Question

Note: This is NOT a question. This is a tutorial!

How to add rational expressions? For example you are given \(\large \frac{2}{x^2 + 2x + 1} + \frac{2x}{x+1}\)

See comment below to see how!

Answer 1

the first thing you need to do is to simplify the denominators. By this, I mean to factor out your denominators (individually) as much as example. Let's take a look at the first denominator. We have \(x^2 + 2x + 1\) using a method of factoring that I have discussed before ( http://openstudy.com/users/lgbasallote#/updates/4f9bd87ee4b000ae9ed11143) we can factor out \(x^2 + 2x + 1\) into \((x+1)(x+1)\).

Now let's take a look at the second denominator. x + 1. This cannot be further simplified so we leave it as is. Now, we find the Least Common Denominator (LCD). Note that the LCD must contain ALL the denominators. Our denominators are \((x+1)(x+1)\) and \((x+1)\). So our LCD must contain these both. \((x+1)(x+1)\) contains both the first denominator and the second denominator, so this becomes our LCD

The next step would be to let the two fractions have this LCD. To do that, we divide the LCD by the ORIGINAL denominator, and the quotient, we multiply to the ORIGINAL numerator. Let's do it on the first fraction.

The original fraction is \(\large \frac{2}{x^2 + 2x + 1}\). Now we divide the LCD by the ORIGINAL denominator so we have \(\LARGE \frac{\cancel{(x+1)(x+1)}}{\cancel{(x+1)(x+1)}}\) what's left is (1) only so we multiply that to the ORIGINAL numerator. (1)(2).

So our new fraction would be \(\LARGE \frac{2(1)}{(x+1)(x+1)}\). We are back to the original equation, but do not worry, this is not a mistake. This is because the first fraction contains the LCD so it is normal that it reverts back to the original.

Now let's do the second fraction. Again, we divide the LCD by the ORIGINAL denominator so \(\large \frac{\cancel{(x+1)}(x+1)}{\cancel{x+1}}\) what's left is (x+1). We multiply this by the numerator so we have 2x(x+1). So our new fraction would be \(\frac{2x(x+1)}{(x+1)(x+1)}\) Note: Do NOT cancel. Cancelling will revert back to the original so we will just simplify the numerator. We have \(2x^2 + 2x\). So our two fractions are now:

\(\large \frac{2}{(x+1)(x+1)} + \frac{2x^2 + 2x}{(x+1)(x+1)}\) we now have the same denominators, so we can now add our numerators then copy the denominator. So we'll have...

\(\LARGE \frac{2 + 2x^2 + 2x}{(x+1)(x+1)} = \frac{2(x^2 + x + 1)}{(x+1)(x+1)}\)

This is now our answer.

Answer 2

Hi lg. I got my account a few days ago

Answer 3

the link for my tutorial on factoring is this: http://openstudy.com/users/lgbasallote#/updates/4f9bd87ee4b000ae9ed11143

it was not linked right

Answer 4

Jst wanted to ay hi

Answer 5

hi to you too :)

Answer 6

how are you

Answer 7

let's not talk here..this post is for the tutorial only..additional irrelevant comments will lead to confusions

Answer 8

oh ok.

Answer 9

can u fan me so we can message

Answer 10

It looks really good lgba :) good job! \m/

Answer 11

You did this for me right ? :D