Easy!

$\quad xy=x+y$

$\quad xy-x-y=0$

By Simon's Favorite Factoring Trick ,

$\quad xy-x-y+1=1$

$\quad (x-1)(y-1)=1$

$\quad \begin{cases}x-1=1\\y-1=1\end{cases}$ or $\begin{cases}x-1=-1\\y-1=-1\end{cases}$

$\quad \begin{cases}x=2\\y=2\end{cases}$ or $\begin{cases}x=0\\y=0\end{cases}$ (rejected)

x=2,y=2 Because only 2 * 2=2+2

2 * 2=2+2

4=4 which is TRUE

PLEASE UPVOTE IF SATISFIED

It can be shown that $\ (x-1)(1-y) = 1$ and the integer solution follows.

2 is the only non-zero integer which produces the same result when added together and when multiplied by itself. $2 \times 2$ = $4$ , $2 + 2$ $= 4$ So both X and Y must be 2.

Since xy=x+y The only number which satisfies this is 2.

Just put the values

There is no need to give options.due to this just put the values and find and.

Sec^2theta = X Csc^2theta = Y Now say ??

As 2x2 and 2+2 are equal and hence satisfy x+y=xy ,we get the answer as x=2 and y=2.

From the given options, the only option that satisfies the given condition (that is multiplication and addition are equal) is the value of x and y is 2. Hence the proof.

xy=x+y x=2 y=2

2*2=2+2

4=4 so the answer is x^2y^2=(xy)^2 x=2, y=2

both are 2... and nothing have not the same result as 2 when it is: x + y x * y x ^ y

NOT a solution. Just an observation. xy=x+y ==> y = x/(x-1) Now take derivative respect to x. y' = -1/(x-1)^2 ==> at x = 1 there is vertical asymptote and no global minimum and maximum values are found.