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Question Number 108295 by bobhans last updated on 16/Aug/20

((BobHans)/(βo♭)) (1) { (((x+y)(x^2 −y^2 ) = 9)),(((x−y)(x^2 +y^2 ) = 5)) :} find the solution (2) x (dy/dx) = x^2 +y^2 when x=1 give y = 2

$$\:\:\:\:\frac{\mathbb{B}\mathrm{ob}\mathbb{H}\mathrm{ans}}{\beta\mathrm{o}\flat} \\ $$$$\:\left(\mathrm{1}\right)\begin{cases}{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{9}}\\{\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{5}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\mathrm{when}\:\mathrm{x}=\mathrm{1}\:\mathrm{give}\:\mathrm{y}\:=\:\mathrm{2}\: \\ $$

Answered by john santu last updated on 16/Aug/20

((⇈ JS ⇈)/♥) ⇒ { (((x+y)^2 (x−y) = 9)),((x−y = (5/(x^2 +y^2 )))) :} ⇒5(x+y)^2 = 9x^2 +9y^2 ⇒5x^2 +10xy+5y^2 = 9x^2 +9y^2 ⇒4x^2 −10xy+4y^2 =0 ⇒2x^2 −5xy+2y^2 =0 ⇒(2x−y)(x−2y)=0 { ((y=2x⇒3x.(−3x^2 )=9→ { ((x=−1)),((y=−2)) :})),((x=2y⇒3y.(3y^2 )=9→ { ((y=1)),((x=2)) :})) :} then solution set is {(−1,−2),(2,1)}

$$\:\:\:\:\:\frac{\upuparrows\:{JS}\:\upuparrows}{\heartsuit} \\ $$$$\Rightarrow\begin{cases}{\left({x}+{y}\right)^{\mathrm{2}} \left({x}−{y}\right)\:=\:\mathrm{9}}\\{{x}−{y}\:=\:\frac{\mathrm{5}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\end{cases} \\ $$$$\Rightarrow\mathrm{5}\left({x}+{y}\right)^{\mathrm{2}} \:=\:\mathrm{9}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{xy}+\mathrm{5}{y}^{\mathrm{2}} \:=\:\mathrm{9}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} −\mathrm{10}{xy}+\mathrm{4}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−{y}\right)\left({x}−\mathrm{2}{y}\right)=\mathrm{0} \\ $$$$\begin{cases}{{y}=\mathrm{2}{x}\Rightarrow\mathrm{3}{x}.\left(−\mathrm{3}{x}^{\mathrm{2}} \right)=\mathrm{9}\rightarrow\begin{cases}{{x}=−\mathrm{1}}\\{{y}=−\mathrm{2}}\end{cases}}\\{{x}=\mathrm{2}{y}\Rightarrow\mathrm{3}{y}.\left(\mathrm{3}{y}^{\mathrm{2}} \right)=\mathrm{9}\rightarrow\begin{cases}{{y}=\mathrm{1}}\\{{x}=\mathrm{2}}\end{cases}}\end{cases} \\ $$$${then}\:{solution}\:{set}\:{is}\:\left\{\left(−\mathrm{1},−\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right)\right\} \\ $$$$\:\: \\ $$

Commented by bemath last updated on 16/Aug/20

cooll

$${cooll} \\ $$

Commented by bobhans last updated on 16/Aug/20

good

$$\mathrm{good} \\ $$

Answered by 1549442205PVT last updated on 16/Aug/20

{ (((x+y)(x^2 −y^2 ) = 9)),(((x−y)(x^2 +y^2 ) = 5)) :} (1) If x−y=0 then LHS =0 ⇒the given system has no solutions,so x−y≠0 From (1)we have 5(x+y)(x^2 −y^2 ) =9(x−y)(x^2 +y^2 ) ⇔(x−y)[5(x+y)^2 −9(x^2 +y^2 )]=0 ⇔5(x+y)^2 −9(x^2 +y^2 )=0 ⇔4(x^2 +y^2 )−10xy=0.Put x=ky(k≠0) we get 4k^2 −10k+4=0⇔2k^2 −5k+2=0 ⇒k∈{2;1/2} i)For k=2⇒x=2y .Replce into first equation we get 3y.3y^2 =9⇔y^3 =1⇔y=1⇒x=2 we have solution (x,y)=(2,1) ii)For k=1/2⇔y=2x, replace into first equation we get −x.(−3x^2 )=9⇔x^3 =1⇔x=1⇒y=2 we get solution (x;y)=(1;2) Thus,given system of equations has two solutions (x;y)∈{(1,2);(2;1)}

