Question Number 163627 by nurtani last updated on 08/Jan/22
Answered by mr W last updated on 08/Jan/22
![u=(1/(x+y)) >0 v=xy (√(21v))(1−u^2 )=8(√2) v=((128)/(21(1−u^2 )^2 )) x=(4/(3(1+u)^2 )) y=((32)/(7(1−u)^2 )) u=(1/(x+y))=(1/((4/(3(1+u)^2 ))+((32)/(7(1−u)^2 ))))=((21(1−u^2 )^2 )/(4[7(1−u)^2 +24(1+u)^2 ])) 4u=((21−42u^2 +21u^4 )/(31u^2 +34u+31)) 21u^4 −124u^3 −178u^2 −124u+21=0 (3u^2 −22u+3)(7u^2 +10u+7)=0 3u^2 −22u+3=0 u=((11−4(√7))/3) x+y=(1/u)=((11+4(√7))/3) xy=v=((128)/(21(1−u^2 )^2 ))=((466+176(√7))/(147)) x,y are roots of z^2 −((11+4(√7))/3)z+((466+176(√7))/(147))=0 x,y=(1/2){((11+4(√7))/3)±(√((((11+4(√7))/3))^2 −4(((466+176(√7))/(147)))))} x=(1/6){11+4(√7)−((5(√(233+88(√7))))/7)}≈1.0278 y=(1/6){11+4(√7)+((5(√(233+88(√7))))/7)}≈6.1666](https://www.tinkutara.com/question/Q163637.png)
$${u}=\frac{\mathrm{1}}{{x}+{y}}\:>\mathrm{0} \\ $$$${v}={xy} \\ $$$$\sqrt{\mathrm{21}{v}}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)=\mathrm{8}\sqrt{\mathrm{2}} \\ $$$${v}=\frac{\mathrm{128}}{\mathrm{21}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${x}=\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}+{u}\right)^{\mathrm{2}} } \\ $$$${y}=\frac{\mathrm{32}}{\mathrm{7}\left(\mathrm{1}−{u}\right)^{\mathrm{2}} } \\ $$$${u}=\frac{\mathrm{1}}{{x}+{y}}=\frac{\mathrm{1}}{\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }+\frac{\mathrm{32}}{\mathrm{7}\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }}=\frac{\mathrm{21}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}\left[\mathrm{7}\left(\mathrm{1}−{u}\right)^{\mathrm{2}} +\mathrm{24}\left(\mathrm{1}+{u}\right)^{\mathrm{2}} \right]} \\ $$$$\mathrm{4}{u}=\frac{\mathrm{21}−\mathrm{42}{u}^{\mathrm{2}} +\mathrm{21}{u}^{\mathrm{4}} }{\mathrm{31}{u}^{\mathrm{2}} +\mathrm{34}{u}+\mathrm{31}} \\ $$$$\mathrm{21}{u}^{\mathrm{4}} −\mathrm{124}{u}^{\mathrm{3}} −\mathrm{178}{u}^{\mathrm{2}} −\mathrm{124}{u}+\mathrm{21}=\mathrm{0} \\ $$$$\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{22}{u}+\mathrm{3}\right)\left(\mathrm{7}{u}^{\mathrm{2}} +\mathrm{10}{u}+\mathrm{7}\right)=\mathrm{0} \\ $$$$\mathrm{3}{u}^{\mathrm{2}} −\mathrm{22}{u}+\mathrm{3}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{11}−\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$${x}+{y}=\frac{\mathrm{1}}{{u}}=\frac{\mathrm{11}+\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$${xy}={v}=\frac{\mathrm{128}}{\mathrm{21}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{466}+\mathrm{176}\sqrt{\mathrm{7}}}{\mathrm{147}}\: \\ $$$${x},{y}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −\frac{\mathrm{11}+\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{3}}{z}+\frac{\mathrm{466}+\mathrm{176}\sqrt{\mathrm{7}}}{\mathrm{147}}=\mathrm{0} \\ $$$${x},{y}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{11}+\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{3}}\pm\sqrt{\left(\frac{\mathrm{11}+\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{466}+\mathrm{176}\sqrt{\mathrm{7}}}{\mathrm{147}}\right)}\right\} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{6}}\left\{\mathrm{11}+\mathrm{4}\sqrt{\mathrm{7}}−\frac{\mathrm{5}\sqrt{\mathrm{233}+\mathrm{88}\sqrt{\mathrm{7}}}}{\mathrm{7}}\right\}\approx\mathrm{1}.\mathrm{0278} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{6}}\left\{\mathrm{11}+\mathrm{4}\sqrt{\mathrm{7}}+\frac{\mathrm{5}\sqrt{\mathrm{233}+\mathrm{88}\sqrt{\mathrm{7}}}}{\mathrm{7}}\right\}\approx\mathrm{6}.\mathrm{1666} \\ $$
Commented by nurtani last updated on 09/Jan/22
$${Thank}\:{you}\:{sir} \\ $$
Commented by Tawa11 last updated on 09/Jan/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by behi834171 last updated on 09/Jan/22
