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x-2-6y-9-y-2-8x-16-x-




Question Number 128477 by Fikret last updated on 07/Jan/21
x^2 +6y=9  y^2 +8x=16  x=?
$${x}^{\mathrm{2}} +\mathrm{6}{y}=\mathrm{9} \\ $$$${y}^{\mathrm{2}} +\mathrm{8}{x}=\mathrm{16} \\ $$$${x}=? \\ $$
Commented by mr W last updated on 07/Jan/21
y=((9−x^2 )/6)  (((9−x^2 )^2 )/(36))+8x=16  x^4 −18x^2 +288x−495=0  (x^2 +6x−15)(x^2 −6x+33)=0  ⇒x=−3±2(√6), 3±2(√6)i
$${y}=\frac{\mathrm{9}−{x}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{36}}+\mathrm{8}{x}=\mathrm{16} \\ $$$${x}^{\mathrm{4}} −\mathrm{18}{x}^{\mathrm{2}} +\mathrm{288}{x}−\mathrm{495}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{15}\right)\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{33}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{6}},\:\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{6}}{i} \\ $$

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