Question

If x,y are integers then find the no. of ordered pairs which​

Answer 1

Question:-

Find number of ordered pairs $(x,y)\:\:\forall\:\: x,y\in\mathbb{Z}$ such that $y = \dfrac{2x-3}{x+2}.$

Answer:-

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,

$y = \dfrac{2x - 3}{x + 2}$

$\longrightarrow y(x + 2) = 2x - 3$

$\longrightarrow xy + 2y = 2x - 3$

$\longrightarrow xy - 2x + 2y = - 3$

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

Now using SFFT we get,

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

$\longrightarrow (x + 2)(y - 2) + 4= - 3$

$\longrightarrow (x + 2)(y - 2) = - 7$

Now, possible pairs include,

• -1 * 7 = -7
• 7 * -1 = -7
• -7 * 1 = -7
• 1 * -7 = -7

1st case:

$x + 2 = - 1$

$\longrightarrow x = -3$

And $y-2=7$

$\longrightarrow y = 9$

So,

$\longrightarrow (x,y) = (-3,9)$

2nd case:

$x + 2 = 7$

$\longrightarrow x = 5$

And $y-2=-1$

$\longrightarrow y = 1$

So,

$\longrightarrow (x,y) = (5,1)$

3rd case:

$x + 2 = - 7$

$\longrightarrow x = -9$

And $y-2=1$

$\longrightarrow y = 3$

So,

$\longrightarrow (x,y) = (-9,3)$

4th case:

$x + 2 = 1$

$\longrightarrow x = -1$

And $y-2 = -7$

$\longrightarrow y=-5$

So,

$\longrightarrow (x,y) = (-1,-5)$

Also, there are two more cases.

Here $x=0$ in 5th case and $y=0$ in 6th case. But after solving we find that $y$ is not an integer in 5th case, and $x$ is not an integer in 6th case.

Hence, the ordered pairs satisfying the above equation are,

$\longrightarrow \underline{\underline{(-3,9),(5,1),(-9,3),(-1,-5)}}$