# Solve the following system of equations - (2).

(a) Solve the following system of equations: $$\begin{cases} x(x + 1) + \dfrac{1}{y}\left(\dfrac{1}{y} + 1\right) = 4\\ \dfrac1{y^2}(x+1) + x^2\left(\dfrac{1}{y} + 1\right) = 4 \end{cases}$$ (b) Solve the following system of equations: $$\begin{cases} x(x + 1) + \dfrac{1}{y}\left(\dfrac{1}{y} + 1\right) = 4\\ \dfrac{1}{y}(x^2 + 1) + x\left(\dfrac{1}{y^2} + 1\right) = 4 \end{cases}$$

This problem comes from a competition I participated this morning (19/3/2019). And the problems were hard. Out of 11 problems, I only did 7. That counts this one.

• So you managed to solve this problem (it seems like it, from your last sentence)? – Minus One-Twelfth Mar 19 at 4:53
• I did manage to solve it. I just want to see another, shorter solution. – Lê Thành Đạt Mar 19 at 4:53
• Ah I see. Congratulations! – Minus One-Twelfth Mar 19 at 4:54
• Thanks! The next question will be the one that I couldn't do. It's geometry. – Lê Thành Đạt Mar 19 at 4:56
• How can we know whether we have a shorter solution than yours, if we don't know yours? – Gerry Myerson Mar 19 at 5:32

We will substitute $$y=tx$$ in the second equation, then we get $$\frac{x}{t^2x^2}+\frac{1}{t^2x^2}+\frac{x^2}{tx}+x^2-4=0$$ and this is
$$(tx^2+2tx+x+1)(tx^2-2tx+1)=0$$ so we get $$tx^2+2tx+x+1=0$$ or
$$tx^2-2tx+1=0$$ which can be solved for $$t$$.