Question

If \(x+ \frac{1}{y}=8\) and \(xy+ \frac{1}{xy}=38\), then the value of \(y+ \frac{1}{x}=....\)

Answer 1

x + 1/y = 8

xy + 1/xy = 38

substitution might help

yx + 1 = 8y

xy = 8y - 1

therefore

8y - 1 + 1/xy = 38

8y^2 - y + 1/x = 38y

8y^2 - y-y +y+ 1/x = 38y

y+ 1/x = 38y +2y-8y^2

Answer 2

was just wondering if dot products might help

a = (x, 1/y)
b = (1, 1)
c = (y, 1/x)

a.b = 8
a.c = 38
b.c = ?

Answer 3

that looks plausible as well :)

Answer 4

not so sure, im stuck actually haha..

Answer 5

its a good thought tho, id never of considered it thats for sure

Answer 6

I have my own stupid trick.\[\dfrac{xy + 1}{y } = 8\tag{1}\]\[\dfrac{xy + 1}{x} = ~?\tag{2}\]\[\dfrac{(xy)^2 + 1}{xy} = 38\tag{3}\]Multiply 1 and 2:\[\dfrac{(xy)^2 + 1 + 2xy}{xy} = (3) + 2\]

Answer 7

latex isnt coding for me so thats just an eyesore :P

Answer 8

This is what I've got so far:

\[xy+\frac{1}{xy}=38\]
\[(xy)^2+1=38xy\]
\[(xy)^2-38xy +1 =0\]

\[xy =\frac{ 38 \pm \sqrt{1444-4}}{2} = \frac{ 38 \pm \sqrt{1440}}{2} = 19 \pm 6\sqrt{10}\]

Answer 9

http://gyazo.com/08afd4459b71491ed74a314e087f263a

Answer 10

@ParthKohli that's great.., how do you get the idea?

Answer 11

OS is acting up.

Anyway,\[(3) + 2 = 40 = 8 \times ?\]\[\Rightarrow ~? = 5\]

Answer 12

Oh, I don't know either. I've noticed that my brain is coming up with ideas it couldn't earlier.

Answer 13

di u get it

Answer 14

Well done @ParthKohli !