# lowest common denominator in fraction(help me with this problems)

• Jul 18th 2008, 11:55 AM
manznitaa5
lowest common denominator in fraction(help me with this problems)
1-(x/x^2-5+4),(x+1/x^2-16)

2-(6/a6^2-7a+6),(3/a^2-36)

3-(7/x^2+4x+4),(5/4-x^2)
• Jul 18th 2008, 03:24 PM
masters
Quote:

Originally Posted by manznitaa5
1-(x/x^2-5x+4),(x+1/x^2-16) I assume you meant that to be a 5x

2-(6/a^2-7a+6),(3/a^2-36) I might assume you meant a^2 instead of a6^2

3-(7/x^2+4x+4),(5/4-x^2)

1)$\displaystyle \frac{x}{x^2-5x+4} \ \ , \ \ \frac{x+1}{x^2-16}$

First, factor the denominators:

$\displaystyle (x-1)(x-4) \ \ , \ \ (x-4)(x+4)$

The LCD is found by using all the different factors the most number of times they appear in any one denominator.

$\displaystyle LCD = (x-1)(x-4)(x+4)$

2)$\displaystyle \frac{6}{a^2-7a+6} \ \ , \ \ \frac{3}{a^2-36}$

Same steps as before. Factor the denominators:

$\displaystyle (a-6)(a-1) \ \ , \ \ (a-6)(a+6)$

$\displaystyle LCD =$

Can you finish? Try #3 and see how far you get.