x+(√y)=7 (√x)+y=11 faind x=? and y=?


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Question Number 138532 by mathlove last updated on 14/Apr/21

$x + \sqrt{y} = 7$ $\sqrt{x} + y = 11$ $f a i n d x = ? a n d y = ?$
	Answered by henderson last updated on 14/Apr/21
	${ ((x + (√(y )) = 7)),(((√x) + y = 11)) :} ⇔ { ((x = 7 − (√y))),(((√(7−(√y))) + y = 11)) :} ⇔ { ((x = 7−(√y))),(((√(7−(√y))) = 11−y)) :}⇔ { ((x = 7−(√y))),((7−(√y) = (11−y)^2 )) :} ⇔ { ((x = 7−(√y))),(((7−(√7))^2 = (121−22y+y^2 )^2 )) :} ⇔ { ((x = 7−(√y))),((y^4 − 44y^3 + 712y^2 − 5017y + 12996 = 0)) :} ⇔ { ((x = 7−(√y))),((y = 9, y ≈ 7,868... (rejected), y = 12,848.... (rejected), y = 14,283... (rejected))) :} { ((x = 7−(√9))),((y = 9)) :} ⇔ determinant (((x = 4 and y = 9.)))$
	${\begin{cases} x + \sqrt{y} = 7 \\ \sqrt{x} + y = 11 \end{cases} \Leftrightarrow {\begin{cases} x = 7 - \sqrt{y} \\ \sqrt{7 - \sqrt{y}} + y = 11 \end{cases} \Leftrightarrow {\begin{cases} x = 7 - \sqrt{y} \\ \sqrt{7 - \sqrt{y}} = 11 - y \end{cases} \Leftrightarrow$ ${\begin{cases} x = 7 - \sqrt{y} \\ 7 - \sqrt{y} = {(11 - y)}^{2} \end{cases} \Leftrightarrow {\begin{cases} x = 7 - \sqrt{y} \\ {(7 - \sqrt{7})}^{2} = {(121 - 22 y + y^{2})}^{2} \end{cases} \Leftrightarrow$ ${\begin{cases} x = 7 - \sqrt{y} \\ y^{4} - 44 y^{3} + 712 y^{2} - 5017 y + 12996 = 0 \end{cases} \Leftrightarrow$ ${\begin{cases} x = 7 - \sqrt{y} \\ y = 9, y \approx 7, 868 . . . (r e j e c t e d), y = 12, 848 . . . . (r e j e c t e d), y = 14, 283 . . . (r e j e c t e d) \end{cases}$ ${\begin{cases} x = 7 - \sqrt{9} \\ y = 9 \end{cases} \Leftrightarrow \begin{matrix} x = 4 and y = 9 . \end{matrix}$
	Commented by mathlove last updated on 14/Apr/21

	$t h a n k y o u$
	Commented by henderson last updated on 14/Apr/21

	${you}^{'} re welcome!$
	Answered by behi83417@gmail.com last updated on 14/Apr/21

	${\begin{cases} (x - 4) + (\sqrt{y} - 3) = 0 \\ (\sqrt{x} - 2) + (y - 9) = 0 \end{cases} \Rightarrow$ $\Rightarrow {\begin{cases} (\sqrt{x} - 2) (\sqrt{x} + 2) + (\sqrt{y} - 3) = 0 \\ (\sqrt{x} - 2) + (\sqrt{y} - 3) (\sqrt{y} + 3) = 0 \end{cases} \Rightarrow$ ${\begin{cases} \sqrt{x} - 2 = - (\sqrt{y} - 3) (\sqrt{y} + 3) \\ \Rightarrow - (\sqrt{y} - 3) (\sqrt{y} + 3) (\sqrt{x} + 2) + (\sqrt{y} - 3) = 0 \end{cases}$ $\Rightarrow - (\sqrt{y} - 3) [(\sqrt{y} + 3) (\sqrt{x} + 2) - 1] = 0$ $\Rightarrow \sqrt{y} - 3 = 0 \Rightarrow \sqrt{y} = 3 \Rightarrow y = 9$ $\Rightarrow \sqrt{x} - 2 = - (\sqrt{y} - 3) (\sqrt{y} + 3) = 0 \Rightarrow \sqrt{x} - 2 = 0$ $\Rightarrow \sqrt{x} = 2 \Rightarrow x = 4 \Rightarrow {\begin{cases} x = 4 \\ y = 9 . ◼ \end{cases}$
	Commented by Rasheed.Sindhi last updated on 15/Apr/21

	$\lor \cap \overset{★}{} \subset\in sir!$
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