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Question-194509

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Question Number 194509 by pascal889 last updated on 08/Jul/23
Answered by horsebrand11 last updated on 09/Jul/23
{: ((u=x+2y)),((v=2x+y)) }⇒3(x+y)=u+v ⇒ {: ((u+v=uv)),(((1/u)+(1/v^2 )=3)) }⇒ { ((u+v^2 =3uv^2 )),((u+v=uv)) :} ⇒ { ((u=(v^2 /(3v^2 −1)))),((u=(v/(v−1)))) :} ⇒(v^2 /(3v^2 −1))=(v/(v−1)) ⇒v((v/(3v^2 −1)) −(1/(v−1)))=0 (1) v=0 (rejected) (2) (v/(3v^2 −1)) = (1/(v−1)) v^2 −v= 3v^2 −1 ⇒2v^2 +v−1=0 (2v−1)(v+1)=0 (3) v+1=0 ⇒ { ((v=−1⇒u=(1/2))),( { ((x+2y=(1/2))),((2x+y=−1)) :}) :} { ((2x+4y=1)),((2x+y=−1)) :}⇒ { ((y=(2/3))),((x=−(5/6))) :} (4)v=(1/2)⇒u=−1 ⇒ { ((x+2y=−1)),((2x+y=(1/2))) :} ⇒ { ((x+2y=−1)),((4x+2y=1)) :}⇒ { ((x=(2/3))),((y=−(5/6))) :} ∴ solution (x ,y)={(−(5/6),(2/3)),((2/3),−(5/6))}
$$\:\:\left.\begin{matrix}{\mathrm{u}=\mathrm{x}+\mathrm{2y}}\\{\mathrm{v}=\mathrm{2x}+\mathrm{y}}\end{matrix}\right\}\Rightarrow\mathrm{3}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{u}+\mathrm{v} \\ $$$$\:\Rightarrow\left.\begin{matrix}{\mathrm{u}+\mathrm{v}=\mathrm{uv}}\\{\frac{\mathrm{1}}{\mathrm{u}}+\frac{\mathrm{1}}{\mathrm{v}^{\mathrm{2}} }=\mathrm{3}}\end{matrix}\right\}\Rightarrow\begin{cases}{\mathrm{u}+\mathrm{v}^{\mathrm{2}} =\mathrm{3uv}^{\mathrm{2}} }\\{\mathrm{u}+\mathrm{v}=\mathrm{uv}}\end{cases} \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{u}=\frac{\mathrm{v}^{\mathrm{2}} }{\mathrm{3v}^{\mathrm{2}} −\mathrm{1}}}\\{\mathrm{u}=\frac{\mathrm{v}}{\mathrm{v}−\mathrm{1}}}\end{cases}\:\Rightarrow\frac{\mathrm{v}^{\mathrm{2}} }{\mathrm{3v}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{v}}{\mathrm{v}−\mathrm{1}} \\ $$$$\:\Rightarrow\mathrm{v}\left(\frac{\mathrm{v}}{\mathrm{3v}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{v}−\mathrm{1}}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{v}=\mathrm{0}\:\left(\mathrm{rejected}\right) \\ $$$$\:\left(\mathrm{2}\right)\:\frac{\mathrm{v}}{\mathrm{3v}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{v}−\mathrm{1}} \\ $$$$\:\:\mathrm{v}^{\mathrm{2}} −\mathrm{v}=\:\mathrm{3v}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\mathrm{2v}^{\mathrm{2}} +\mathrm{v}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\left(\mathrm{2v}−\mathrm{1}\right)\left(\mathrm{v}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\left(\mathrm{3}\right)\:\mathrm{v}+\mathrm{1}=\mathrm{0}\:\Rightarrow\begin{cases}{\mathrm{v}=−\mathrm{1}\Rightarrow\mathrm{u}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\begin{cases}{\mathrm{x}+\mathrm{2y}=\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{2x}+\mathrm{y}=−\mathrm{1}}\end{cases}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{2x}+\mathrm{4y}=\mathrm{1}}\\{\mathrm{2x}+\mathrm{y}=−\mathrm{1}}\end{cases}\Rightarrow\begin{cases}{\mathrm{y}=\frac{\mathrm{2}}{\mathrm{3}}}\\{\mathrm{x}=−\frac{\mathrm{5}}{\mathrm{6}}}\end{cases} \\ $$$$\:\left(\mathrm{4}\right)\mathrm{v}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{u}=−\mathrm{1}\:\Rightarrow\begin{cases}{\mathrm{x}+\mathrm{2y}=−\mathrm{1}}\\{\mathrm{2x}+\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\:\:\Rightarrow\begin{cases}{\mathrm{x}+\mathrm{2y}=−\mathrm{1}}\\{\mathrm{4x}+\mathrm{2y}=\mathrm{1}}\end{cases}\Rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{2}}{\mathrm{3}}}\\{\mathrm{y}=−\frac{\mathrm{5}}{\mathrm{6}}}\end{cases} \\ $$$$\:\:\therefore\:\mathrm{solution}\:\left(\mathrm{x}\:,\mathrm{y}\right)=\left\{\left(−\frac{\mathrm{5}}{\mathrm{6}},\frac{\mathrm{2}}{\mathrm{3}}\right),\left(\frac{\mathrm{2}}{\mathrm{3}},−\frac{\mathrm{5}}{\mathrm{6}}\right)\right\} \\ $$
Commented by pascal889 last updated on 10/Jul/23
please sir how is 3(x+y)=u+v i dont understand the side sir
$$\mathrm{please}\:\mathrm{sir}\:\mathrm{how}\:\mathrm{is}\:\mathrm{3}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)=\boldsymbol{\mathrm{u}}+\boldsymbol{\mathrm{v}} \\ $$$$\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{dont}}\:\boldsymbol{\mathrm{understand}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{sir}} \\ $$
Answered by Frix last updated on 09/Jul/23
Let y=px ⇒ x≠0 and transform (1) (p+2)(2p+1)x^2 −3(p+1)x=0 (2) 3(p+2)^2 (2p+1)x^3 −(p+2)^2 x^2 −(2p+1)x=0 (1) x=((3(p+1))/((p+2)(2p+1))) ⇒ (2) p^2 +((41p)/(20))+1=0 ⇒ p=−(5/4)∨p=−(4/5) ⇒ x=(2/3)∧y=−(5/6) ∨ x=−(5/6)∧y=(2/3)
$$\mathrm{Let}\:{y}={px}\:\Rightarrow\:{x}\neq\mathrm{0}\:\mathrm{and}\:\mathrm{transform} \\ $$$$\left(\mathrm{1}\right)\:\:\left({p}+\mathrm{2}\right)\left(\mathrm{2}{p}+\mathrm{1}\right){x}^{\mathrm{2}} −\mathrm{3}\left({p}+\mathrm{1}\right){x}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{3}\left({p}+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{2}{p}+\mathrm{1}\right){x}^{\mathrm{3}} −\left({p}+\mathrm{2}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\left(\mathrm{2}{p}+\mathrm{1}\right){x}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\:{x}=\frac{\mathrm{3}\left({p}+\mathrm{1}\right)}{\left({p}+\mathrm{2}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\left(\mathrm{2}\right)\:\:{p}^{\mathrm{2}} +\frac{\mathrm{41}{p}}{\mathrm{20}}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{p}=−\frac{\mathrm{5}}{\mathrm{4}}\vee{p}=−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{3}}\wedge{y}=−\frac{\mathrm{5}}{\mathrm{6}}\:\vee\:{x}=−\frac{\mathrm{5}}{\mathrm{6}}\wedge{y}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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