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Question Number 27507 by NECx last updated on 07/Jan/18

2(√(x ))+y=9....(1) x+ 2(√y)=3....(2) solve the simultaneous equation

$$\mathrm{2}\sqrt{{x}\:}+{y}=\mathrm{9}....\left(\mathrm{1}\right) \\ $$$${x}+\:\mathrm{2}\sqrt{{y}}=\mathrm{3}....\left(\mathrm{2}\right) \\ $$$$ \\ $$$${solve}\:{the}\:{simultaneous}\:{equation} \\ $$

Commented by Rasheed.Sindhi last updated on 08/Jan/18

(1)+(2): x+y+2((√x)+(√y))=12.....(3) (1)−(2): −(x−y)+2((√x)−(√y))=6...(4) Let (√x)+(√y)=u & (√x)−(√y)=v (a+b)^2 −(a−b)^2 =4ab u^2 −v^2 =4(√(xy)) ^• (√(xy))=((u^2 −v^2 )/4) ((√x)+(√y))^2 =u^2 x+y+2(√(xy))=u^2 ^• x+y=u^2 −2(((u^2 −v^2 )/4)) =((u^2 +v^2 )/2) (a−b)^2 =(a+b)^2 −4ab ^• (x−y)^2 =(((u^2 +v^2 )/2))^2 −4(((u^2 −v^2 )/4))^2 =(((u^2 +v^2 )^2 )/4)−(((u^2 −v^2 )^2 )/4) =((u^4 +2u^2 v^2 +v^4 −u^4 +2u^2 v^2 −v^4 )/4) (x−y)^2 =u^2 v^2 x−y=±uv (3):x+y+2((√x)+(√y))=12 ((u^2 +v^2 )/2)+2u=12 u^2 +4u+v^2 =24.......(5) (4):−(x−y)+2((√x)−(√y))=6 −(±uv)+2v=6 ∓uv+2v=6 v=(6/(∓u+2)) (5): u^2 +4u+((6/(∓u+2)))^2 =24 (5): u(u+4)+((36)/((∓u+2)^2 ))=24 u(u+4)(∓u+2)^2 +36=24(∓u+2)^2 Continue

$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right): \\ $$$$\:\mathrm{x}+\mathrm{y}+\mathrm{2}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)=\mathrm{12}.....\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right): \\ $$$$\:−\left(\mathrm{x}−\mathrm{y}\right)+\mathrm{2}\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\right)=\mathrm{6}...\left(\mathrm{4}\right) \\ $$$$\mathrm{Let}\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\mathrm{u}\:\&\:\:\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}=\mathrm{v} \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{4ab} \\ $$$$\:\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} =\mathrm{4}\sqrt{\mathrm{xy}} \\ $$$$\:^{\bullet} \sqrt{\mathrm{xy}}=\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)^{\mathrm{2}} =\mathrm{u}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}}=\mathrm{u}^{\mathrm{2}} \\ $$$$\:\:\:\:^{\bullet} \:\mathrm{x}+\mathrm{y}=\mathrm{u}^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} =\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{4ab}\:\:\:\: \\ $$$$\:\:^{\bullet} \:\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} =\left(\frac{\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} \:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}}−\frac{\left(\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{u}^{\mathrm{4}} +\mathrm{2u}^{\mathrm{2}} \mathrm{v}^{\mathrm{2}} +\mathrm{v}^{\mathrm{4}} −\mathrm{u}^{\mathrm{4}} +\mathrm{2u}^{\mathrm{2}} \mathrm{v}^{\mathrm{2}} −\mathrm{v}^{\mathrm{4}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} =\mathrm{u}^{\mathrm{2}} \mathrm{v}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}−\mathrm{y}=\pm\mathrm{uv} \\ $$$$\left(\mathrm{3}\right):\mathrm{x}+\mathrm{y}+\mathrm{2}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)=\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2u}=\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{v}^{\mathrm{2}} =\mathrm{24}.......\left(\mathrm{5}\right) \\ $$$$\left(\mathrm{4}\right):−\left(\mathrm{x}−\mathrm{y}\right)+\mathrm{2}\left(\sqrt{\mathrm{x}}−\sqrt{\mathrm{y}}\right)=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:−\left(\pm\mathrm{uv}\right)+\mathrm{2v}=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mp\mathrm{uv}+\mathrm{2v}=\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{v}=\frac{\mathrm{6}}{\mp\mathrm{u}+\mathrm{2}} \\ $$$$\:\left(\mathrm{5}\right):\:\:\:\:\mathrm{u}^{\mathrm{2}} +\mathrm{4u}+\left(\frac{\mathrm{6}}{\mp\mathrm{u}+\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{24} \\ $$$$\:\left(\mathrm{5}\right):\:\:\:\:\mathrm{u}\left(\mathrm{u}+\mathrm{4}\right)+\frac{\mathrm{36}}{\left(\mp\mathrm{u}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{24} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{u}\left(\mathrm{u}+\mathrm{4}\right)\left(\mp\mathrm{u}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{36}=\mathrm{24}\left(\mp\mathrm{u}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{Continue} \\ $$

Commented by prakash jain last updated on 08/Jan/18

2(√x)+y=9 y_1 =9−2(√x) x+2(√y)=3 2(√y)=3−x for real y,x≤3 y_2 =(1/4)(3−x)^2 u=y_1 −y_2 =9−2(√x)−(1/4)(3−x)^2 for real solution x≤3 x≤3⇒9−2(√x)≥9−2(√3)>5 (1/4)(3−x)^2 ≤(9/4) u≥5−(9/4)>0⇒no real solution

$$\mathrm{2}\sqrt{{x}}+{y}=\mathrm{9} \\ $$$${y}_{\mathrm{1}} =\mathrm{9}−\mathrm{2}\sqrt{{x}}\: \\ $$$${x}+\mathrm{2}\sqrt{{y}}=\mathrm{3} \\ $$$$\mathrm{2}\sqrt{{y}}=\mathrm{3}−{x} \\ $$$$\mathrm{for}\:\mathrm{real}\:{y},{x}\leqslant\mathrm{3} \\ $$$${y}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}−{x}\right)^{\mathrm{2}} \\ $$$${u}={y}_{\mathrm{1}} −{y}_{\mathrm{2}} =\mathrm{9}−\mathrm{2}\sqrt{{x}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}−{x}\right)^{\mathrm{2}} \\ $$$${for}\:{real}\:{solution}\:{x}\leqslant\mathrm{3} \\ $$$${x}\leqslant\mathrm{3}\Rightarrow\mathrm{9}−\mathrm{2}\sqrt{{x}}\geqslant\mathrm{9}−\mathrm{2}\sqrt{\mathrm{3}}>\mathrm{5} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}−{x}\right)^{\mathrm{2}} \leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${u}\geqslant\mathrm{5}−\frac{\mathrm{9}}{\mathrm{4}}>\mathrm{0}\Rightarrow{no}\:{real}\:{solution} \\ $$

Commented by Rasheed.Sindhi last updated on 08/Jan/18

New way for me to attack such problem!

$$\mathrm{New}\:\mathrm{way}\:\mathrm{for}\:\mathrm{me}\:\mathrm{to}\:\mathrm{attack}\:\mathrm{such} \\ $$$$\mathrm{problem}! \\ $$

Commented by prakash jain last updated on 08/Jan/18

Actually finding the complex solutions is tough.

Answered by jota@ last updated on 09/Jan/18

there is not a real solution.

$${there}\:{is}\:{not}\:{a}\:{real}\:{solution}. \\ $$

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