Find distinct real numbers satisfying $\frac{xy}{x-y} = \frac{1}{30}$ and $\frac{x^2y^2}{x^2+y^2} = \frac{1}{2018}$

Question

Figure out with distinct real numbers the system of equations.

$$\frac{xy}{x-y} = \frac{1}{30}$$ $$\frac{x^2y^2}{x^2+y^2} = \frac{1}{2018}$$

I multiplied x-y both side on the first equation and square on both side, and I stucked.

Help me...

Answer 1

From the first equation we get: $$30xy=x-y$$ or $$y(30x+1)=x$$ so $$y=\frac{x}{30x+1}$$ substituting this in the second equation we get $${\frac {1}{2018}}\,{\frac { \left( 43\,x+1 \right) \left( 13\,x-1 \right) }{450\,{x}^{2}+30\,x+1}} =0$$ Can you finish?

Answer 2

Inverting the equations gives $$ 30 = \frac{x-y}{xy} = \frac 1y - \frac 1x \\ 2018 = \frac{x^2+x^2}{x^2y^2} = \frac 1{y^2} + \frac 1{x^2} $$ This suggests to substitute $a= \frac 1x, b= \frac 1y$: $$ 30 = b - a \\ 2018 = b^2 + a^2 \, $$ This leads to a simple quadratic equation for $a$ (or $b$) which has two different real (actually integral) solutions.

Answer 3

Since $x,y\ne0$, we can use $u=1/x$, $v=1/y$: \begin{align} v-u &= 30 \tag{1}\label{1} ,\\ v^2+u^2 &= 2018 \tag{2}\label{2} . \end{align}

\begin{align} \eqref{2}-\eqref{1}^2:\quad 2uv&=1118 ,\\ v^2+u^2+2uv &= 2018+1118 ,\\ (u+v)^2&=56^2 ,\\ u+v&=\pm56 \tag{3}\label{3} . \end{align}

Combination of \eqref{3} with \eqref{1} provides two cases:

Case 1.

\begin{align} v-u &= 30 ,\\ v+u&=56 ,\\ u&=13 \quad v=43 ,\\ x&=\frac1{13},\quad y=\frac1{43} . \end{align}

Case 2.

\begin{align} v-u &= 30 ,\\ v+u&=-56 ,\\ u&=-43 \quad v=-13 ,\\ x&=-\frac1{43},\quad y=-\frac1{13} . \end{align}

Answer 4

$$ \left\{ \begin{array}{rcl} \frac{xy}{x-y} & = & \frac{1}{30}\\ \frac{x^2y^2}{x^2+y^2} & = & \frac{1}{2018} \end{array} \right. \Rightarrow \left\{ \begin{array}{rcl} 900 x^2y^2 & = & x^2+y^2-2 xy\\ 2018x^2y^2 & = & x^2+y^2 \end{array} \right. \Rightarrow x y = \frac{1}{559} $$

etc.