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Question Number 144148 by bobhans last updated on 22/Jun/21

{ ((2ln x+ln y = 2)),((x^2 + y = e^2 +1)) :}

$$\begin{cases}{\mathrm{2ln}\:\mathrm{x}+\mathrm{ln}\:\mathrm{y}\:=\:\mathrm{2}}\\{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{2}} +\mathrm{1}}\end{cases} \\ $$

Answered by EDWIN88 last updated on 22/Jun/21

{ ((x^2 y = e^2 )),((x^2 +y = e^2 +1)) :} ⇒ (e^2 /y)+y−(e^2 +1)=0 ⇒y^2 −(e^2 +1)y+e^2 = 0 ⇒y = ((e^2 +1+ (√((e^2 +1)^2 −4e^2 )))/2) ⇒y=((e^2 +1+(√(e^4 −2e^2 +1)))/2)=((e^2 +1+e^2 −1)/2)=e^2 ⇒x^2 =(e^2 /e^2 ) =1 ; x=1

$$\:\begin{cases}{\mathrm{x}^{\mathrm{2}} \mathrm{y}\:=\:\mathrm{e}^{\mathrm{2}} }\\{\mathrm{x}^{\mathrm{2}} +\mathrm{y}\:=\:\mathrm{e}^{\mathrm{2}} +\mathrm{1}}\end{cases}\:\Rightarrow\:\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{y}}+\mathrm{y}−\left(\mathrm{e}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{y}^{\mathrm{2}} −\left(\mathrm{e}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{y}+\mathrm{e}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{y}\:=\:\frac{\mathrm{e}^{\mathrm{2}} +\mathrm{1}+\:\sqrt{\left(\mathrm{e}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4e}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{y}=\frac{\mathrm{e}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{e}^{\mathrm{4}} −\mathrm{2e}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}=\frac{\mathrm{e}^{\mathrm{2}} +\mathrm{1}+\mathrm{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}=\mathrm{e}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{e}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{2}} }\:=\mathrm{1}\:;\:\mathrm{x}=\mathrm{1} \\ $$$$ \\ $$

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