Problem
ALG-B1-M11-P011 Hidden sum and product
#11
★★★☆☆ Level 3 of 5
Solve \(x+y+xy=29\), \(x^2+y^2=41\).
Set \(s=x+y\), \(p=xy\).
\(s+p=29\), \(s^2-2p=41\). We get \(s^2+2s-99=0\), so \(s=9\) or \(s=-11\). For \(s=9\), \(p=20\), \(\{x,y\}=\{4,5\}\); the second case gives no real roots.
Strategy: reduce the number of variables.