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2x-1-2-8-2xy-4-4y-8xy-1-1-

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Question Number 89344 by jagoll last updated on 17/Apr/20
(2x−1)^2 +8(√(2xy)) = 4 4y−(√(8xy−1)) = 1
$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{8}\sqrt{\mathrm{2}{xy}}\:=\:\mathrm{4} \\ $$$$\mathrm{4}{y}−\sqrt{\mathrm{8}{xy}−\mathrm{1}}\:=\:\mathrm{1} \\ $$
Commented by john santu last updated on 17/Apr/20
8xy−1 = 16y^2 −8y+1 8xy = 16y^2 −8y+2 2x = ((16y^2 −8y+2)/(4y)) ⇒2xy = ((16y^2 −8y+2)/4) 2x−1 = ((16y^2 −12y+2)/(4y)) (((16y^2 −12y+2)/(4y)))^2 +4(√(16y^2 −8y+2)) = 4 let (√(16y^2 −8y+2)) = t
$$\mathrm{8}{xy}−\mathrm{1}\:=\:\mathrm{16}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{1} \\ $$$$\mathrm{8}{xy}\:=\:\mathrm{16}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{2} \\ $$$$\mathrm{2}{x}\:=\:\frac{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{2}}{\mathrm{4}{y}}\:\Rightarrow\mathrm{2}{xy}\:=\:\frac{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{2}}{\mathrm{4}} \\ $$$$\mathrm{2}{x}−\mathrm{1}\:=\:\frac{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{12}{y}+\mathrm{2}}{\mathrm{4}{y}} \\ $$$$\left(\frac{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{12}{y}+\mathrm{2}}{\mathrm{4}{y}}\right)^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{2}}\:=\:\mathrm{4} \\ $$$${let}\:\sqrt{\mathrm{16}{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{2}}\:=\:{t} \\ $$$$ \\ $$$$ \\ $$
Commented by john santu last updated on 17/Apr/20
then we get x = (1/2) ∧ y = (1/4)
$${then}\:{we}\:{get}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:{y}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by MJS last updated on 17/Apr/20
8xy−1≥0 ⇒ xy≥(1/8) 8(√(2xy))=4−(2x−1)^2 ⇒ 4−(2x−1)^2 ≥0 ⇒ ⇒ −(1/2)≤x≤(3/2) (√(8xy−1))=4y−1 ⇒ 4y−1≥0 ⇒ y≥(1/4) xy≥(1/8)∧y≥(1/4)∧−(1/2)≤x≤(3/2) ⇒ 0≤x≤(1/2)∧y≥(1/4) now I first tried the borders...
$$\mathrm{8}{xy}−\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow\:{xy}\geqslant\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{8}\sqrt{\mathrm{2}{xy}}=\mathrm{4}−\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow\:\mathrm{4}−\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{8}{xy}−\mathrm{1}}=\mathrm{4}{y}−\mathrm{1}\:\Rightarrow\:\mathrm{4}{y}−\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow\:{y}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$${xy}\geqslant\frac{\mathrm{1}}{\mathrm{8}}\wedge{y}\geqslant\frac{\mathrm{1}}{\mathrm{4}}\wedge−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{0}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\wedge{y}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{now}\:\mathrm{I}\:\mathrm{first}\:\mathrm{tried}\:\mathrm{the}\:\mathrm{borders}… \\ $$
Commented by john santu last updated on 17/Apr/20
what the exact solution sir?
$${what}\:{the}\:{exact}\:{solution}\:{sir}? \\ $$
Commented by john santu last updated on 17/Apr/20
if x = (1/2) ∧ y = 1 ⇒(2.(1/2)−1)^2 +8(√(2.(1/2).1)) ≠ 4 sir
$${if}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\wedge\:{y}\:=\:\mathrm{1}\: \\ $$$$\Rightarrow\left(\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{8}\sqrt{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{1}}\:\neq\:\mathrm{4}\:{sir} \\ $$
Commented by MJS last updated on 17/Apr/20
your solution is ok, I wanted to say that I found the same by trying the borders for x and y: x=(1/2)∧y=(1/4)
$$\mathrm{your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{ok},\:\mathrm{I}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{say}\:\mathrm{that}\:\mathrm{I} \\ $$$$\mathrm{found}\:\mathrm{the}\:\mathrm{same}\:\mathrm{by}\:\mathrm{trying}\:\mathrm{the}\:\mathrm{borders}\:\mathrm{for} \\ $$$${x}\:\mathrm{and}\:{y}:\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\wedge{y}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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