Problem
ALG-B1-M06-P006 A symmetric system
#6
★★☆☆☆ Level 2 of 5
Solve the system \[x+y+xy=7,\qquad x^2+y^2=10.\]
Denote \(s=x+y\), \(p=xy\).
We get \(s+p=7\) and \(s^2-2p=10\). Since \(p=7-s\), it follows that \(s^2+2s-24=0\), so \(s=4\) or \(s=-6\).
If \(s=4\), then \(p=3\), and \(x,y\) are the roots of \(t^2-4t+3=0\), hence \(\{x,y\}=\{1,3\}\). If \(s=-6\), then \(p=13\), but the discriminant \(s^2-4p=36-52<0\), so no real \(x,y\) exist.
Answer: \((1,3)\) and \((3,1)\).
It is important to finish the substitution by checking whether real roots exist.