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Mathematics 53 Online
OpenStudy (theviper):

If x & y are positive with x-y=2 & xy=24, then\[\large{\frac{1}{x}+\frac{1}{y}=?}\]is equal to?? @Hero @ParthKohli @lgbasallote @apoorvk :)

Parth (parthkohli):

Trial and error gives me that x = 6 and y = 4.

OpenStudy (lgbasallote):

me too

Parth (parthkohli):

You just have to find 1/6 + 1/4 :D

OpenStudy (lgbasallote):

do you want legit explanation though lol...i can probably try...

Parth (parthkohli):

6 x 4 = 24 6 - 4 = 2 Trial and error > Algebra in this case.

OpenStudy (lgbasallote):

x = 2 + y xy = 24 (2+y)y = 24 2y + y^2 = 24 y^2 + 2y - 24 = 0 (y+6)(y-4) = 0 y = - 6 y = 4 it says x and y are positive so take y = 4

OpenStudy (lgbasallote):

now sub in to orig equation

OpenStudy (lgbasallote):

x - 4 = 2 x = 2 + 4 x = 6

OpenStudy (theviper):

Answer:-\[\Huge{\color{green}{\frac{5}{12}}}\]

OpenStudy (lgbasallote):

so then you do that thingy 1/6 + 1/4 = 10/24 = 5/12

OpenStudy (theviper):

thanx all f u:)

OpenStudy (anonymous):

\[(\frac{1}{x}+\frac{1}{y})^2-(\frac{1}{x}-\frac{1}{y})^2=\frac{4}{xy}\\(\frac{1}{x}+\frac{1}{y})^2=(\frac{1}{x}-\frac{1}{y})^2+\frac{4}{xy}=(\frac{x-y}{xy})^2+\frac{4}{xy}\]

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