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Question Number 87561 by jagoll last updated on 05/Apr/20
for x,y ∈R ,   x ,y > 0 satisfy the equation    { ((x^3 y−xy^3  = 24)),((x^2 +y^2  = 10 )) :}  find the possible value of x+y?  (a) 6   (b) 5   (c) 4   (d) 3(√2)  (e) 2
$$\mathrm{for}\:\mathrm{x},\mathrm{y}\:\in\mathbb{R}\:,\: \\ $$$$\mathrm{x}\:,\mathrm{y}\:>\:\mathrm{0}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \mathrm{y}−\mathrm{xy}^{\mathrm{3}} \:=\:\mathrm{24}}\\{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{10}\:}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{6}\:\:\:\left(\mathrm{b}\right)\:\mathrm{5}\:\:\:\left(\mathrm{c}\right)\:\mathrm{4}\:\:\:\left(\mathrm{d}\right)\:\mathrm{3}\sqrt{\mathrm{2}}\:\:\left(\mathrm{e}\right)\:\mathrm{2} \\ $$
Commented by danger last updated on 05/Apr/20
c)4  x=3     y=1
$$\left.{c}\right)\mathrm{4} \\ $$$${x}=\mathrm{3}\:\:\:\:\:{y}=\mathrm{1} \\ $$
Commented by jagoll last updated on 05/Apr/20
yes
$$\mathrm{yes} \\ $$
Commented by john santu last updated on 05/Apr/20

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