Shouldn't the last be $ay^3+bz=cx^5$? MS_Kekas wrote: For any given real $a, b, c$ solve the following system of equations: $$\left\{\begin{array}{l}ax^3+by=cz^5,\\az^3+bx=cy^5,\\az^3+bx=cy^5.\end{array}\right.$$Proposed by Oleksiy Masalitin, Bogdan Rublov

NO_SQUARES wrote: Shouldn't the last be $ay^3+bz=cx^5$? MS_Kekas wrote: For any given real $a, b, c$ solve the following system of equations: $$\left\{\begin{array}{l}ax^3+by=cz^5,\\az^3+bx=cy^5,\\az^3+bx=cy^5.\end{array}\right.$$Proposed by Oleksiy Masalitin, Bogdan Rublov Yes, thanks!