solve for x and y: x^2 +y^2 =10 (1/x)+(1/y)=(4/3)


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Question Number 171825 by Mikenice last updated on 21/Jun/22

$${solve}\:{for}\:{x}\:{and}\:{y}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{10} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$
	Answered by Rasheed.Sindhi last updated on 21/Jun/22
	$x^2 +y^2 =10.......(i) (1/x)+(1/y)=(4/3)........(ii) (i)⇒(1/x^2 )+(1/y^2 )=((10)/(x^2 y^2 )) (ii)⇒(1/x^2 )+(1/y^2 )+(2/(xy))=((16)/9) ((10)/(x^2 y^2 ))=((16)/9)−(2/(xy)) 10((1/(xy)))^2 +2((1/(xy)))−((16)/9)=0 45a^2 +9a−8=0; (1/(xy))=a (3a−1)(15a+8)=0 a=(1/3) ∣ a=−(8/(15)) xy=3 ∣ xy=−((15)/8) (i)⇒(x+y)^2 −2xy=10 (x+y)^2 −2(3)=10 { (((x+y)^2 −2(3)=10⇒x+y=±4...(iii))),(((x+y)^2 −2(−((15)/8))=10⇒x+y=±(5/2)...(iv))) :} (iii)⇒x+(3/x)=±4 x^2 ∓4x+3=0 x=((±4±(√(16−12)))/2)=((±4±2)/2)=3,−3 y=(3/x)=1,−1 (x,y)_1 ={(3,1),(−3,−1)} x+y=±(5/2)⇒x−((15)/(8x))=±(5/2) 8x^2 ∓20x−15=0 x=((±20±(√(400+480)))/(16))=((±20±4(√(55)))/(16)) x=((±5±(√(55)))/4) y=−((15)/(8x))=((15)/(8(((±5±(√(55)))/4))))=((15)/(2(±5±(√(55)))))$
	$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{10}.......\left({i}\right) \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{4}}{\mathrm{3}}........\left({ii}\right) \\ $$$$\left({i}\right)\Rightarrow\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=\frac{\mathrm{10}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\left({ii}\right)\Rightarrow\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\frac{\mathrm{2}}{{xy}}=\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$\frac{\mathrm{10}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }=\frac{\mathrm{16}}{\mathrm{9}}−\frac{\mathrm{2}}{{xy}} \\ $$$$\mathrm{10}\left(\frac{\mathrm{1}}{{xy}}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{{xy}}\right)−\frac{\mathrm{16}}{\mathrm{9}}=\mathrm{0} \\ $$$$\mathrm{45}{a}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{8}=\mathrm{0};\:\:\:\frac{\mathrm{1}}{{xy}}={a} \\ $$$$\left(\mathrm{3}{a}−\mathrm{1}\right)\left(\mathrm{15}{a}+\mathrm{8}\right)=\mathrm{0} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}}\:\mid\:{a}=−\frac{\mathrm{8}}{\mathrm{15}} \\ $$$${xy}=\mathrm{3}\:\mid\:{xy}=−\frac{\mathrm{15}}{\mathrm{8}} \\ $$$$\left({i}\right)\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}\right)=\mathrm{10} \\ $$$$\:\:\:\:\begin{cases}{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}\right)=\mathrm{10}\Rightarrow{x}+{y}=\pm\mathrm{4}...\left({iii}\right)}\\{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}\left(−\frac{\mathrm{15}}{\mathrm{8}}\right)=\mathrm{10}\Rightarrow{x}+{y}=\pm\frac{\mathrm{5}}{\mathrm{2}}...\left({iv}\right)}\end{cases} \\ $$$$\left({iii}\right)\Rightarrow{x}+\frac{\mathrm{3}}{{x}}=\pm\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \mp\mathrm{4}{x}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{\pm\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{12}}}{\mathrm{2}}=\frac{\pm\mathrm{4}\pm\mathrm{2}}{\mathrm{2}}=\mathrm{3},−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{y}=\frac{\mathrm{3}}{{x}}=\mathrm{1},−\mathrm{1} \\ $$$$\left({x},{y}\right)_{\mathrm{1}} =\left\{\left(\mathrm{3},\mathrm{1}\right),\left(−\mathrm{3},−\mathrm{1}\right)\right\} \\ $$$${x}+{y}=\pm\frac{\mathrm{5}}{\mathrm{2}}\Rightarrow{x}−\frac{\mathrm{15}}{\mathrm{8}{x}}=\pm\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{8}{x}^{\mathrm{2}} \mp\mathrm{20}{x}−\mathrm{15}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\frac{\pm\mathrm{20}\pm\sqrt{\mathrm{400}+\mathrm{480}}}{\mathrm{16}}=\frac{\pm\mathrm{20}\pm\mathrm{4}\sqrt{\mathrm{55}}}{\mathrm{16}} \\ $$$$\:\:\:\:\:{x}=\frac{\pm\mathrm{5}\pm\sqrt{\mathrm{55}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:{y}=−\frac{\mathrm{15}}{\mathrm{8}{x}}=\frac{\mathrm{15}}{\mathrm{8}\left(\frac{\pm\mathrm{5}\pm\sqrt{\mathrm{55}}}{\mathrm{4}}\right)}=\frac{\mathrm{15}}{\mathrm{2}\left(\pm\mathrm{5}\pm\sqrt{\mathrm{55}}\right)} \\ $$
	Commented by Mikenice last updated on 21/Jun/22

	$${thanks}\:{sir} \\ $$
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