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Question Number 83756 by naka3546 last updated on 05/Mar/20

Find  all  real  solutions  of  (x, y)  such  that           x + 3y + (4/(x + y))  =  5           y + 3x + (5/(x + y))  =  7

$${Find}\:\:{all}\:\:{real}\:\:{solutions}\:\:{of}\:\:\left({x},\:{y}\right)\:\:{such}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:{x}\:+\:\mathrm{3}{y}\:+\:\frac{\mathrm{4}}{{x}\:+\:{y}}\:\:=\:\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:{y}\:+\:\mathrm{3}{x}\:+\:\frac{\mathrm{5}}{{x}\:+\:{y}}\:\:=\:\:\mathrm{7} \\ $$

Commented by Prithwish Sen 1 last updated on 05/Mar/20

adding both  4(x+y) + (9/((x+y))) = 12  4a^2 −12a +9 = 0   putting x+y = a  a=(3/2)  ⇒x+y = (3/2)...(i)  substracting  2(x−y)+(1/(x+y)) = 2  x−y=(2/3) ....(ii)  from (i) and (ii) we get  x=((13)/(12)) and y = (5/(12))  please check

$$\mathrm{adding}\:\mathrm{both} \\ $$$$\mathrm{4}\left(\mathrm{x}+\mathrm{y}\right)\:+\:\frac{\mathrm{9}}{\left(\mathrm{x}+\mathrm{y}\right)}\:=\:\mathrm{12} \\ $$$$\mathrm{4a}^{\mathrm{2}} −\mathrm{12a}\:+\mathrm{9}\:=\:\mathrm{0}\:\:\:\boldsymbol{\mathrm{putting}}\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{a}} \\ $$$$\boldsymbol{\mathrm{a}}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\Rightarrow\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}...\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\mathrm{substracting} \\ $$$$\mathrm{2}\left(\mathrm{x}−\mathrm{y}\right)+\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}\:=\:\mathrm{2} \\ $$$$\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}=\frac{\mathrm{2}}{\mathrm{3}}\:....\left(\boldsymbol{\mathrm{ii}}\right) \\ $$$$\boldsymbol{\mathrm{from}}\:\left(\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{and}}\:\left(\boldsymbol{\mathrm{ii}}\right)\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}} \\ $$$$\boldsymbol{\mathrm{x}}=\frac{\mathrm{13}}{\mathrm{12}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}\:=\:\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

Commented by naka3546 last updated on 06/Mar/20

thank you , sir .

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

Commented by Prithwish Sen 1 last updated on 06/Mar/20

welcome

$$\mathrm{welcome} \\ $$

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