Question

Find solutions in x and y if real number solutions exist for the system of equations given.

Answer 1

$\quad \quad method \: 1$



Given,
x + y = x / y
x - y = xy
___________
2x = x/y + xy
___________

= > 2x = x( 1 / y + y )
= > 2 = ( 1 + y² ) / y
= > 2y = 1 + y²
= > 0 = 1 + y² - 2y
= > 0 = ( 1 - y )²
= > 1 = y


Now, substitute the value y,
= > x - y = xy
= > x - 1 = x( 1 )
= > x - 1 = x

But the solution of this equation can't be true, as 1 ≠ 0.
From this, we didn't get any solution.


$\quad \quad method \: 2$
Given, x + y = x/y
x - y = xy

Multiply both,

= > ( x + y )( x - y ) = (x/y)(xy)
= > x² - y² = x²
= > y = 0

Substituting the value of y

= > x + y = x/y
= > x + 0 = x/0
= > x = x/0

Anything divided by 0, is undefined.

hence, value of x is undefined and value of y is 0.


$\mathsf{ Result : There \: is \: no \: real \: solution.}$

Answer 2

Solution is given below:-------------

x + y = x/y
x - y = xy
Multiply both,
= > ( x + y )( x - y ) = (x/y)(xy)
= > x² - y² = x²
= > y = 0
Substituting the value of y
= > x + y = x/y
= > x + 0 = x/0
= > x = x/0

Anything divided by 0, is undefined.
hence, value of x is undefined and value of y is 0