
Arceus02: (-3,2) , (1, -2) , (3,8) , (7,4)
Arceus02: (0,1/2) (1/3,0)
Arceus02: 6 solutions
Anonymous: find alpha and beta belongs to natural number. satisfying 3alpha+2beta=alpha.beta+1. this the right question

Answers

Answered by Arceus02
14

[Note: Here \alpha is denoted by x and \beta is denoted by y. ]

\\

Complete question:

3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+.

\\

Here we have to use a variant of Simon's Favourite Factorizing Trick (SFFT). It's just the name of a technique which is used to solve Diophantine Equations.

\\

Simon's Favourite Factorizing Trick:

(x + 1)(y + 1) = xy + x + y + 1

It's the same if you put  a = b = 1 in the following equation:

{(x + a)(y + b) = xy + xb + ay + ab}

If you just put a = b = 1 it becomes SFFT, and it's incredibly useful in solving a lot of Diophantine Equations which has a "xy" term in it.

\\

Now, this SFFT has a lot of variants, and can be used in a number of ways to solve Diophantine Equations.

One of the conclusion/variants is:

 xy + kx + ly = (x + l)(y + k) - kl

\\

Using that variant in the given question,

3x + 2y = xy + 1

 \longrightarrow xy - 3x - 2y =  - 1

 \longrightarrow xy  + ( -  3)x  + ( -  2)y =  - 1

 {\longrightarrow  \{x + ( - 2) \} \{y +  ( - 3) \} - ( - 3 \times  - 2)=  - 1}

 \longrightarrow (x - 2)(y - 3) - 6 =  - 1

 \longrightarrow (x - 2)(y - 3)  = 5

Now, product of two integers is 5, possible pairs include,

  • (1, 5)
  • (5, 1)
  • (-1, 5)
  • (1, -5)

So,

First pair:

x - 2 = 1

 \longrightarrow x  = 3

And, y-3 = 5

 \longrightarrow y  = 8

\longrightarrow (x,y) = (3,8)

Second pair:

x - 2 = 5

 \longrightarrow x  = 7

And y - 3 = 1

 \longrightarrow y  = 4

\longrightarrow (x,y) = (7,4)

For third and fourth pairs, on solving them for x and y we find y to be negative in the third pair, and x to be negative in the fourth pair.

But since accroding to the question, x and y are positive integers, we will not consider them.

\\

The possible pairs of (x,y) satisfying 3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+ are

\longrightarrow \underline{\underline{(3,8),(7,4)}}


Arceus02: See that I simplified the problem and then found factors and then possible (x,y) pairs. If you directly apply hit and trial without simplifying it will be mostly impractical
Anonymous: Yes , Thats True But Its Not In the Syllabus of NCERT for Classes 11 and 12th or even R.D. Sharma donot gave much information on this topic . Its an advanced level question . By the way thanks For the answer and also for the new Concept .
Arceus02: welcome :)
Arceus02: welcome :)
Anonymous: A grade! (:
Arceus02: thanks :)
michaelgimmy: Fab! :)
Arceus02: thanks :)
BaroodJatti12: fabulous:-)
Arceus02: thanks :)
Answered by Anonymous
2

Step-by-step explanation:

Note: Here \alphaα is denoted by xx and \betaβ is denoted by y.y. ]

\begin{gathered}\\\end{gathered}

Complete question:

3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+.3x+2y=xy+1∀x,y∈Z

+

.

\begin{gathered}\\\end{gathered}

Here we have to use a variant of Simon's Favourite Factorizing Trick (SFFT). It's just the name of a technique which is used to solve Diophantine Equations.

\begin{gathered}\\\end{gathered}

Simon's Favourite Factorizing Trick:

(x + 1)(y + 1) = xy + x + y + 1(x+1)(y+1)=xy+x+y+1

It's the same if you put a = b = 1a=b=1 in the following equation:

{(x + a)(y + b) = xy + xb + ay + ab}(x+a)(y+b)=xy+xb+ay+ab

If you just put a = b = 1a=b=1 it becomes SFFT, and it's incredibly useful in solving a lot of Diophantine Equations which has a "xy""xy" term in it.

\begin{gathered}\\\end{gathered}

Now, this SFFT has a lot of variants, and can be used in a number of ways to solve Diophantine Equations.

One of the conclusion/variants is:

xy + kx + ly = (x + l)(y + k) - klxy+kx+ly=(x+l)(y+k)−kl

\begin{gathered}\\\end{gathered}

Using that variant in the given question,

3x + 2y = xy + 13x+2y=xy+1

\longrightarrow xy - 3x - 2y = - 1⟶xy−3x−2y=−1

\longrightarrow xy + ( - 3)x + ( - 2)y = - 1⟶xy+(−3)x+(−2)y=−1

{\longrightarrow \{x + ( - 2) \} \{y + ( - 3) \} - ( - 3 \times - 2)= - 1}⟶{x+(−2)}{y+(−3)}−(−3×−2)=−1

\longrightarrow (x - 2)(y - 3) - 6 = - 1⟶(x−2)(y−3)−6=−1

\longrightarrow (x - 2)(y - 3) = 5⟶(x−2)(y−3)=5

Now, product of two integers is 5, possible pairs include,

(1, 5)

(5, 1)

(-1, 5)

(1, -5)

So,

First pair:

x - 2 = 1x−2=1

\longrightarrow x = 3⟶x=3

And, y-3 = 5y−3=5

\longrightarrow y = 8⟶y=8

\longrightarrow (x,y) = (3,8)⟶(x,y)=(3,8)

Second pair:

x - 2 = 5x−2=5

\longrightarrow x = 7⟶x=7

And y - 3 = 1y−3=1

\longrightarrow y = 4⟶y=4

\longrightarrow (x,y) = (7,4)⟶(x,y)=(7,4)

For third and fourth pairs, on solving them for xx and yy we find y to be negative in the third pair, and x to be negative in the fourth pair.

But since accroding to the question, xx and yy are positive integers, we will not consider them.

\begin{gathered}\\\end{gathered}

The possible pairs of (x,y)(x,y) satisfying 3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+3x+2y=xy+1∀x,y∈Z

+

are

\longrightarrow \underline{\underline{(3,8),(7,4)}}⟶

(3,8),(7,4)

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