{ (((1/(2x−y)) + (√y) = 1)),((((√y)/(2x−y)) = −6)) :}


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Question Number 100954 by bobhans last updated on 29/Jun/20

${\begin{cases} \frac{1}{2 x - y} + \sqrt{y} = 1 \\ \frac{\sqrt{y}}{2 x - y} = - 6 \end{cases}$
Commented by Dwaipayan Shikari last updated on 29/Jun/20
$((−6)/(√y))=(1/(2x−y)) →(1) ((−6)/(√y))+(√y)=1 ⇒t^2 −t−6=0⇒t= 3 or −2 {suppose (√y)=t} ((−6)/3)=(1/(2x−9 ))⇒−4x+18=1 ⇒x=((17)/4) { ((x=((17)/4))),((y=9)) :}$
$\frac{- 6}{\sqrt{y}} = \frac{1}{2 x - y} \to (1)$ $\frac{- 6}{\sqrt{y}} + \sqrt{y} = 1 \Rightarrow t^{2} - t - 6 = 0 \Rightarrow t = 3 o r - 2 {s u p p o s e \sqrt{y} = t}$ $\frac{- 6}{3} = \frac{1}{2 x - 9} \Rightarrow - 4 x + 18 = 1 \Rightarrow x = \frac{17}{4}$ ${\begin{cases} x = \frac{17}{4} \\ y = 9 \end{cases}$
	Answered by ajfour last updated on 29/Jun/20

	$s = \frac{1}{2 x - y}, t = \sqrt{y}$ $s + t = 1, s t = - 6 \Rightarrow t > 0, s < 0$ $s, t = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 6}$ $t = \frac{1}{2} + \frac{5}{2} = 3, s = - 2$ $\Rightarrow y = t^{2} = 9,$ $s = \frac{1}{2 x - y} = \frac{1}{2 x - 9} = - 2$ $\Rightarrow 2 x = 9 - \frac{1}{2} \Rightarrow x = \frac{17}{4} .$
	Answered by john santu last updated on 29/Jun/20

	$\Rightarrow \frac{1}{2 x - y} = - \frac{6}{\sqrt{y}} (1)$ $\Rightarrow - \frac{6}{\sqrt{y}} + \sqrt{y} = 1; {(\sqrt{y})}^{2} - \sqrt{y} - 6 = 0$ $(\sqrt{y} - 3) (\sqrt{y} + 2) = 0; \sqrt{y} = 3$ $y = 9 \Rightarrow \frac{1}{2 x - 9} = - 2; 2 x - 9 = - \frac{1}{2}$ $2 x = \frac{17}{2}, x = \frac{17}{4}$
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