$$\:\begin{cases}{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{9}}\\{\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{5}}\end{cases}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{If}\:\mathrm{x}−\mathrm{y}=\mathrm{0}\:\mathrm{then}\:\mathrm{LHS}\:=\mathrm{0}\:\Rightarrow\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{system}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solutions},\mathrm{so}\:\mathrm{x}−\mathrm{y}\neq\mathrm{0} \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\mathrm{we}\:\mathrm{have}\:\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)\: \\ $$$$=\mathrm{9}\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\Leftrightarrow\left(\mathrm{x}−\mathrm{y}\right)\left[\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} −\mathrm{9}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\right]=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} −\mathrm{9}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)−\mathrm{10xy}=\mathrm{0}.\mathrm{Put}\:\mathrm{x}=\mathrm{ky}\left(\mathrm{k}\neq\mathrm{0}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{4k}^{\mathrm{2}} −\mathrm{10k}+\mathrm{4}=\mathrm{0}\Leftrightarrow\mathrm{2k}^{\mathrm{2}} −\mathrm{5k}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{k}\in\left\{\mathrm{2};\mathrm{1}/\mathrm{2}\right\} \\ $$$$\left.\mathrm{i}\right)\mathrm{For}\:\mathrm{k}=\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{2y}\:.\mathrm{Replce}\:\mathrm{into}\:\mathrm{first} \\ $$$$\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{3y}.\mathrm{3y}^{\mathrm{2}} =\mathrm{9}\Leftrightarrow\mathrm{y}^{\mathrm{3}} =\mathrm{1}\Leftrightarrow\mathrm{y}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{2}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{solution}\:\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{2},\mathrm{1}\right) \\ $$$$\left.\mathrm{ii}\right)\mathrm{For}\:\mathrm{k}=\mathrm{1}/\mathrm{2}\Leftrightarrow\mathrm{y}=\mathrm{2x},\:\mathrm{replace}\:\mathrm{into}\:\mathrm{first}\:\mathrm{equation} \\ $$$$\mathrm{we}\:\mathrm{get}\:−\mathrm{x}.\left(−\mathrm{3x}^{\mathrm{2}} \right)=\mathrm{9}\Leftrightarrow\mathrm{x}^{\mathrm{3}} =\mathrm{1}\Leftrightarrow\mathrm{x}=\mathrm{1}\Rightarrow\mathrm{y}=\mathrm{2} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{solution}\:\left(\mathrm{x};\mathrm{y}\right)=\left(\mathrm{1};\mathrm{2}\right) \\ $$$$\mathrm{Thus},\mathrm{given}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{has}\: \\ $$$$\mathrm{two}\:\mathrm{solutions}\:\left(\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}\right)\in\left\{\left(\mathrm{1},\mathrm{2}\right);\left(\mathrm{2};\mathrm{1}\right)\right\} \\ $$$$ \\ $$

Commented by bemath last updated on 16/Aug/20

if (1,2)⇒(1+2)(1−4)=−9≠9 wrong sir

$${if}\:\left(\mathrm{1},\mathrm{2}\right)\Rightarrow\left(\mathrm{1}+\mathrm{2}\right)\left(\mathrm{1}−\mathrm{4}\right)=−\mathrm{9}\neq\mathrm{9} \\ $$$${wrong}\:{sir} \\ $$

