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Question Number 111611 by mathdave last updated on 04/Sep/20
solve for x ad y in 5x(1+(1/(x^2 +y^2 )))=12.........(1) 5y(1−(1/(x^2 +y^2 )))=4..........(2)
$${solve}\:{for}\:{x}\:{ad}\:{y}\:{in} \\ $$$$\mathrm{5}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{12}………\left(\mathrm{1}\right) \\ $$$$\mathrm{5}{y}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{4}……….\left(\mathrm{2}\right) \\ $$
Answered by Her_Majesty last updated on 04/Sep/20
y=px (1) 5x^2 −12x+(5/(p^2 +1))=0 (2) 5x^2 −(4/p)x−(5/(p^2 +1))=0 (1)−(2) ⇒ x=((5p)/(2(3p−1)(p^2 +1))) (1)+(2) ⇒ x=((2(3p+1))/(5p)) ⇒ ((5p)/(2(3p−1)(p^2 +1)))=((2(3p+1))/(5p)) p^4 +(7/(36))p^2 −(1/9)=0 ⇒ p=±(1/2)∨p=±(2/3)i ⇒ (x; y)∈{((2/5); −(1/5)); (2; 1); ((6/5)±(3/5)i; (2/5)∓(4/5)i)}
$${y}={px} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{5}{x}^{\mathrm{2}} −\mathrm{12}{x}+\frac{\mathrm{5}}{{p}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{5}{x}^{\mathrm{2}} −\frac{\mathrm{4}}{{p}}{x}−\frac{\mathrm{5}}{{p}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\:\Rightarrow\:{x}=\frac{\mathrm{5}{p}}{\mathrm{2}\left(\mathrm{3}{p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\Rightarrow\:{x}=\frac{\mathrm{2}\left(\mathrm{3}{p}+\mathrm{1}\right)}{\mathrm{5}{p}} \\ $$$$\Rightarrow\:\frac{\mathrm{5}{p}}{\mathrm{2}\left(\mathrm{3}{p}−\mathrm{1}\right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{2}\left(\mathrm{3}{p}+\mathrm{1}\right)}{\mathrm{5}{p}} \\ $$$${p}^{\mathrm{4}} +\frac{\mathrm{7}}{\mathrm{36}}{p}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow\:{p}=\pm\frac{\mathrm{1}}{\mathrm{2}}\vee{p}=\pm\frac{\mathrm{2}}{\mathrm{3}}\mathrm{i} \\ $$$$\Rightarrow\:\left({x};\:{y}\right)\in\left\{\left(\frac{\mathrm{2}}{\mathrm{5}};\:−\frac{\mathrm{1}}{\mathrm{5}}\right);\:\left(\mathrm{2};\:\mathrm{1}\right);\:\left(\frac{\mathrm{6}}{\mathrm{5}}\pm\frac{\mathrm{3}}{\mathrm{5}}\mathrm{i};\:\frac{\mathrm{2}}{\mathrm{5}}\mp\frac{\mathrm{4}}{\mathrm{5}}\mathrm{i}\right)\right\} \\ $$
Answered by mathdave last updated on 04/Sep/20
solution to 5x(1+(1/(x^2 +y^2 )))=12........(1) and 5y(1−(1/(x^2 +y^2 )))=4.........(2) let decompose the equations to make them similar that is 5x(1+(1/(x^2 +y^2 )))=5•(2.4)........(3) 5y(1−(1/(x^2 +y^2 )))=5•(0.8)..........(4) comparing decomposed equations 5x=5.........(5) x=1 substitute the value of x in (6) below 1+(1/(x^2 +y^2 ))=2.4..........(6) 1+(1/(1+y^2 ))=2.4⇒2+y^2 =2.4+2.4y^2 −1.4y^2 =0.4 y=0.535i(3d.p) and from equation (4) comparing sides 5y=5........(7) y=1 substituting for the value of y below 1−(1/(x^2 +y^2 ))=0.8.........(8) x^2 +1−1=0.8x^2 +0.8 0.2x^2 =0.8 x=2 ∵(x_1 ,y_1 ) and (x_2 ,y_2 )=(1,0.535i) and (2,1) by mathdave(04/09/2020)
$${solution}\:{to}\:\:\:\:\: \\ $$$$\mathrm{5}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{12}……..\left(\mathrm{1}\right)\:\:{and} \\ $$$$\mathrm{5}{y}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{4}………\left(\mathrm{2}\right) \\ $$$${let}\:{decompose}\:{the}\:{equations}\:{to}\:{make}\:{them}\:{similar} \\ $$$${that}\:{is} \\ $$$$\mathrm{5}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{5}\bullet\left(\mathrm{2}.\mathrm{4}\right)……..\left(\mathrm{3}\right) \\ $$$$\mathrm{5}{y}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{5}\bullet\left(\mathrm{0}.\mathrm{8}\right)……….\left(\mathrm{4}\right) \\ $$$${comparing}\:{decomposed}\:{equations} \\ $$$$\mathrm{5}{x}=\mathrm{5}………\left(\mathrm{5}\right) \\ $$$${x}=\mathrm{1} \\ $$$${substitute}\:{the}\:{value}\:{of}\:{x}\:{in}\:\left(\mathrm{6}\right)\:{below} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{2}.\mathrm{4}……….\left(\mathrm{6}\right) \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} }=\mathrm{2}.\mathrm{4}\Rightarrow\mathrm{2}+{y}^{\mathrm{2}} =\mathrm{2}.\mathrm{4}+\mathrm{2}.\mathrm{4}{y}^{\mathrm{2}} \\ $$$$−\mathrm{1}.\mathrm{4}{y}^{\mathrm{2}} =\mathrm{0}.\mathrm{4}\:\:\:\: \\ $$$${y}=\mathrm{0}.\mathrm{535}{i}\left(\mathrm{3}{d}.{p}\right) \\ $$$${and}\:{from}\:{equation}\:\:\left(\mathrm{4}\right)\:\:{comparing}\:{sides} \\ $$$$\mathrm{5}{y}=\mathrm{5}……..\left(\mathrm{7}\right) \\ $$$${y}=\mathrm{1} \\ $$$${substituting}\:{for}\:{the}\:{value}\:{of}\:{y}\:{below}\: \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{0}.\mathrm{8}………\left(\mathrm{8}\right) \\ $$$${x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}=\mathrm{0}.\mathrm{8}{x}^{\mathrm{2}} +\mathrm{0}.\mathrm{8} \\ $$$$\mathrm{0}.\mathrm{2}{x}^{\mathrm{2}} =\mathrm{0}.\mathrm{8} \\ $$$${x}=\mathrm{2} \\ $$$$\because\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:\:{and}\:\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)=\left(\mathrm{1},\mathrm{0}.\mathrm{535}{i}\right)\:\:{and}\:\:\left(\mathrm{2},\mathrm{1}\right) \\ $$$${by}\:{mathdave}\left(\mathrm{04}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$

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