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Question Number 165881 by bounhome last updated on 09/Feb/22
{ (((√(x/y))−(√(y/x))=(3/2))),((x+y+xy=9)) :}
$$\begin{cases}{\sqrt{\frac{{x}}{{y}}}−\sqrt{\frac{{y}}{{x}}}=\frac{\mathrm{3}}{\mathrm{2}}}\\{{x}+{y}+{xy}=\mathrm{9}}\end{cases} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Feb/22
{ (((√(x/y))−(√(y/x))=(3/2))),((x+y+xy=9)) :} (√(x/y)) =a: a−(1/a)=(3/2) 2a^2 −3a−2=0 (a−2)(2a+1)=0 a=2 ∣ a=−(1/2) (√(x/y)) =2 ∣ (√(x/y)) =−(1/2) (Rejected) (x/y)=4⇒x=4y x+y+xy=9⇒4y+y+4y×y=9 4y^2 +5y−9=0 (y−1)(4y+9)=0 y=1 ∣ y=−(9/4) x=4 ∣ x=−9 S_1 ={(4,1),(−9,−(9/4))}
$$\begin{cases}{\sqrt{\frac{{x}}{{y}}}−\sqrt{\frac{{y}}{{x}}}=\frac{\mathrm{3}}{\mathrm{2}}}\\{{x}+{y}+{xy}=\mathrm{9}}\end{cases} \\ $$$$\sqrt{\frac{{x}}{{y}}}\:={a}: \\ $$$${a}−\frac{\mathrm{1}}{{a}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{3}{a}−\mathrm{2}=\mathrm{0} \\ $$$$\left({a}−\mathrm{2}\right)\left(\mathrm{2}{a}+\mathrm{1}\right)=\mathrm{0} \\ $$$${a}=\mathrm{2}\:\mid\:{a}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sqrt{\frac{{x}}{{y}}}\:=\mathrm{2}\:\mid\:\sqrt{\frac{{x}}{{y}}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\left({Rejected}\right) \\ $$$$\:\:\:\frac{{x}}{{y}}=\mathrm{4}\Rightarrow{x}=\mathrm{4}{y} \\ $$$${x}+{y}+{xy}=\mathrm{9}\Rightarrow\mathrm{4}{y}+{y}+\mathrm{4}{y}×{y}=\mathrm{9} \\ $$$$\mathrm{4}{y}^{\mathrm{2}} +\mathrm{5}{y}−\mathrm{9}=\mathrm{0} \\ $$$$\left({y}−\mathrm{1}\right)\left(\mathrm{4}{y}+\mathrm{9}\right)=\mathrm{0} \\ $$$${y}=\mathrm{1}\:\mid\:{y}=−\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${x}=\mathrm{4}\:\mid\:{x}=−\mathrm{9} \\ $$$${S}_{\mathrm{1}} =\left\{\left(\mathrm{4},\mathrm{1}\right),\left(−\mathrm{9},−\frac{\mathrm{9}}{\mathrm{4}}\right)\right\} \\ $$
Answered by MJS_new last updated on 09/Feb/22
let t=(√(x/y))∧t≥0 t−(1/t)=(3/2) ⇒ t=2 (√(x/y))=2 ⇒ y=(x/4) x+(x/4)+x×(x/4)=9 ⇔ x^2 +5x−36=0 ⇒ x=−9∧y=−(9/4) ∨ x=4∧y=1
$$\mathrm{let}\:{t}=\sqrt{\frac{{x}}{{y}}}\wedge{t}\geqslant\mathrm{0} \\ $$$${t}−\frac{\mathrm{1}}{{t}}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{t}=\mathrm{2} \\ $$$$\sqrt{\frac{{x}}{{y}}}=\mathrm{2}\:\Rightarrow\:{y}=\frac{{x}}{\mathrm{4}} \\ $$$${x}+\frac{{x}}{\mathrm{4}}+{x}×\frac{{x}}{\mathrm{4}}=\mathrm{9}\:\Leftrightarrow\:{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{36}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\mathrm{9}\wedge{y}=−\frac{\mathrm{9}}{\mathrm{4}}\:\vee\:{x}=\mathrm{4}\wedge{y}=\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 10/Feb/22
