Please help with this Maths problem?

Question

Please could you help me with this problem. I need to get to a quadratic from this question but please give all working also as I need to understand this for an exam.

The question is as follows

y=1+1/2+1/2+1/y

BUT the equation is set up as follows from this link

http://marauder.millersville.edu/~bikenaga/numbertheory/percfrac/percfrac31.png

Cheers

This has been solved

The correct answer was

5y² - 5y - 3 = 0

and now I understand how to work it out so thanks everybody :) I will choose a best answer later today

Answer 1

First take the final 1/[2 + 1/y]

Multiply top and bottom by y

y/y(2 + 1/y)
= y/(2y + 1)

Now take the next fraction, with the replacement value

1/[2+ y/(2y + 1)]

Multiply top and bottom by 2y + 1

(2y + 1)/(2y + 1)[2+ y/(2y + 1)]
= (2y + 1)/(4y + 2) + y
= (2y + 1)/(5y + 2)

So, your equation is now

y = 1 + (2y + 1)/(5y + 2)

Multiply through by 5y + 2
(5y + 2)y = (5y + 2) + (2y + 1)
5y² + 2y = 7y + 3
5y² - 5y - 3 = 0

Using the pythagorean formula

y = [-5 ± √(-5^2 - (4*5*-3))]/2*5
= (-5 ± √(25 + 60))/10
= (-5 ± √85)/10

This approximately equals either 0.42 or - 1.42

Answer 2

The best way to write this is:

y = 1 + 1/[2 + 1/(2 + {1/y})]

Multiply both sides by [2 + 1/(2 + {1/y})], and we get

y[2 + 1/(2 + {1/y})] = 2 + 1/(2 + {1/y}) + 1

Expanding the left hand side, and simplifying the right hand side, we get

2y + y/(2 + {1/y}) = 3 + 1/(2 + {1/y})

Now, we multiply both sides by (2 + {1/y}), which gives us

2y(2 + {1/y}) + y = 3(2 + {1/y}) + 1

Expanding the left hand side and right hand side,

4y + 2 + y = 6 + 3/y + 1

Simplifying,

5y + 2 = 7 + 3/y

Multiply both sides by y,

5y^2 + 2y = 7y + 3

Simplifying

5y^2 - 5y - 3 = 0

Answer 3

First, you must rationalize the denominator.

y = 1 + 1 / (2 * (2 + 1/y) + 1/ (2 + 1/y))
y = 1 + (2 + 1/y) / (5 + 2/y). Trust me on this one, it's magic.
5y + 2 = 5 + 2/y + 2 + 1/y
5y = 5 + 3/y
5y^2 - 5y - 3 = 0

Cheers.

Answer 4

I don't think anyone looked at the original equation. I do not think that what you have equals the original equation. From the original equation do:

1)2+1/y...2y/y+1/y=2y+1/y
2)1/((2y+1/y)=y/2y+y
3)2+((y/2y+y)=(2(2y+y)/2y+y)+y/(2y+y) this gives you
3b)(4y+2y+y)/(2y+y)=7y/3y

keep solving the equation from the bottom up and from right to left...

finally, you get 10y2=4y3
4y3-10y2=0
2y2(2y-5)=0

Answer 5

5y^2-7y-1=0....
this is the soln of the eqn.first solve the last fraction then the above and so on...

Answer 6

i think it is 2 and 8

Answer 7

2+y superscript -1
It basically depends on what y is

Answer 8

i am with jeremy's