Problem
ALG-B1-M03-P025 A System with Product and Sum
#25
★★★★★ Level 5 of 5
Solve the system \(x+y+xy=11\), \(x^2+y^2=25\).
Set \(s=x+y\), \(p=xy\).
Let \(s=x+y\), \(p=xy\). Then \(s+p=11\), and \(x^2+y^2=s^2-2p=25\).
Since \(p=11-s\), we get \(s^2-2(11-s)=25\), so \(s^2+2s-47=0\). Thus \(s=-1\pm4\sqrt3\), \(p=11-s\).
Real \(x,y\) require \(s^2-4p\ge0\), which holds for both values. Then \(x,y\) are roots of \(t^2-st+p=0\). Thus for each \(s=-1\pm4\sqrt3\), \(p=11-s\), the pairs \((x,y)\) are the two permutations of the roots of \(t^2-st+p=0\).
Not every answer must be a neat integer pair.