Alternative forms { (((√x)+(√y)=((23)/(12)))),((9x+16y=29)) :}


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Question Number 117555 by dw last updated on 12/Oct/20

$A l t e r n a t i v e f o r m s$ ${\begin{cases} \sqrt{x} + \sqrt{y} = \frac{23}{12} \\ 9 x + 16 y = 29 \end{cases}$
	Answered by Olaf last updated on 12/Oct/20
	$(√y) = ((23)/(12))−(√x) 9x+16(((23)/(12))−(√x))^2 = 29 9x+((529)/9)−((184)/3)(√x)+16x = 29 25x−((184)/3)(√x)+((268)/9) = 0 Δ = (((184)/3))^2 −4×25×((268)/9) = 784 = 28^2 (√x) = ((((184)/3)±28)/(50)) = (2/3) or ((134)/(75)) ⇒ x = (4/9) or ((17956)/(5625)) y = ((29−9x)/(16)) = ((23)/(16)) or ((169)/(10000)) S = {((2/3);((23)/(16))),(((134)/(75));((169)/(10000)))}$
	$\sqrt{y} = \frac{23}{12} - \sqrt{x}$ $9 x + 16 {(\frac{23}{12} - \sqrt{x})}^{2} = 29$ $9 x + \frac{529}{9} - \frac{184}{3} \sqrt{x} + 16 x = 29$ $25 x - \frac{184}{3} \sqrt{x} + \frac{268}{9} = 0$ $Δ = {(\frac{184}{3})}^{2} - 4 \times 25 \times \frac{268}{9} = 784 = 28^{2}$ $\sqrt{x} = \frac{\frac{184}{3} \pm 28}{50} = \frac{2}{3} or \frac{134}{75}$ $\Rightarrow x = \frac{4}{9} or \frac{17956}{5625}$ $y = \frac{29 - 9 x}{16} = \frac{23}{16} or \frac{169}{10000}$ $S = {(\frac{2}{3}; \frac{23}{16}), (\frac{134}{75}; \frac{169}{10000})}$
	Answered by Olaf last updated on 12/Oct/20

	Commented by Olaf last updated on 12/Oct/20

	$2 solutions .$
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