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Question Number 215180 by hardmath last updated on 30/Dec/24

$$\{\begin{array}{l}{\mathrm{x}}^2+\mathrm{y}=31\\ {\mathrm{y}}^2+\mathrm{x}=41\end{array}\Rightarrow (\mathrm{x};\mathrm{y})=?$$

Commented by Ghisom last updated on 31/Dec/24

$$\mathrm{obviously}$$$$5^2+6=31$$$$6^2+5=41$$

Answered by Ghisom last updated on 31/Dec/24

$$y=31-x^2$$$${(31-x^2)}^2+x-41=0$$$$x^4-62x^2+x+920=0$$$$(x-5)(x^3+5x^2-37x-184)=0$$$$x_1=5\Rightarrow y_1=6$$$$\mathrm{it}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{make}\mathrm{much}\mathrm{sense}\mathrm{to}\mathrm{exactly}$$$$\mathrm{solve}\mathrm{the}\mathrm{remaining}{3}^{\mathrm{rd}}\mathrm{degree}$$$$x_2\approx -6.15360\Rightarrow y_2\approx -6.86685$$$$x_3\approx -4.92173\Rightarrow y_3\approx 6.77656$$$$x_4\approx 6.07534\Rightarrow y_4\approx -5.90971$$

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