Problem
ALG-B1-M10-P006 Symmetric system
#6
★★☆☆☆ Level 2 of 5
Solve \(x+y+xy=19\), \(x^2+y^2=25\).
Set \(s=x+y\), \(p=xy\).
We get \(s+p=19\), \(s^2-2p=25\). Then \(s^2+2s-63=0\), so \(s=7\) or \(s=-9\). For \(s=7\), \(p=12\), so \(\{x,y\}=\{3,4\}\). For \(s=-9\), \(p=28\), there are no real roots. Answer: \((3,4),(4,3)\).
A mixed problem on systems and discriminant checking.