Solve $x+\frac{1}{y-x}=1$, $y+\frac{1}{x-y}=2$

Question

I've got this problem:

Solve for pairs of reals, $$ \left \{ \begin{array}{rcl} x+\dfrac{1}{y-x} & = & 1 \\ y+\dfrac{1}{x-y} & = & 2 \end{array} \right. $$

I've tried many different approaches and I tried on Wolfram Mathematica, which gives me the solutions, but I don't understand how to prove that there are no other.

Solution pairs $(x,y)=(2,1)$ and $(\frac{1}{2}, \frac{5}{2})$.

Answer 1

$y-x= a $ ⇒ $y=x+a$

$y-1/a=2$

$x+a -1/a=2 $

$x+1/a=1$

Subtracting two relations we get:

⇒$x+1/a-x-a+1/a=1-2 ⇒ a^2-a-2=0 ⇒ a=2,.. a=-1$

$a=2$ gives $y-x=2$ and summing two initial relations gives $x+y=3$. The solution of this system of equation is $x=0,5 $ and $y=2.5$

$a=-1$ gives $y-x=-1$this with $x+y=3$give another system of equation which solution is $x=2$ and $y=1$

Answer 2

Hint: if you add both equations you get $x+y=3$. Can you proceed ?