{ (((√(x/y))−(√(y/x))=(3/2))),((x+y+xy=9)) :} An Alternate { ((((x−y)/( (√(xy))))=(3/2).........(i))),((x+y=9−xy.....(ii))) :} ⇒ { (((x−y)^2 =(((3(√(xy)) )/2))^2 ......(iii))),(((x+y)^2 =(9−xy)^2 .....(iv))) :} (iv)−(iii):4xy=(9−xy)^2 −(((3(√(xy)) )/2))^2 4xy=81−18xy+(xy)^2 −((9xy)/4) 4(xy)^2 −72xy−9xy−16xy+324=0 4(xy)^2 −97xy+324=0 xy=((−(−97)±(√((−97)^2 −4∙4∙324)))/(2∙4)) xy=((97±65)/8)=((162)/8) , ((32)/8)=((81)/4) , 4 xy=81/4 : (i): ((x−y)/( (√((81)/4))))=(3/2)⇒x−y=((27)/4) (ii): x+y=9−((81)/4)=−((45)/4) x=−(9/4),y=−9 xy=4 : (i): ((x−y)/( (√4)))=(3/2)⇒x−y=3 (ii): x+y=9−4=5 x=4,y=1
$$\begin{cases}{\sqrt{\frac{{x}}{{y}}}−\sqrt{\frac{{y}}{{x}}}=\frac{\mathrm{3}}{\mathrm{2}}}\\{{x}+{y}+{xy}=\mathrm{9}}\end{cases}\: \\ $$$$\mathbb{A}\mathrm{n}\:\mathbb{A}\mathrm{lternate} \\ $$$$\begin{cases}{\frac{{x}−{y}}{\:\sqrt{{xy}}}=\frac{\mathrm{3}}{\mathrm{2}}………\left({i}\right)}\\{{x}+{y}=\mathrm{9}−{xy}…..\left({ii}\right)}\end{cases}\: \\ $$$$\Rightarrow\begin{cases}{\left({x}−{y}\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}\sqrt{{xy}}\:}{\mathrm{2}}\right)^{\mathrm{2}} ……\left({iii}\right)}\\{\left({x}+{y}\right)^{\mathrm{2}} =\left(\mathrm{9}−{xy}\right)^{\mathrm{2}} …..\left({iv}\right)}\end{cases} \\ $$$$\left({iv}\right)−\left({iii}\right):\mathrm{4}{xy}=\left(\mathrm{9}−{xy}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}\sqrt{{xy}}\:}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{xy}=\mathrm{81}−\mathrm{18}{xy}+\left({xy}\right)^{\mathrm{2}} −\frac{\mathrm{9}{xy}}{\mathrm{4}} \\ $$$$\mathrm{4}\left({xy}\right)^{\mathrm{2}} −\mathrm{72}{xy}−\mathrm{9}{xy}−\mathrm{16}{xy}+\mathrm{324}=\mathrm{0} \\ $$$$\mathrm{4}\left({xy}\right)^{\mathrm{2}} −\mathrm{97}{xy}+\mathrm{324}=\mathrm{0} \\ $$$${xy}=\frac{−\left(−\mathrm{97}\right)\pm\sqrt{\left(−\mathrm{97}\right)^{\mathrm{2}} −\mathrm{4}\centerdot\mathrm{4}\centerdot\mathrm{324}}}{\mathrm{2}\centerdot\mathrm{4}} \\ $$$${xy}=\frac{\mathrm{97}\pm\mathrm{65}}{\mathrm{8}}=\frac{\mathrm{162}}{\mathrm{8}}\:,\:\frac{\mathrm{32}}{\mathrm{8}}=\frac{\mathrm{81}}{\mathrm{4}}\:,\:\mathrm{4} \\ $$$$\underline{{xy}=\mathrm{81}/\mathrm{4}\::} \\ $$$$\:\:\:\left({i}\right):\:\:\frac{{x}−{y}}{\:\sqrt{\frac{\mathrm{81}}{\mathrm{4}}}}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{x}−{y}=\frac{\mathrm{27}}{\mathrm{4}} \\ $$$$\:\:\:\left({ii}\right):\:{x}+{y}=\mathrm{9}−\frac{\mathrm{81}}{\mathrm{4}}=−\frac{\mathrm{45}}{\mathrm{4}} \\ $$$$\:\:\:{x}=−\frac{\mathrm{9}}{\mathrm{4}},{y}=−\mathrm{9} \\ $$$$\underline{{xy}=\mathrm{4}\::} \\ $$$$\:\:\:\:\left({i}\right):\:\:\frac{{x}−{y}}{\:\sqrt{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{x}−{y}=\mathrm{3} \\ $$$$\:\:\:\:\left({ii}\right):\:{x}+{y}=\mathrm{9}−\mathrm{4}=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{4},{y}=\mathrm{1} \\ $$

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