Adding the first two equations, we get, $y(x+z)$+ $z$ $=$ $0$ $(y+1)(x+z)$ $=$ $0$ So, either $y$ $=$ $-1$ or $x$ $=$ $-z$ If $y$ $=$ $-1$, then from the 3rd equation we have $zx$ $=$ $-17$.....(1) And from the second equation we have, $x -$z $=$ $30$....(2) We can very easily solve (1) and (2). Now, if $x$ $=$ $-z$ Then, $y$ $=$ $z^2$ -$18$ And, $z(y-1)$ $=$ $30$ So, $z(z^2-19)$ $=$ 30 So, we see that the above equation has $z$ $=$ $5$ as one of the roots. So, we can easily get the other two roots of $z$ and thereafter we will obtain the respective values of $x$ and $y$.

I highly doubt if this even was TST??

IMO2019 wrote: I highly doubt if this even was TST?? This was just P1.

IMO2019 wrote: I highly doubt if this even was TST?? Well, it wasn't. It was the first problem in the Final Round of the German National Olympiad for Grade 12. Yes, an easy problem, but that's intended.

oh! nice so they had to do this by another approach(like we had to find min. using some weird constraints by simplex method), or the one as stated.