Commented by bemath last updated on 16/Aug/20

yes sir. thank you

$${yes}\:{sir}.\:{thank}\:{you} \\ $$

Commented by 1549442205PVT last updated on 16/Aug/20

Thank you.I mistaked .The cases k=(1/2) should correct as:y=2x,so replace into first eqn.⇒3x.(−3x^2 )=9⇔x^3 =−1 ⇔x=−1⇒y=−2⇒(x,y)=(−1,−2)

$$\mathrm{Thank}\:\mathrm{you}.\mathrm{I}\:\mathrm{mistaked}\:.\mathrm{The}\:\mathrm{cases}\:\mathrm{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{should}\:\mathrm{correct}\:\mathrm{as}:\mathrm{y}=\mathrm{2x},\mathrm{so}\:\mathrm{replace}\:\mathrm{into} \\ $$$$\mathrm{first}\:\mathrm{eqn}.\Rightarrow\mathrm{3x}.\left(−\mathrm{3x}^{\mathrm{2}} \right)=\mathrm{9}\Leftrightarrow\mathrm{x}^{\mathrm{3}} =−\mathrm{1} \\ $$$$\Leftrightarrow\boldsymbol{\mathrm{x}}=−\mathrm{1}\Rightarrow\boldsymbol{\mathrm{y}}=−\mathrm{2}\Rightarrow\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)=\left(−\mathrm{1},−\mathrm{2}\right) \\ $$

Answered by Dwaipayan Shikari last updated on 16/Aug/20

(((x+y)(x^2 −y^2 ))/((x−y)(x^2 +y^2 )))=(9/5) (((x+y)^2 )/((x^2 +y^2 )))=(9/5) 1+((2xy)/(x^2 +y^2 ))=(9/5) ((2xy)/(x^2 +y^2 ))=(4/5) ((x^2 +y^2 )/(2xy))=(5/4) ((x^2 +y^2 +2xy)/(x^2 +y^2 −2xy))=(9/1) (((x+y)/(x−y)))^2 =9 ((x+y)/(x−y))=3 or ((x+y)/(x−y))=−3 (x/y)=2 or (x/y)=(1/2) , x=k_0 y=2k_(0 ) so 3k_0 .(−3k_0 ^2 )=9 ⇒k_0 =−1 x=2k y=k (x+y)(x^2 −y^2 )=9 3k.3k^2 =9 k^3 =1 k=1 { ((x=2)),((y=1)) :} or { ((x=−1)),((y=−2)) :}

$$\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}{\left({x}−{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\frac{\left({x}+{y}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\mathrm{1}+\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}{xy}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}}=\frac{\mathrm{9}}{\mathrm{1}} \\ $$$$\left(\frac{{x}+{y}}{{x}−{y}}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\frac{{x}+{y}}{{x}−{y}}=\mathrm{3}\:\:\:\:\:\:\:{or}\:\:\:\:\:\frac{{x}+{y}}{{x}−{y}}=−\mathrm{3} \\ $$$$\frac{{x}}{{y}}=\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:{or}\:\:\:\:\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:,\:\:{x}={k}_{\mathrm{0}} \:\:{y}=\mathrm{2}{k}_{\mathrm{0}\:\:} \:\:\:{so}\:\:\mathrm{3}{k}_{\mathrm{0}} .\left(−\mathrm{3}{k}_{\mathrm{0}} ^{\mathrm{2}} \right)=\mathrm{9}\:\Rightarrow{k}_{\mathrm{0}} =−\mathrm{1} \\ $$$${x}=\mathrm{2}{k}\:\:\:\:\: \\ $$$${y}={k} \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{9} \\ $$$$\mathrm{3}{k}.\mathrm{3}{k}^{\mathrm{2}} =\mathrm{9} \\ $$$${k}^{\mathrm{3}} =\mathrm{1} \\ $$$${k}=\mathrm{1} \\ $$$$\begin{cases}{{x}=\mathrm{2}}\\{{y}=\mathrm{1}}\end{cases}\:\:\:\:\:\:\:\:{or\begin{cases}{{x}=−\mathrm{1}}\\{{y}=−\mathrm{2}}\end{cases}} \\ $$

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