Another math problem

[Nixona]

I cant believe how this guy assigns problems without any basis in the text or lecture. PLEASE HELP...I know there are some math geniuses out there.

Let a be a real number different from -1. Prove that:

1
---- +
1+1
-----
1+1
----
1+a


2+a
--------
3+2a


I hope that comes out clearly. I can kill the complex fraction, but I end up with something like

1
---- +
2+a


2+a
------
3+2a


Anywhere close?

[vass0922]

one tip.. you can use the [ code ] tag [ /code ] tag so the html won't strip the spaces so your fractions will look like fractions and not ascii art
(remove the spaces from within the brackets of the code tag though )

1      2+a 
---- = --------  
1+1    3+2a
-----
1+1
----
1+a


I hope that comes out clearly. I can kill the complex fraction, but I end up with something like 

1       2+a
---- = -------
2+a     3+2a

not sure if I did the complex fraction on the top correctly...

[Nixona]

Blah...UBB code. I'll learn it one day.

[Nixona]

Close enough

I can prolly crack it in 10 minutes but don't know what you are saying.

Use a*b for "a times b"
a**b for "a to the power b"
a/b for "a divided by b"
And use brackets - [{()}] - generously (but accurately) for easier reading.

Using brackets correctly is easy and helps in solving the problem for than 50%. Tip - number of open brackets must match number of closing brackets.

If you post it real fast, I'll crack it like last time - I have an important 1pm meeting.

[Nixona]

It's complex fractions. Let me see if I can make it any better. There is only one multiplication, no powers.

1/1+(1/1+(1)/(1)+a) = 2+a/3+2*a

or

1/1+1/1+1/1+a = 2+a/3+2*a

Vass's coding of the problem is closer to how it looks on paper and may be better than either of these. The objective is: Let a be a real number different from -1, Prove that: (the problem)

1/1+1/1+1/1+a = 2+a/3+2*a

Do you mean:

2+1/(1+a) = (2+a)/3 + 2*a

or

2+ 1/(1+a) = 2 + a/3 + 2*a

Either case, if you put a=1, it doesn't work out so maybe I'm not getting the equality right.

[Nixona]

Look at Vass' code in the second post. There is a left complex fraction, which actually contains three fractions, and a right simple fraction. The plus between them is actually an equal. (=)
I typed it wrong.

[Nixona]

Bump...HEEEEEEEEEEELP.

[Banti]

   1
 -----
 1+1
--------
 1+1
 -----
 1+a


doesn't this equal

  1
 ---
  2
-----
  2
 ---
 1+a

??

so we multiply top and bottom by 1+a/2

and we get

1+a      2+a
-----  = -------
  4        3+2a


multiply both sides of equation by (4)(3+2a)

any you get

(1+a)(3+2a) = (4)(2+a)

3+5a+2a^2 = 8 + 4a

2a^2 + a - 5 = 0

quad it and the answers are

1.31 and -1.81

Banti