![x≠0,y≠0 (2/( (√(3x))))+((4(√2))/( (√(7y))))=2⇒((2(√2))/( (√(7y))))=1−(1/( (√(3x))))⇒ ⇒(8/(7y))=1−(2/( (√(3x))))+(1/(3x))⇒y=(8/(7(1−(2/( (√(3x))))+(1/(3x))))) ⇒(√(3x))(1+(1/(x+(8/(7(1−(2/( (√(3x))))+(1/(3x))))))))=2⇒^((1/( (√(3x))))=m) (1/m)(1+(1/((1/(3m^2 ))+(8/(7−14m+7m^2 )))))=2⇒ (1/(3m^2 ))+(8/(7−14m+7m^2 ))=(1/(2m−1))⇒ (8/(7−14m+7m^2 ))=((3m^2 −2m+1)/(3m^2 (2m−1)))⇒ ⇒24m^2 (2m−1)=7(m^2 −2m+1)(3m^2 −2m+1)⇒ ⇒48m^3 −24m^2 =7(3m^4 −8m^3 +8m^2 −4m+1)⇒ ⇒21m^4 −104m^3 +80m^2 −28m+7=0 ⇒m=0.57,m=4.1 ⇒(1/(3x))=m^2 ⇒x=(1/(3m^2 ))⇒[x=1.03 , 0.021] ⇒(8/(7y))=1−2m+m^2 ⇒ [y=6.2 , 0.12] .■](https://www.tinkutara.com/question/Q163643.png)
$${x}\neq\mathrm{0},{y}\neq\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}{x}}}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}{y}}}=\mathrm{2}\Rightarrow\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}{y}}}=\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}{x}}}\Rightarrow \\ $$$$\Rightarrow\frac{\mathrm{8}}{\mathrm{7}{y}}=\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}{x}}}+\frac{\mathrm{1}}{\mathrm{3}{x}}\Rightarrow{y}=\frac{\mathrm{8}}{\mathrm{7}\left(\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}{x}}}+\frac{\mathrm{1}}{\mathrm{3}{x}}\right)} \\ $$$$\Rightarrow\sqrt{\mathrm{3}{x}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}+\frac{\mathrm{8}}{\mathrm{7}\left(\mathrm{1}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}{x}}}+\frac{\mathrm{1}}{\mathrm{3}{x}}\right)}}\right)=\mathrm{2}\overset{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}{x}}}={m}} {\Rightarrow} \\ $$$$\frac{\mathrm{1}}{{m}}\left(\mathrm{1}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{3}{m}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{7}−\mathrm{14}{m}+\mathrm{7}{m}^{\mathrm{2}} }}\right)=\mathrm{2}\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{3}{m}^{\mathrm{2}} }+\frac{\mathrm{8}}{\mathrm{7}−\mathrm{14}{m}+\mathrm{7}{m}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{m}−\mathrm{1}}\Rightarrow \\ $$$$\frac{\mathrm{8}}{\mathrm{7}−\mathrm{14}{m}+\mathrm{7}{m}^{\mathrm{2}} }=\frac{\mathrm{3}{m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{1}}{\mathrm{3}{m}^{\mathrm{2}} \left(\mathrm{2}{m}−\mathrm{1}\right)}\Rightarrow \\ $$$$\Rightarrow\mathrm{24}{m}^{\mathrm{2}} \left(\mathrm{2}{m}−\mathrm{1}\right)=\mathrm{7}\left({m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{1}\right)\left(\mathrm{3}{m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{1}\right)\Rightarrow \\ $$$$\Rightarrow\mathrm{48}{m}^{\mathrm{3}} −\mathrm{24}{m}^{\mathrm{2}} =\mathrm{7}\left(\mathrm{3}{m}^{\mathrm{4}} −\mathrm{8}{m}^{\mathrm{3}} +\mathrm{8}{m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{1}\right)\Rightarrow \\ $$$$\Rightarrow\mathrm{21}{m}^{\mathrm{4}} −\mathrm{104}{m}^{\mathrm{3}} +\mathrm{80}{m}^{\mathrm{2}} −\mathrm{28}{m}+\mathrm{7}=\mathrm{0} \\ $$$$\Rightarrow{m}=\mathrm{0}.\mathrm{57},{m}=\mathrm{4}.\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}{x}}={m}^{\mathrm{2}} \Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}{m}^{\mathrm{2}} }\Rightarrow\left[{x}=\mathrm{1}.\mathrm{03}\:,\:\mathrm{0}.\mathrm{021}\right] \\ $$$$\Rightarrow\frac{\mathrm{8}}{\mathrm{7}{y}}=\mathrm{1}−\mathrm{2}{m}+{m}^{\mathrm{2}} \Rightarrow\:\:\:\:\left[{y}=\mathrm{6}.\mathrm{2}\:,\:\mathrm{0}.\mathrm{12}\right]\:\:\:\:\:\:\:\:.\blacksquare \\ $$
Commented by nurtani last updated on 09/Jan/22
$${thank}\:{you}\:{sir} \\ $$
Commented by mr W last updated on 09/Jan/22
$${only}\:{one}\:{solution}\:{fulfills}\:{the} \\ $$$${original}\:{equations}! \\ $$