Solving two equations for $p$ and $q$

Question

How we obtain from the equations $z^2-pqxy=0$ and $z-a(xp+yq)=0$ that $p=\frac{z}{cx}$ and $q=\frac{cz}{y}$, where $a(c+\frac{1}{c})=1$?

Answer 1

Using the first equation then the second, $$a(xp)^2-z(xp)+a(xp)(yq)=0\quad\Rightarrow\quad a(xp)^2-z(xp)+az^2=0\ .$$ You can easily show that $yq$ satisfies the same quadratic, so $$xp,yq=z\left(\frac{1\pm\sqrt{1-4a^2}}{2a}\right)\ .$$ Therefore $$p=\frac{z}{cx}\quad\hbox{where}\quad c=\frac{2a}{1\pm\sqrt{1-4a^2}}\ .$$ Unless you have more information, it is impossible to tell which sign you should take. Whichever it is, you get $$q=\frac{z}{cy}$$ with the other sign in $c$. In both cases it is easy to show that $$c+\frac{1}{c}=\frac{1}{a}\ .$$