Satisfied

Multiplying both equations yields:

$(x+y)(x-y) = \frac{x}{y}\cdot xy$

$x^2 - y^2 = x^2$

$y^2 = 0$

$y = 0$

This will however not give a valid solution to the first equation, therefore no solutions are possible.

$\begin{cases} x+y = \dfrac xy & ...(1) \\ x - y = xy & ...(2) \end{cases}$

$\begin{aligned} (1) \times y: \quad {\color{#3D99F6} xy} + y^2 & = x & \small \color{#3D99F6} (2): \implies xy = x-y \\ {\color{#3D99F6}x-y} + y^2 & = x \\ y(y-1) & = 0 & \small \color{#3D99F6} \text{Since }(1) \text{ is not defined when }y = 0 \\ \implies y & = 1 \end{aligned}$

$\implies \begin{cases} (1): \quad x + 1 = x & \small \color{#D61F06} \implies \text{No solution.} \\ (2): \quad x - 1 = x & \small \color{#D61F06} \implies \text{No solution.} \end{cases}$

First note that $y\ne 0$ , otherwise the first equation would not make sense.

1. Dividing the second equation by $y$ , yields $\frac{x}{y}=x+1$ .

2. Replacing in the first equation gives $x+y=x+1$ , from which we get $y=1$ .

3. But then, in the original first equation we would get $x+1=x$ which is impossible.

Adding the two equations together gives $2x = x/y + xy$

$2 = 1/y + y$ [see footnote]

$y^{2}- 2y+1=0$

$(y-1)^{2}=0$

which solves to $y=1$ but that gives $x+1=x$ and $x-1=x$ which makes no sense. Hence real solutions are impossible.

footnote: The step of cancelling x on each side of the equation assumes that x is non-zero. But if $x=0$ then $0+y=0$ so $y=0$ and $x/y$ becomes undefined.

postscript: I believe the above argument proves that no complex solutions are possible either.

x+y=x/y xy+y^2=x That is x=(y^2)/(y-1) x-y=xy (y^2)/(y-1)-y=(y^2)(y)/(y-1) -y=y^3 But y isn’t 0 So, y^2=-1 y isn’t real number not have (x,y) in real number.

Rearranging the first equation: $x + y = \frac{x}{y}$

$xy + y^2 = x$

$y^2 = x -x y$

$\frac{y^2}{1 - y} = x$

Rearranging the second equation: $x- y = xy$

$x - xy = y$

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

$\frac{y}{1 - y} = x = \frac{y^2}{1 - y}$

$y = y^2$

$0 = y(y-1)$

$y=0,1$

However, we would divide by 0 in both cases. Hence, there are no solutions to the system of equations.

I tried looking for every number and no numbers work for these equations

Rearranging the second equation gives $x + xy = y \; \Rightarrow \; x = xy + y = y(x+1) \; \Rightarrow \; y = \dfrac x{x+1}.$ Substitute this into the first equation gives

$\begin{aligned} x + \dfrac x{x+1} &=& \dfrac x{x/(x+1)} \\ x + \dfrac {x+1-1}{x+1}&=& x+1 \\ x + 1 - \dfrac1{x+1} &=& x+1 \\ - \dfrac1{x+1} &=& 0 \end{aligned}$

But there is no real $x$ satisfying $\dfrac1{x+1} = 0$ , so there is no solution.

Assuming x and y are positive. x+y must be larger than x-y, yet as y increases x+y increases but x/y must decrease while xy increases. Intuitively a solution is impossible. As negative x or y affects both sides of the equations the same is true for negative numbers. Therefore, no real answer.

Adding both equations we get:

$2xy = x + y^2$

We can write this in the following form, which is a quadratic: $xy^2 - 2xy + x = 0$

Using the quadratic equation: $y = \frac{-(-2x) \pm \sqrt{(2x)^2-4 \cdot x \cdot x}}{2x}$

Simplifying, we get: $y = \frac{2x}{2x} = 1$

Substituting 1 as our y value into the first equation gives: $x+1 = x$ which gives no valid solution.

To make sure, substituting the same value into the 2nd equation, we get $x-1 = x$ which also gives no valid solution.

x+y=x/y (1)

x-y=xy (2)

When you eliminate x you get

x=(x+y)*y=xy+y^2 and

x=xy+y

So y^2=y

The only solutions are y=1 or y=-1

(y=0 is no solution because it says x/y in (1) and you can't divide by 0)

Now insert this into (1)

x+1=x/1 => x+1=x or

x-1=-1

Which can't be true.

This is only possible if x is infinity 😀, and y is 1

we can conclude that x - y + y*y - x = 0 ,so that x - 1 = x, it is impossible

If x + y = x/y, multiply by y giving xy + y^2 = x.But xy = x - y, and substituting, x - y + y^2 = x. cancelling x, y^2 = y. So y = 0 or y = 1. y = 1 means x + 1 = x, which is impossible. y = 0 means x = 0, and 0/0 is not defined. Ed Gray

If you graph both equations using aymptotes, the intersection point is (0, 0) and for the top equation, you will get 0/0 on the right-hand​ side, so it is impossible to solve so there are no real solutions.