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

Commented by tabata last updated on 02/Oct/21

$$\mathit{S}\mathit{o}\mathit{l}\mathit{v}\mathit{e}::\sqrt{\mathit{x}}+\mathit{y}=5,\sqrt{\mathit{y}}+\mathit{x}=3$$$$$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}::$$$$$$$$\mathit{l}\mathit{e}\mathit{t}:{\mathit{a}}^2=\mathit{x},{\mathit{b}}^2=\mathit{y}$$$$$$$$\mathit{a}+{\mathit{b}}^2=5\to \left(1\right)$$$${\mathit{a}}^2+\mathit{b}=3\to \left(2\right)\times \left(\mathit{b}\right)$$$${\mathit{a}}^2\mathit{b}+{\mathit{b}}^2=3\mathit{b}\to \left(3\right)$$$$\left(3\right)-\left(1\right)\Rightarrow {\mathit{a}}^2\mathit{b}-\mathit{a}=3\mathit{b}-5\Rightarrow \mathit{b}=\frac{\mathit{a}-5}{{\mathit{a}}^2-3}$$$${\mathit{a}}^2+\frac{\mathit{a}-5}{{\mathit{a}}^2-3}=3\Rightarrow {\mathit{a}}^2({\mathit{a}}^2-3)+(\mathit{a}-5)=3({\mathit{a}}^2-3)$$$$\Rightarrow {\mathit{a}}^4-6{\mathit{a}}^2+\mathit{a}+4=0\Rightarrow (\mathit{a}-1)({\mathit{a}}^3+{\mathit{a}}^2-5\mathit{a}-4)=0\Rightarrow \mathit{a}=1$$$$$$$$\Rightarrow \mathit{b}=\frac{1-5}{1-3}=\frac{-4}{-2}=2$$$$$$$$\because \mathit{x}={\mathit{a}}^2\Rightarrow \mathit{x}={\left(1\right)}^2=1$$$$\because \mathit{y}={\mathit{b}}^2\Rightarrow \mathit{y}={\left(2\right)}^2=4$$$$$$$$\therefore (\mathit{x},\mathit{y})=(1,4)$$$$$$$$⟨\mathit{M}.\mathit{T}⟩$$

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

$$\mathrm{nice}$$

Answered by amin96 last updated on 02/Oct/21

$$\{\begin{array}{l}\sqrt{x}+y=5\\ \sqrt{y}+x=3\end{array}\Rightarrow \{\begin{array}{l}\sqrt{y}=b\\ \sqrt{x}=a\end{array}\Rightarrow \{\begin{array}{l}y=b^2\\ x=a^2\end{array}\Rightarrow$$$$\{\begin{array}{l}{b}^2+a=5\\ a^2+b=3\end{array}\Rightarrow (b-a)(b+a)+(a-b)=2$$$$(b-a)(b+a-1)=2\Rightarrow \{\begin{array}{l}b-a=1\\ b+a-1=2\end{array}\Rightarrow \{\begin{array}{l}b-a=1\\ b+a=3\end{array}$$$$2b=4b=2$$$$a=1\Rightarrow \{\begin{array}{l}y=b^2\\ x=a^2\end{array}\Rightarrow y=4x=1$$

Commented by 0731619 last updated on 02/Oct/21

$$thanks$$

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