Question-159469

Question Number 159469 by 0731619 last updated on 17/Nov/21

Commented by cortano last updated on 17/Nov/21

(√x) = u and (√y) = v { ((u^3 +v^3 = 35)),((u^2 v+ uv^2 = 30)) :} { ((u^3 +v^3 = 35⇒(u+v)^3 −3uv(u+v)=35)),((uv (u+v)= 30)) :} ⇒(u+v)^3 = 125 ⇒u+v=5 ⇒uv = 6 ⇒u(5−u)=6 ⇒u^2 −5u+6=0 ⇒ { ((u=3 ; v=2)),((u=2 ; v=3)) :} ⇒ { ((x=9 ; y=4)),((x=4 ; y=9)) :}

$$\:\sqrt{{x}}\:=\:{u}\:{and}\:\sqrt{{y}}\:=\:{v}\: \\ $$$$\:\begin{cases}{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:=\:\mathrm{35}}\\{{u}^{\mathrm{2}} \:{v}+\:{uv}^{\mathrm{2}} \:=\:\mathrm{30}}\end{cases} \\ $$$$\:\begin{cases}{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} \:=\:\mathrm{35}\Rightarrow\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}{uv}\left({u}+{v}\right)=\mathrm{35}}\\{{uv}\:\left({u}+{v}\right)=\:\mathrm{30}}\end{cases} \\ $$$$\Rightarrow\left({u}+{v}\right)^{\mathrm{3}} \:=\:\mathrm{125}\:\Rightarrow{u}+{v}=\mathrm{5} \\ $$$$\Rightarrow{uv}\:=\:\mathrm{6}\:\Rightarrow{u}\left(\mathrm{5}−{u}\right)=\mathrm{6} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −\mathrm{5}{u}+\mathrm{6}=\mathrm{0}\:\Rightarrow\begin{cases}{{u}=\mathrm{3}\:;\:{v}=\mathrm{2}}\\{{u}=\mathrm{2}\:;\:{v}=\mathrm{3}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{x}=\mathrm{9}\:;\:{y}=\mathrm{4}}\\{{x}=\mathrm{4}\:;\:{y}=\mathrm{9}}\end{cases} \\ $$

Answered by FongXD last updated on 17/Nov/21

{ (((√x^3 )+(√y^3 )=35 (1))),(((√(x^2 y))+(√(xy^2 ))=30 (2))) :} • take (1)+3(2): ⇔ (√x^3 )+3(√(x^2 y))+3(√(xy^2 ))+(√y^3 )=125 ⇔ ((√x)+(√y))^3 =5^3 , ⇒ (√x)+(√y)=5 (3) square both sides, ⇔ x+y+2(√(xy))=25 ⇒ x+y−(√(xy))=25−3(√(xy)) (1): (√x^3 )+(√y^3 )=35 ⇔ ((√x)+(√y))(x−(√(xy))+y)=35 ⇔ 5(25−3(√(xy)))=35 ⇔ (√(xy))=6, ⇒ xy=36 if you want to find the values of x and y which satisfy the system of Eq. above, just substitute (√x)=(6/( (√y))) into (3).

$$\begin{cases}{\sqrt{\mathrm{x}^{\mathrm{3}} }+\sqrt{\mathrm{y}^{\mathrm{3}} }=\mathrm{35}\:\:\:\left(\mathrm{1}\right)}\\{\sqrt{\mathrm{x}^{\mathrm{2}} \mathrm{y}}+\sqrt{\mathrm{xy}^{\mathrm{2}} }=\mathrm{30}\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\bullet\:\mathrm{take}\:\left(\mathrm{1}\right)+\mathrm{3}\left(\mathrm{2}\right): \\ $$$$\Leftrightarrow\:\sqrt{\mathrm{x}^{\mathrm{3}} }+\mathrm{3}\sqrt{\mathrm{x}^{\mathrm{2}} \mathrm{y}}+\mathrm{3}\sqrt{\mathrm{xy}^{\mathrm{2}} }+\sqrt{\mathrm{y}^{\mathrm{3}} }=\mathrm{125} \\ $$$$\Leftrightarrow\:\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)^{\mathrm{3}} =\mathrm{5}^{\mathrm{3}} ,\:\Rightarrow\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}=\mathrm{5}\:\:\:\left(\mathrm{3}\right) \\ $$$$\mathrm{square}\:\mathrm{both}\:\mathrm{sides},\:\Leftrightarrow\:\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}}=\mathrm{25} \\ $$$$\Rightarrow\:\mathrm{x}+\mathrm{y}−\sqrt{\mathrm{xy}}=\mathrm{25}−\mathrm{3}\sqrt{\mathrm{xy}} \\ $$$$\left(\mathrm{1}\right):\:\sqrt{\mathrm{x}^{\mathrm{3}} }+\sqrt{\mathrm{y}^{\mathrm{3}} }=\mathrm{35} \\ $$$$\Leftrightarrow\:\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\right)\left(\mathrm{x}−\sqrt{\mathrm{xy}}+\mathrm{y}\right)=\mathrm{35} \\ $$$$\Leftrightarrow\:\mathrm{5}\left(\mathrm{25}−\mathrm{3}\sqrt{\mathrm{xy}}\right)=\mathrm{35} \\ $$$$\Leftrightarrow\:\sqrt{\mathrm{xy}}=\mathrm{6},\:\Rightarrow\:\mathrm{xy}=\mathrm{36} \\ $$$$\:\:\:\mathrm{if}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{which} \\ $$$$\mathrm{satisfy}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{Eq}.\:\mathrm{above},\:\mathrm{just}\:\mathrm{substitute}\:\sqrt{\mathrm{x}}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{y}}}\:\mathrm{into}\:\left(\mathrm{3}\right). \\ $$

Commented by Rasheed.Sindhi last updated on 17/Nov/21

$$\mathcal{N}{ice}\:{approach}! \\ $$

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