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Question Number 143328 by Rankut last updated on 13/Jun/21
x+(√(xy))+y=14 and x^2 +xy+y^2 =84 , find x and y
$${x}+\sqrt{{xy}}+{y}=\mathrm{14}\:\:{and}\: \\ $$$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{84}\:, \\ $$$${find}\:{x}\:{and}\:{y} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jun/21
x+(√(xy))+y=14...(i) x^2 +xy+y^2 =84...(ii) (x+(√(xy))+y)^2 =14^2 x^2 +xy+y^2 +2x(√(xy))+2y(√(xy))+2xy=196 84+2x(√(xy))+2y(√(xy))+2xy=196 x(√(xy))+y(√(xy))+xy=56 (√(xy))(x+y+(√(xy)))=56 (√(xy))(14)=56 (√(xy))=4 xy=16⇒y=((16)/x) (i)⇒x+y=10⇒x+((16)/x)=10 x^2 −10x+16=0 (x−8)(x−2)=0 x=8,2⇒y=2,8 (x,y)={(8,2),(2,8)}
$${x}+\sqrt{{xy}}+{y}=\mathrm{14}…\left({i}\right)\: \\ $$$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{84}…\left({ii}\right) \\ $$$$\left({x}+\sqrt{{xy}}+{y}\right)^{\mathrm{2}} =\mathrm{14}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{xy}}+\mathrm{2}{y}\sqrt{{xy}}+\mathrm{2}{xy}=\mathrm{196} \\ $$$$\mathrm{84}+\mathrm{2}{x}\sqrt{{xy}}+\mathrm{2}{y}\sqrt{{xy}}+\mathrm{2}{xy}=\mathrm{196} \\ $$$${x}\sqrt{{xy}}+{y}\sqrt{{xy}}+{xy}=\mathrm{56} \\ $$$$\sqrt{{xy}}\left({x}+{y}+\sqrt{{xy}}\right)=\mathrm{56} \\ $$$$\sqrt{{xy}}\left(\mathrm{14}\right)=\mathrm{56} \\ $$$$\sqrt{{xy}}=\mathrm{4} \\ $$$$\:\:\:\:{xy}=\mathrm{16}\Rightarrow{y}=\frac{\mathrm{16}}{{x}} \\ $$$$\left({i}\right)\Rightarrow{x}+{y}=\mathrm{10}\Rightarrow{x}+\frac{\mathrm{16}}{{x}}=\mathrm{10} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{16}=\mathrm{0} \\ $$$$\left({x}−\mathrm{8}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:{x}=\mathrm{8},\mathrm{2}\Rightarrow{y}=\mathrm{2},\mathrm{8} \\ $$$$\left({x},{y}\right)=\left\{\left(\mathrm{8},\mathrm{2}\right),\left(\mathrm{2},\mathrm{8}\right)\right\} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jun/21
{ ((x+(√(xy))+y=14⇒(x+y)^2 =(14−(√(xy)))^2 )),((x^2 +xy+y^2 =84⇒(x+y)^2 =84+xy)) :} ⇒(14−(√(xy)))^2 =84+xy 196+xy^(×) −28(√(xy))=84+xy^(×) 28(√(xy))=196−84=112 xy=16 x+(√(xy))+y=14⇒x+y=10 (x,y)={(2,8),(8,2)}
$$\begin{cases}{{x}+\sqrt{{xy}}+{y}=\mathrm{14}\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\left(\mathrm{14}−\sqrt{{xy}}\right)^{\mathrm{2}} }\\{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{84}\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{84}+{xy}}\end{cases} \\ $$$$\Rightarrow\left(\mathrm{14}−\sqrt{{xy}}\right)^{\mathrm{2}} =\mathrm{84}+{xy} \\ $$$$\:\:\:\:\:\mathrm{196}+\overset{×} {{xy}}−\mathrm{28}\sqrt{{xy}}=\mathrm{84}+\overset{×} {{xy}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{28}\sqrt{{xy}}=\mathrm{196}−\mathrm{84}=\mathrm{112} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xy}=\mathrm{16} \\ $$$${x}+\sqrt{{xy}}+{y}=\mathrm{14}\Rightarrow{x}+{y}=\mathrm{10} \\ $$$$\left({x},{y}\right)=\left\{\left(\mathrm{2},\mathrm{8}\right),\left(\mathrm{8},\mathrm{2}\right)\right\} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jun/21
x+(√(xy))+y=14^((i)) ∧ x^2 +xy+y^2 =84^((ii)) (i)⇒xy=(14−x−y)^2 (ii)⇒xy=84−x^2 −y^2 (i)&(ii)⇒(14−x−y)^2 =84−x^2 −y^2 196+x^2 +y^2 −28x+2xy−28y=84−x^2 −y^2 2x^2 +2y^2 −28x+2xy−28y=84−196 x^2 +y^2 −14x+xy−14y=−56 84−xy−14x+xy−14y=−56 −14x−14y=−56−84=−140 x+y=10 (i):x+(√(xy))+y=14 ⇒(√(xy))=14−10=4 xy=16 x+y=10∧xy=16⇒(x,y)={(2,8),(8,2)}
$$\overset{\left({i}\right)} {{x}+\sqrt{{xy}}+{y}=\mathrm{14}}\:\wedge\:\overset{\left({ii}\right)} {{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{84}} \\ $$$$\left({i}\right)\Rightarrow{xy}=\left(\mathrm{14}−{x}−{y}\right)^{\mathrm{2}} \\ $$$$\left({ii}\right)\Rightarrow{xy}=\mathrm{84}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow\left(\mathrm{14}−{x}−{y}\right)^{\mathrm{2}} =\mathrm{84}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\mathrm{196}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{2}{xy}−\mathrm{28}{y}=\mathrm{84}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{28}{x}+\mathrm{2}{xy}−\mathrm{28}{y}=\mathrm{84}−\mathrm{196} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{14}{x}+{xy}−\mathrm{14}{y}=−\mathrm{56} \\ $$$$\mathrm{84}−{xy}−\mathrm{14}{x}+{xy}−\mathrm{14}{y}=−\mathrm{56} \\ $$$$−\mathrm{14}{x}−\mathrm{14}{y}=−\mathrm{56}−\mathrm{84}=−\mathrm{140} \\ $$$$\:\:\:\:\:\:\:\:\:{x}+{y}=\mathrm{10} \\ $$$$\left({i}\right):{x}+\sqrt{{xy}}+{y}=\mathrm{14}\:\Rightarrow\sqrt{{xy}}=\mathrm{14}−\mathrm{10}=\mathrm{4} \\ $$$${xy}=\mathrm{16} \\ $$$${x}+{y}=\mathrm{10}\wedge{xy}=\mathrm{16}\Rightarrow\left({x},{y}\right)=\left\{\left(\mathrm{2},\mathrm{8}\right),\left(\mathrm{8},\mathrm{2}\right)\right\} \\ $$

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