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Question Number 107999 by Don08q last updated on 13/Aug/20

Commented by udaythool last updated on 13/Aug/20

⇒4x^2 +3xy−y^2 =0 ⇒(x+y)(4x−y)=0 ⇒4x=y; (∵ x+y≠0) ⇒4x^2 =xy ⇒5xy=5 ⇒xy=1

$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{xy}−{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({x}+{y}\right)\left(\mathrm{4}{x}−{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{x}={y};\:\:\:\left(\because\:{x}+{y}\neq\mathrm{0}\right) \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} ={xy} \\ $$$$\Rightarrow\mathrm{5}{xy}=\mathrm{5} \\ $$$$\Rightarrow{xy}=\mathrm{1} \\ $$

Commented by Don08q last updated on 13/Aug/20

Thanks Sir

$${Thanks}\:{Sir} \\ $$

Commented by Rasheed.Sindhi last updated on 13/Aug/20

Why x+y≠0? Such condition is not included in question.

$${Why}\:{x}+{y}\neq\mathrm{0}?\:{Such}\:{condition}\:{is} \\ $$$${not}\:{included}\:{in}\:{question}. \\ $$

Commented by Sarah85 last updated on 13/Aug/20

if x+y=0 ⇔ y=−x ⇔ x=−y 4x^2 −4x^2 =5 y^2 −y^2 =5 both are impossible

$$\mathrm{if}\:{x}+{y}=\mathrm{0}\:\Leftrightarrow\:{y}=−{x}\:\Leftrightarrow\:{x}=−{y} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} =\mathrm{5} \\ $$$${y}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{5} \\ $$$$\mathrm{both}\:\mathrm{are}\:\mathrm{impossible} \\ $$

Commented by Rasheed.Sindhi last updated on 14/Aug/20

THANKS madam!

$$\mathcal{THANKS}\:{madam}! \\ $$

Answered by Sarah85 last updated on 13/Aug/20

x=α−β∧y=α+β 8α^2 −8αβ=5 ⇒ β=((8α^2 −5)/(8α)) 2α^2 +2αβ=5 ⇒ β=((5−2α^2 )/(2α)) ((8α^2 −5)/(8α))=((5−2α^2 )/(2α)) 8α^2 −5=20−8α^2 16α^2 =25 α=±(5/4) ⇒ β=±(3/4) ⇒ x=±(1/2)∧y=±2 or y=αx 4x^2 +4αx^2 =5 α^2 x^2 +αx^2 =5 x^2 =(5/(4(1+α))) x^2 =(5/(α(1+α))) ⇒ α=4 ⇒ y=4x 4x^2 +16x^2 =5 ⇒ x^2 =(1/4) ⇒ y^2 =4 or just test if y=(1/x) solves both equations 4x^2 +4=5 ⇒ x^2 =(1/4) (1/x^2 )+1=5 ⇒ x^2 =(1/4) ⇒ y^2 =4

$${x}=\alpha−\beta\wedge{y}=\alpha+\beta \\ $$$$\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{8}\alpha\beta=\mathrm{5}\:\Rightarrow\:\beta=\frac{\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{5}}{\mathrm{8}\alpha} \\ $$$$\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{2}\alpha\beta=\mathrm{5}\:\Rightarrow\:\beta=\frac{\mathrm{5}−\mathrm{2}\alpha^{\mathrm{2}} }{\mathrm{2}\alpha} \\ $$$$\frac{\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{5}}{\mathrm{8}\alpha}=\frac{\mathrm{5}−\mathrm{2}\alpha^{\mathrm{2}} }{\mathrm{2}\alpha} \\ $$$$\mathrm{8}\alpha^{\mathrm{2}} −\mathrm{5}=\mathrm{20}−\mathrm{8}\alpha^{\mathrm{2}} \\ $$$$\mathrm{16}\alpha^{\mathrm{2}} =\mathrm{25} \\ $$$$\alpha=\pm\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow\:\beta=\pm\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\wedge{y}=\pm\mathrm{2} \\ $$$$ \\ $$$$\mathrm{or}\:{y}=\alpha{x} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}\alpha{x}^{\mathrm{2}} =\mathrm{5} \\ $$$$\alpha^{\mathrm{2}} {x}^{\mathrm{2}} +\alpha{x}^{\mathrm{2}} =\mathrm{5} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}\left(\mathrm{1}+\alpha\right)} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{5}}{\alpha\left(\mathrm{1}+\alpha\right)} \\ $$$$\Rightarrow\:\alpha=\mathrm{4}\:\Rightarrow\:{y}=\mathrm{4}{x} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{16}{x}^{\mathrm{2}} =\mathrm{5}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:{y}^{\mathrm{2}} =\mathrm{4} \\ $$$$ \\ $$$$\mathrm{or}\:\mathrm{just}\:\mathrm{test}\:\mathrm{if}\:{y}=\frac{\mathrm{1}}{{x}}\:\mathrm{solves}\:\mathrm{both}\:\mathrm{equations} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}=\mathrm{5}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}=\mathrm{5}\:\Rightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} =\mathrm{4} \\ $$

