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Question Number 155562 by 0731619 last updated on 02/Oct/21

Commented by tabata last updated on 02/Oct/21

Solve :: (√x) + y = 5 , (√y) + x = 3 Solution:: let: a^2 = x , b^2 = y a+b^2 = 5 →(1) a^2 + b = 3 →(2) ×(b ) a^2 b + b^2 = 3 b →(3) (3) − (1) ⇒ a^2 b − a = 3 b − 5 ⇒b = ((a − 5)/(a^2 −3)) a^2 + ((a−5)/(a^2 −3)) = 3 ⇒ a^2 (a^2 −3) + (a−5) = 3 (a^2 −3) ⇒a^4 −6a^2 + a+ 4 = 0 ⇒ (a−1) (a^3 +a^2 −5a−4)= 0 ⇒ a = 1 ⇒ b = ((1−5)/(1−3)) = ((−4)/(−2)) = 2 ∵ x = a^2 ⇒ x = (1)^2 = 1 ∵ y = b^2 ⇒ y = (2)^2 = 4 ∴ (x,y) = (1,4) ⟨ M . T ⟩

$$\boldsymbol{{Solve}}\:::\:\sqrt{\boldsymbol{{x}}}\:+\:\boldsymbol{{y}}\:=\:\mathrm{5}\:\:,\:\sqrt{\boldsymbol{{y}}}\:+\:\boldsymbol{{x}}\:=\:\mathrm{3}\: \\ $$$$ \\ $$$$\boldsymbol{{Solution}}:: \\ $$$$ \\ $$$$\boldsymbol{{let}}:\:\boldsymbol{{a}}^{\mathrm{2}} \:=\:\boldsymbol{{x}}\:\:\:\:,\:\:\:\boldsymbol{{b}}^{\mathrm{2}} \:=\:\boldsymbol{{y}}\: \\ $$$$ \\ $$$$\boldsymbol{{a}}+\boldsymbol{{b}}^{\mathrm{2}} =\:\mathrm{5}\:\rightarrow\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}\:=\:\mathrm{3}\:\rightarrow\left(\mathrm{2}\right)\:×\left(\boldsymbol{{b}}\:\right) \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} \:\boldsymbol{{b}}\:+\:\boldsymbol{{b}}^{\mathrm{2}} \:=\:\mathrm{3}\:\boldsymbol{{b}}\:\rightarrow\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{3}\right)\:−\:\left(\mathrm{1}\right)\:\Rightarrow\:\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}\:−\:\boldsymbol{{a}}\:=\:\mathrm{3}\:\boldsymbol{{b}}\:−\:\mathrm{5}\:\Rightarrow\boldsymbol{{b}}\:=\:\frac{\boldsymbol{{a}}\:−\:\mathrm{5}}{\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}} \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} \:+\:\frac{\boldsymbol{{a}}−\mathrm{5}}{\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}}\:=\:\mathrm{3}\:\Rightarrow\:\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}\right)\:+\:\left(\boldsymbol{{a}}−\mathrm{5}\right)\:=\:\mathrm{3}\:\left(\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}\right) \\ $$$$\Rightarrow\boldsymbol{{a}}^{\mathrm{4}} −\mathrm{6}\boldsymbol{{a}}^{\mathrm{2}} +\:\boldsymbol{{a}}+\:\mathrm{4}\:=\:\mathrm{0}\:\Rightarrow\:\left(\boldsymbol{{a}}−\mathrm{1}\right)\:\left(\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{a}}−\mathrm{4}\right)=\:\mathrm{0}\:\Rightarrow\:\boldsymbol{{a}}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:\boldsymbol{{b}}\:=\:\frac{\mathrm{1}−\mathrm{5}}{\mathrm{1}−\mathrm{3}}\:=\:\frac{−\mathrm{4}}{−\mathrm{2}}\:=\:\mathrm{2}\: \\ $$$$ \\ $$$$\because\:\boldsymbol{{x}}\:=\:\boldsymbol{{a}}^{\mathrm{2}} \:\Rightarrow\:\boldsymbol{{x}}\:=\:\left(\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\because\:\boldsymbol{{y}}\:=\:\boldsymbol{{b}}^{\mathrm{2}} \:\Rightarrow\:\boldsymbol{{y}}\:=\:\left(\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$ \\ $$$$\therefore\:\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)\:=\:\left(\mathrm{1},\mathrm{4}\right) \\ $$$$ \\ $$$$\langle\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}}\:\:\rangle \\ $$

Commented by otchereabdullai@gmail.com last updated on 08/Oct/21

nice

$$\mathrm{nice} \\ $$

Answered by amin96 last updated on 02/Oct/21

{ (((√x)+y=5)),(((√y)+x=3)) :} ⇒ { (((√y)=b)),(((√x)=a)) :}⇒ { ((y=b^2 )),((x=a^2 )) :} ⇒ { ((b^2 +a=5)),((a^2 +b=3)) :} ⇒ (b−a)(b+a)+(a−b)=2 (b−a)(b+a−1)=2 ⇒ { ((b−a=1)),((b+a−1=2)) :} ⇒ { ((b−a=1)),((b+a=3)) :} 2b=4 b=2 a=1 ⇒ { ((y=b^2 )),((x=a^2 )) :} ⇒ y=4 x=1

$$\begin{cases}{\sqrt{{x}}+{y}=\mathrm{5}}\\{\sqrt{{y}}+{x}=\mathrm{3}}\end{cases}\:\:\Rightarrow\:\:\begin{cases}{\sqrt{{y}}={b}}\\{\sqrt{{x}}={a}}\end{cases}\Rightarrow\:\begin{cases}{{y}={b}^{\mathrm{2}} }\\{{x}={a}^{\mathrm{2}} }\end{cases}\:\:\Rightarrow \\ $$$$\begin{cases}{{b}^{\mathrm{2}} +{a}=\mathrm{5}}\\{{a}^{\mathrm{2}} +{b}=\mathrm{3}}\end{cases}\:\:\:\Rightarrow\:\left({b}−{a}\right)\left({b}+{a}\right)+\left({a}−{b}\right)=\mathrm{2} \\ $$$$\left({b}−{a}\right)\left({b}+{a}−\mathrm{1}\right)=\mathrm{2}\:\:\Rightarrow\:\begin{cases}{{b}−{a}=\mathrm{1}}\\{{b}+{a}−\mathrm{1}=\mathrm{2}}\end{cases}\:\Rightarrow\begin{cases}{{b}−{a}=\mathrm{1}}\\{{b}+{a}=\mathrm{3}}\end{cases} \\ $$$$\mathrm{2}{b}=\mathrm{4}\:\:{b}=\mathrm{2} \\ $$$${a}=\mathrm{1}\:\:\:\:\:\:\Rightarrow\:\begin{cases}{{y}={b}^{\mathrm{2}} }\\{{x}={a}^{\mathrm{2}} }\end{cases}\:\:\Rightarrow\:\:{y}=\mathrm{4}\:\:\:{x}=\mathrm{1}\:\:\: \\ $$

Commented by 0731619 last updated on 02/Oct/21

thanks

$${thanks} \\ $$

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