# Simplifying complicated fraction.

• Jan 31st 2010, 07:28 AM
Anemori
Simplifying complicated fraction.
show
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{x+\frac{2}{x}}} and \displaystyle \frac{3x^2+2x+6}{x^2+2x+2} are equal when x = 2 and x = -1. True or False?/ They are equivalent expressions on (-infinity,infinity) I'm lost. How do you do this? Thank you! • Jan 31st 2010, 08:09 AM 1005 Quote: Originally Posted by Anemori show \displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{x+\frac{2}{x}2}}$ and $\displaystyle \frac{3x^2+2x+6}{x^2+2x+2}$ are equal when x = 2 and x = -1.

True or False?/ They are equivalent expressions on (-infinity,infinity)

I'm lost. How do you do this?

Thank you!

Start simplifying the left hand equation.
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}} \displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2+2}{x}}}$
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2x}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2}{x^2+2} + \frac{2x}{x^2+2}}$
$\displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2x+2}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2(x^2+2)}{x^2+2x+2}$
$\displaystyle \frac{x^2+2x+2}{x^2+2x+2}+\frac{2x^2+4}{x^2+2x+2}$
$\displaystyle \frac{3x^2+2x+6}{x^2+2x+2}$

• Jan 31st 2010, 08:14 AM
1005
edit: your post is gone now i look silly :P
• Jan 31st 2010, 08:20 AM
Anemori
It looks false (Wondering). on the fraction i will evaluate the bottom part first going up?
• Jan 31st 2010, 08:25 AM
1005
Quote:

Originally Posted by Anemori
It looks false (Wondering). on the fraction i will evaluate the bottom part first going up?

No, it's true. The math I showed was for when you had that extra two multiplied in. I have corrected my work to reflect your correction in the problem given.
• Jan 31st 2010, 08:54 AM
Anemori
Quote:

Originally Posted by 1005
No, it's true. The math I showed was for when you had that extra two multiplied in. I have corrected my work to reflect your correction in the problem given.

let me see it (Evilgrin)
• Jan 31st 2010, 09:51 AM
1005
Quote:

Originally Posted by Anemori
let me see it (Evilgrin)

Quote:

Originally Posted by 1005
Start simplifying the left hand equation.
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}} \displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2}{\frac{x^2+2}{x}}}$
$\displaystyle 1+$$\displaystyle \frac {2}{1+\frac{2x}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2}{x^2+2} + \frac{2x}{x^2+2}}$
$\displaystyle 1+$$\displaystyle \frac{2}{\frac{x^2+2x+2}{x^2+2}} \displaystyle 1+$$\displaystyle \frac{2(x^2+2)}{x^2+2x+2}$
$\displaystyle \frac{x^2+2x+2}{x^2+2x+2}+\frac{2x^2+4}{x^2+2x+2}$
$\displaystyle \frac{3x^2+2x+6}{x^2+2x+2}$

After simplifying the equation of the left, you see that they are exactly the same equation.
• Jan 31st 2010, 10:46 AM
Anemori
$\displaystyle 1+\frac {2}{1+\frac{2}{x+\frac{2}{x}}}$

how did you do it from that to this:

$\displaystyle 1+\frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}}$
• Jan 31st 2010, 01:58 PM
1005
Quote:

Originally Posted by Anemori
$\displaystyle 1+\frac {2}{1+\frac{2}{x+\frac{2}{x}}}$

how did you do it from that to this:

$\displaystyle 1+\frac {2}{1+\frac{2}{\frac{x^2}{x}+\frac{2}{x}}}$

I multiplied by $\displaystyle \frac{x}{x}$ which simplifies to 1. You can do that, because 1 X anything = that same anything. I did this so the two fractions at the bottom would have the same denominator, allowing them to be added. If you continue to look at what I did, I used that method about five times to add the fractions to 1.