Answered by 1549442205PVT last updated on 14/Aug/20

4x^2 +4xy=y^2 +xy⇒4x^2 +3xy−y^2 =0 Δ=(3y)^2 −4.4.(−y^2 )=25y^2 ⇒x=((−3y±5y)/8)∈{−2y;(y/4)} i)If x=−2y then replace into the second equation we get y^2 +(−2y)y=5⇔−y^2 =5 (rejected) ii)If x=(y/4)⇔y=4x then we get (4x)^2 +x(4x)=5⇔20x^2 =5⇔x^2 =(1/4) ⇔x=±(1/2)⇒y=4x=±2 Clearly we have xy=(±(1/2))×(±2)=1 Hence y is the multiplicative inverse of x (q.e.d)

$$\mathrm{4x}^{\mathrm{2}} +\mathrm{4xy}=\mathrm{y}^{\mathrm{2}} +\mathrm{xy}\Rightarrow\mathrm{4x}^{\mathrm{2}} +\mathrm{3xy}−\mathrm{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(\mathrm{3y}\right)^{\mathrm{2}} −\mathrm{4}.\mathrm{4}.\left(−\mathrm{y}^{\mathrm{2}} \right)=\mathrm{25y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}=\frac{−\mathrm{3y}\pm\mathrm{5y}}{\mathrm{8}}\in\left\{−\mathrm{2y};\frac{\mathrm{y}}{\mathrm{4}}\right\} \\ $$$$\left.\mathrm{i}\right)\mathrm{If}\:\mathrm{x}=−\mathrm{2y}\:\mathrm{then}\:\mathrm{replace}\:\mathrm{into}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{y}^{\mathrm{2}} +\left(−\mathrm{2y}\right)\mathrm{y}=\mathrm{5}\Leftrightarrow−\mathrm{y}^{\mathrm{2}} =\mathrm{5}\:\left(\mathrm{rejected}\right) \\ $$$$\left.\mathrm{ii}\right)\mathrm{If}\:\mathrm{x}=\frac{\mathrm{y}}{\mathrm{4}}\Leftrightarrow\mathrm{y}=\mathrm{4x}\:\:\mathrm{then}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{4x}\right)^{\mathrm{2}} +\mathrm{x}\left(\mathrm{4x}\right)=\mathrm{5}\Leftrightarrow\mathrm{20x}^{\mathrm{2}} =\mathrm{5}\Leftrightarrow\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{y}=\mathrm{4x}=\pm\mathrm{2} \\ $$$$\mathrm{Clearly}\:\mathrm{we}\:\mathrm{have}\:\mathrm{xy}=\left(\pm\frac{\mathrm{1}}{\mathrm{2}}\right)×\left(\pm\mathrm{2}\right)=\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{y}\:\mathrm{is}\:\mathrm{the}\:\mathrm{multiplicative}\:\mathrm{inverse} \\ $$$$\mathrm{of}\:\mathrm{x}\:\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

Commented by Don08q last updated on 14/Aug/20

Thanks Sir

$${Thanks}\:{Sir} \\ $$

Answered by Rasheed.Sindhi last updated on 14/Aug/20

y is inverse of x⇒xy=1.So now the question in other words is: “To show the following system consistent ”: { ((4x^2 +4xy=5)),((y^2 +xy=5)),((xy=1)) :} I-E the system has common solution. ⇒ { ((4x^2 +4(1)=5⇒x=±(1/2))),((y^2 +(1)=5⇒y=±2)) :} Since ((1/2),2) & (−(1/2),−2) are solutions satisfying the system so the system is consistent. (Evident from solutions that y is multiplicative inverse of x.)

$${y}\:{is}\:{inverse}\:{of}\:{x}\Rightarrow{xy}=\mathrm{1}.\mathrm{So}\:\mathrm{now} \\ $$$$\mathrm{the}\:\mathrm{question}\:\mathrm{in}\:\mathrm{other}\:\mathrm{words}\:\mathrm{is}: \\ $$$$``{To}\:{show}\:{the}\:{following}\:{system} \\ $$$$\:\:\:\boldsymbol{{consistent}}\:'': \\ $$$$\:\begin{cases}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{xy}=\mathrm{5}}\\{{y}^{\mathrm{2}} +{xy}=\mathrm{5}}\\{{xy}=\mathrm{1}}\end{cases} \\ $$$${I}-{E}\:\:{the}\:{system}\:{has}\:{common}\: \\ $$$${solution}. \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{1}\right)=\mathrm{5}\Rightarrow{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}}\\{{y}^{\mathrm{2}} +\left(\mathrm{1}\right)=\mathrm{5}\Rightarrow{y}=\pm\mathrm{2}}\end{cases} \\ $$$${Since}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{2}\right)\:\&\:\left(−\frac{\mathrm{1}}{\mathrm{2}},−\mathrm{2}\right)\:{are} \\ $$$${solutions}\:{satisfying}\:{the}\:{system} \\ $$$${so}\:{the}\:{system}\:{is}\:\boldsymbol{{consistent}}. \\ $$$$\left({Evident}\:{from}\:{solutions}\:{that}\right. \\ $$$$\left.{y}\:{is}\:{multiplicative}\:{inverse}\:{of}\:{x}.\right) \\ $$

Commented by Don08q last updated on 14/Aug/20

Thanks Sir

$${Thanks}\:{Sir} \\ $$

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