Just a simple question #2

from the first two equations it is evident that y^2=1 so y=+/- 1 let this series be equal to k, then x=ky ,k+y these two values are equal checking the values of +1 and -1 we see that there is no solution at y=+1 so plugging values of-1 and then solve for x

Separate it as 2 equations :

$xy = \frac{x}{y}$ . . . . . . . . . . . . . . . . . 1

$xy = x - y$ . . . . . . . . . . . . 2

From the first equation, we can get the value of $y$ :

$xy = \frac{x}{y}$

$y = \frac{1}{y}$

$y^2 = 1 => (y = 1)$ or $(y = -1)$

From the second equation, we can get the value of $x$

$xy = x - y$

$xy + y = x\ => y( x + 1 ) = x$

• when $y=1$ then, $1( x + 1 ) = x => x -x = -1$ (undefined)

• when $y=-1$ then, $-1( x + 1 ) = x => -2x = 1$ $\boxed{y = -1} ; \boxed{x = -\frac{1}{2}}$

So, $x + y = -\frac{3}{2} = -1.5$