Adding rational expressions with different denominators: Quadratic

[Orangey159]

I'm having a lot of trouble understanding anything in algebra, and this is my biggest hump in the road.
It's called"Adding rational expressions with different denominators: Quadratic"


This is one of them
2/x(^2)+4x+3 - 1/x^(2)+3x+2

Will you go through all the steps, and how you come up with certain numbers?
All it looks like is 23 magically becomes 1.34...

 

[tkhunny]

Why is it different from adding other fractions? Find a common denominator and add away! I recommend factoring those denominators, first.

 

[soroban]

Hello, Orangey159!

\(\displaystyle \text{Simplify: }\:\dfrac{2}{x^2+4x+3} - \dfrac{1}{x^2+3x+2}\)

To add two fractions, they must have the same denominator.


First, we find the Least Common Denominator.

. . Factor the denominators: .\(\displaystyle \dfrac{2}{(x+1)(x+3)} - \dfrac{1}{(x+1)(x+2)}\)

. . The LCD is \(\displaystyle (x+1)(x+2)(x+3).\)


Convert the fractions to the LCD.

. . \(\displaystyle \displaystyle\frac{2}{(x+1)(x+3)}\cdot \frac{x+2}{x+2} - \frac{1}{(x+1)(x+2)}\cdot\frac{x+3}{x+3}\)


We have:

. . \(\displaystyle \displaystyle \frac{2(x+2)}{(x+1)(x+2)(x+3)} - \frac{x+3}{(x+1)(x+2)(x+3)} \)


They have the same denominator.
Combine the numerators; "keep" the denominator.

. . \(\displaystyle \displaystyle \frac{2(x+2)-(x+3)}{(x+1)(x+2)(x+3)} \;=\;\frac{2x+4 - x - 3}{(x+1)(x+2)(x+3)} \;=\;\frac{x+1}{(x+1)(x+2)(x+2)}\)


Reduce.

. . \(\displaystyle \dfrac{\rlap{/////}x+1}{(\rlap{/////}x+1)(x+2)(x+3)} \;=\;\dfrac{1}{(x+2)(x+3)}\)

 

[lookagain]

This is one of them
2/x(^2)+4x+3 - 1/x^(2)+3x+2

Orangey159, you put those parentheses in the wrong places. Type it like this: 2/(x^2 + 4x + 3) - 1/(x^2 + 3x + 2).