# Reducing Fractions

• Aug 2nd 2006, 09:11 PM
pashah
Reducing Fractions
Reduce to the lowest terms.

_x + 4__ __x + 4__
x^2-x-12 + x^2 + 5x + 6

__3_ __c - 4___ __5__
4c+2 - 2c^2+c - 6c
• Aug 3rd 2006, 12:07 AM
earboth
Quote:

Originally Posted by pashah
Reduce to the lowest terms.

_x + 4__ __x + 4__
x^2-x-12 + x^2 + 5x + 6

__3_ __c - 4___ __5__
4c+2 - 2c^2+c - 6c

Hello, pashah,

1rst problem:

Factorize the denominators and you'll get:
$\displaystyle \frac{x+4}{(x-4)(x+3)}+\frac{x+4}{(x+2)(x+3)}$. Put the common factor in front of a bracket:
$\displaystyle \frac{1}{x+3}\cdot \left(\frac{x+4}{x-4}+\frac{x+4}{x+2}\right)$. Now rearrange the terms in the bracket. You can factorize the numerator again. The final result should be:
$\displaystyle \frac{2}{x+3}\cdot \left(\frac{(x-1)(x+4)}{(x-4)(x+2)}\right)$. But as you see, you can't simplify this fraction (maybe there is a typo?)

2nd problem
Factorize (if possible) the denominators:
$\displaystyle \frac{3}{2(2c+1)}-\frac{c-4}{c(2c+1)}-\frac{5}{6c}$
The common denominator is 6c(2c+1). Expand all fractions:
$\displaystyle \frac{3\cdot 2c}{6c(2c+1)}-\frac{6(c-4)}{6c(2c+1)}-\frac{5(2c+1)}{6c(2c+1)}$
Add all terms of the numerators:
$\displaystyle \frac{-10c+19}{6c(2c+1)}$

Greetings

EB