Exercise 955Algebra › Equations and radicalsL1Added 3 Nov 2026

A symmetric system

#symmetric system#substitution#Vieta

Statement

Solve in $\mathbb{R}^2$ the system
$\begin{cases}x+y+xy=5\\ x^2+y^2+xy=7\end{cases}$

Hint

Set $S=x+y$ and $P=xy$ . Rewrite both equations in terms of $S$ and $P$ .

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Rewrite. $x^2+y^2=S^2-2P$ , so the second equation becomes $S^2-P=7$ . The first is $S+P=5$ .

Solve. From the first, $P=5-S$ . Substituting: $S^2-(5-S)=7$ , i.e. $S^2+S-12=0$ . Roots: $S=3$ or $S=-4$ .

Case $S=3$ : $P=2$ . Then $x,y$ are roots of $t^2-3t+2=0$ : $t=1,2$ . Solutions $(1,2)$ and $(2,1)$ .

Case $S=-4$ : $P=9$ . Then $t^2+4t+9=0$ , discriminant $16-36<0$ , no real solutions.

Conclusion. $(x,y)\in\{(1,2),(2,1)\}$ .