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 :)

Question

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 :)

parthkohli · Answer

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

lgbasallote · Answer

me too

parthkohli · Answer

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

lgbasallote · Answer

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

parthkohli · Answer

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

lgbasallote · Answer

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

lgbasallote · Answer

now sub in to orig equation

lgbasallote · Answer

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

theviper · Answer

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

lgbasallote · Answer

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

theviper · Answer

thanx all f u:)

anonymous · Answer

$$(\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}$$