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 :)
Trial and error gives me that x = 6 and y = 4.
me too
You just have to find 1/6 + 1/4 :D
do you want legit explanation though lol...i can probably try...
6 x 4 = 24 6 - 4 = 2 Trial and error > Algebra in this case.
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
now sub in to orig equation
x - 4 = 2 x = 2 + 4 x = 6
Answer:-\[\Huge{\color{green}{\frac{5}{12}}}\]
so then you do that thingy 1/6 + 1/4 = 10/24 = 5/12
thanx all f u:)
\[(\frac{1}{x}+\frac{1}{y})^2-(\frac{1}{x}-\frac{1}{y})^2=\frac{4}{xy}\\(\frac{1}{x}+\frac{1}{y})^2=(\frac{1}{x}-\frac{1}{y})^2+\frac{4}{xy}=(\frac{x-y}{xy})^2+\frac{4}{xy}\]
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