Hi sorry about the over flow of questions today just going over some stuff for and exam tomorrow.

how would you simplify the following fraction into a single fraction?

$\displaystyle \frac{x}{x^2+4x+3}-\frac{2}{x^2+3x+2}$

cheers

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- May 10th 2006, 04:05 AMdadonsimplifying two fractions
Hi sorry about the over flow of questions today just going over some stuff for and exam tomorrow.

how would you simplify the following fraction into a single fraction?

$\displaystyle \frac{x}{x^2+4x+3}-\frac{2}{x^2+3x+2}$

cheers - May 10th 2006, 07:43 AMearbothQuote:

Originally Posted by**dadon**

1. Factorize the denominators. You'll get:

$\displaystyle x^2+4x+3 = (x+1)(x+3)$

$\displaystyle x^2+3x+2 = (x+1)(x+2)$

2. Thus the lcd = (x+1)(x+2)(x+3).

3. Transform the fractions and you'll get:

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

$\displaystyle \frac{x^2+2x-2x-6}{(x+1)(x+2)(x+3)}$ = $\displaystyle \frac{x^2-6}{(x+1)(x+2)(x+3)} $

Greetings

EB - May 10th 2006, 09:01 AMdadon
thanks earboth :)