{ ((((15)/(x^2 +xy)) + (3/(y^2 +xy)) = 5,5)),((((12)/(x^2 +xy))


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Question Number 146535 by mathdanisur last updated on 13/Jul/21
${ ((((15)/(x^2 +xy)) + (3/(y^2 +xy)) = 5,5)),((((12)/(x^2 +xy)) - (6/(y^2 +xy)) = 3)) :} ⇒ ∣x+y∣=?$
$$\begin{cases}{\frac{\mathrm{15}}{{x}^{\mathrm{2}} +{xy}}\:+\:\frac{\mathrm{3}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{5},\mathrm{5}}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}\:-\:\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{3}}\end{cases}\:\:\:\Rightarrow\:\:\mid{x}+{y}\mid=? \\ $$
	Answered by liberty last updated on 14/Jul/21
	${ ((((30)/(x^2 +xy))+(6/(y^2 +xy))=11)),((((12)/(x^2 +xy))−(6/(y^2 +xy))=3)) :} (1)+(2)→((42^3 )/(x^2 +xy))=14^1 ⇒x^2 +xy=3 ⇒y^2 +xy=6 ⇒(1)+(2) (x+y)^2 =9 { ((x=3−y or)),((x=−3−y)) :}⇒∣x+y∣=(√((x+y)^2 ))=3$
	$$\:\begin{cases}{\frac{\mathrm{30}}{{x}^{\mathrm{2}} +{xy}}+\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}=\mathrm{11}}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}−\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}=\mathrm{3}}\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\rightarrow\frac{\cancel{\mathrm{42}}\:^{\mathrm{3}} }{{x}^{\mathrm{2}} +{xy}}=\cancel{\mathrm{14}}\:^{\mathrm{1}} \: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{xy}=\mathrm{3}\:\: \\ $$$$\Rightarrow{y}^{\mathrm{2}} +{xy}=\mathrm{6}\: \\ $$$$\Rightarrow\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{9}\: \\ $$$$\begin{cases}{{x}=\mathrm{3}−{y}\:{or}}\\{{x}=−\mathrm{3}−{y}}\end{cases}\Rightarrow\mid{x}+{y}\mid=\sqrt{\left({x}+{y}\right)^{\mathrm{2}} }=\mathrm{3} \\ $$
	Commented by mathdanisur last updated on 14/Jul/21

	$${thanks}\:{Ser}\:{cool} \\ $$
	Answered by Olaf_Thorendsen last updated on 14/Jul/21
	${ ((((15)/(x^2 +xy))+(3/(y^2 +xy)) = 5,5 (×2))),((((12)/(x^2 +xy))−(6/(y^2 +xy)) = 3)) :} { ((((30)/(x^2 +xy))+(6/(y^2 +xy)) = 11 (1))),((((12)/(x^2 +xy))−(6/(y^2 +xy)) = 3 (2))) :} (1)+(2) : ((42)/(x^2 +xy)) = 14 (1/(x^2 +xy)) = (1/3) x^2 +xy = 3 (3) (1) : (6/(y^2 +xy)) = 11−((30)/(x^2 +xy)) (1) : (6/(y^2 +xy)) = 11−30×(1/3) = 1 y^2 +xy = 6 (4) (3)+(4) : x^2 +xy+y^2 +xy = 3+6 (x+y)^2 = 9 ∣x+y∣ = 3$
	$$\begin{cases}{\frac{\mathrm{15}}{{x}^{\mathrm{2}} +{xy}}+\frac{\mathrm{3}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{5},\mathrm{5}\:\:\:\:\:\left(×\mathrm{2}\right)}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}−\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{3}}\end{cases} \\ $$$$\begin{cases}{\frac{\mathrm{30}}{{x}^{\mathrm{2}} +{xy}}+\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{11}\:\:\:\:\:\left(\mathrm{1}\right)}\\{\frac{\mathrm{12}}{{x}^{\mathrm{2}} +{xy}}−\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{3}\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\::\:\frac{\mathrm{42}}{{x}^{\mathrm{2}} +{xy}}\:=\:\mathrm{14} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{xy}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} +{xy}\:=\:\mathrm{3}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)\::\:\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{11}−\frac{\mathrm{30}}{{x}^{\mathrm{2}} +{xy}} \\ $$$$\left(\mathrm{1}\right)\::\:\frac{\mathrm{6}}{{y}^{\mathrm{2}} +{xy}}\:=\:\mathrm{11}−\mathrm{30}×\frac{\mathrm{1}}{\mathrm{3}}\:=\:\mathrm{1} \\ $$$${y}^{\mathrm{2}} +{xy}\:=\:\mathrm{6}\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)+\left(\mathrm{4}\right)\::\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} +{xy}\:=\:\mathrm{3}+\mathrm{6} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} \:=\:\mathrm{9} \\ $$$$\mid{x}+{y}\mid\:=\:\mathrm{3} \\ $$$$ \\ $$
	Commented by mathdanisur last updated on 14/Jul/21

	$${thanks}\:{Ser}\:{cool} \\ $$
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