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Question Number 143130 by liberty last updated on 10/Jun/21

Commented by liberty last updated on 10/Jun/21

Answered by EDWIN88 last updated on 10/Jun/21

$$\mathrm{let}\mathrm{y}=\mathrm{px}$$$$\{\begin{array}{l}\mathrm{x}+\mathrm{px}+\frac{{\mathrm{x}}^2}{{\mathrm{p}}^2{\mathrm{x}}^2}=7\\ (\mathrm{x}-\mathrm{px})\frac{{\mathrm{x}}^2}{{\mathrm{p}}^2{\mathrm{x}}^2}=12\end{array}\Rightarrow \{\begin{array}{l}\mathrm{x}(1+\mathrm{p})+\frac{1}{{\mathrm{p}}^2}=7\\ \mathrm{x}\left(\frac{1-\mathrm{p}}{{\mathrm{p}}^2}\right)=12\end{array}$$$$\{\begin{array}{l}\mathrm{x}=\frac{{7\mathrm{p}}^2-1}{{\mathrm{p}}^2(1+\mathrm{p})}\\ \mathrm{x}=\frac{{12\mathrm{p}}^2}{1-\mathrm{p}}\end{array}\Rightarrow \frac{{12\mathrm{p}}^2}{1-{\mathrm{p}}^2}=\frac{{7\mathrm{p}}^2-1}{{\mathrm{p}}^2(1+\mathrm{p})}$$$$\Rightarrow {12\mathrm{p}}^4(1+\mathrm{p})=({7\mathrm{p}}^2-1)(1-{\mathrm{p}}^2)$$$$\Rightarrow {12\mathrm{p}}^4+{12\mathrm{p}}^5=-{7\mathrm{p}}^4+{8\mathrm{p}}^2-1$$$$\Rightarrow {12\mathrm{p}}^5+{19\mathrm{p}}^4-{8\mathrm{p}}^2+1=0$$$$\Rightarrow (\mathrm{p}+1)({12\mathrm{p}}^4+{7\mathrm{p}}^3-{7\mathrm{p}}^2-\mathrm{p}+1)=0$$$$\Rightarrow (\mathrm{p}+1)(\mathrm{p}+1)({12\mathrm{p}}^3-{5\mathrm{p}}^2-2\mathrm{p}+1)=0$$$$\Rightarrow {(\mathrm{p}+1)}^2({12\mathrm{p}}^3-{5\mathrm{p}}^2-2\mathrm{p}+1)=0$$$$\mathrm{we}\mathrm{get}\mathrm{p}=-1\left(\mathrm{rejected}\right)$$$${12\mathrm{p}}^3-{5\mathrm{p}}^2-2\mathrm{p}+1=0$$$$\Rightarrow \mathrm{p}\approx -0.4279$$$$\mathrm{then}\mathrm{y}=-0.4279\mathrm{x}$$$$\Rightarrow \mathrm{x}=\frac{12{(-0.4279)}^2}{1-(-0.4279)}=1.386$$$$\Rightarrow \mathrm{y}=-0.43\times 1.55=-0.558$$

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