Exercise 1680Algebra › Equations and radicalsOlympiadAdded 3 Nov 2026

A quadratic system

#nonlinear system#elimination#olympiad

Statement

Solve in $\mathbb{R}^2$ the system
$\begin{cases}x^2+y^2=5\\ x^3+y^3=9(x+y)\end{cases}$

Hint

Use $x^3+y^3=(x+y)(x^2-xy+y^2)$ .

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First equation. $x^2+y^2=5$ .

Second equation. $x^3+y^3=(x+y)(x^2-xy+y^2)=9(x+y)$ .

Case $x+y=0$ . Then $y=-x$ , so $x^2+x^2=5$ , i.e. $x=\pm\sqrt{5/2}$ . Solutions $(\sqrt{5/2},-\sqrt{5/2})$ and $(-\sqrt{5/2},\sqrt{5/2})$ .

Case $x+y\neq0$ . Divide: $x^2-xy+y^2=9$ , i.e. $(x^2+y^2)-xy=9$ , so $5-xy=9$ , $xy=-4$ .

Now $x+y$ and $xy=-4$ determine $x,y$ . From $x^2+y^2=5$ and $xy=-4$ : $(x+y)^2=5-8=-3<0$ . Impossible in $\mathbb{R}$ .

Conclusion. $(x,y)\in\{(\sqrt{5/2},-\sqrt{5/2}),(-\sqrt{5/2},\sqrt{5/2})\}$ .