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Question Number 176316 by Matica last updated on 16/Sep/22

  xy+xz=255   xz−yz=224.  find x y and z

$$\:\:{xy}+{xz}=\mathrm{255} \\ $$$$\:{xz}−{yz}=\mathrm{224}. \\ $$$${find}\:{x}\:{y}\:{and}\:{z} \\ $$

Answered by Frix last updated on 16/Sep/22

x∈R\{0}  y=(x/2)+((255)/(2x))±((√(x^4 +386x^2 +65025))/(2x))  z=−(x/2)+((255)/(2x))∓((√(x^4 +386x^2 +65025))/(2x))

$${x}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$${y}=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{255}}{\mathrm{2}{x}}\pm\frac{\sqrt{{x}^{\mathrm{4}} +\mathrm{386}{x}^{\mathrm{2}} +\mathrm{65025}}}{\mathrm{2}{x}} \\ $$$${z}=−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{255}}{\mathrm{2}{x}}\mp\frac{\sqrt{{x}^{\mathrm{4}} +\mathrm{386}{x}^{\mathrm{2}} +\mathrm{65025}}}{\mathrm{2}{x}} \\ $$

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