# Adding More than Two Fractions

• Aug 13th 2008, 12:28 PM
cmf0106
Im just reading through a basic pre-algebra book; however, I am not certain what the author is trying to get across here.

Text preceding the problem: Finding the LCD for three or more fractions is pretty much the same as finding the LCD for two fractions. Factor each denominator into its prime factorization and list the primes that appear in each. Divide the LCD by each denominator. Multiply each fraction by this number over itself.

Example
$\displaystyle \frac{3}{10} + \frac{5}{12} + \frac{1}{18}$

The Author then shows the prime factorization of each denominator

10 = 2 * 5
12 = 2 * 2 * 3
18 = 2 * 3 * 3

LCD = 2 * 2 * 3 * 3 * 5 = 180

I will stop the explanation here. I understand doing the prime factorization of each fractions denominator. But as to which numbers from the prime factorization to use (the bolded ones) is beyond me could some one please clarify how one is supposed to know which numbers to multiply together to result in 180?

Many thanks
• Aug 13th 2008, 01:09 PM
Quick
Quote:

Originally Posted by cmf0106
Im just reading through a basic pre-algebra book; however, I am not certain what the author is trying to get across here.

Text preceding the problem: Finding the LCD for three or more fractions is pretty much the same as finding the LCD for two fractions. Factor each denominator into its prime factorization and list the primes that appear in each. Divide the LCD by each denominator. Multiply each fraction by this number over itself.

Example
$\displaystyle \frac{3}{10} + \frac{5}{12} + \frac{1}{18}$

The Author then shows the prime factorization of each denominator

10 = 2 * 5
12 = 2 * 2 * 3
18 = 2 * 3 * 3

LCD = 2 * 2 * 3 * 3 * 5 = 180

I will stop the explanation here. I understand doing the prime factorization of each fractions denominator. But as to which numbers from the prime factorization to use (the bolded ones) is beyond me could some one please clarify how one is supposed to know which numbers to multiply together to result in 180?

Many thanks

Actually, you've bolded the wrong numbers, it should be:

10 = 2 * 5
12 = 2 * 2 * 3
18 = 2 * 3 * 3

12 has 2 for a prime factor twice, which means at the very least, the lcm must have two 2's

18 has two 3's, so the lcm must have at least two 3's as well.

10 has one 5, so the lcm must have at least one 5.

If we looked at ten, we would see that it has one 2, so the lcm must have at least one 2, but because of the number twelve, we already have included two 2's into the lcm.

I hope that cleared things up.
• Aug 13th 2008, 01:19 PM
masters
Quote:

Originally Posted by Quick
Actually, you've bolded the wrong numbers, it should be:

10 = 2 * 5
12 = 2 * 2 * 3
18 = 2 * 3 * 3

Quick is correct. But as a general rule I like to put the prime factors in exponential form and use this rule:

LCD = the product of all the different factors with the largest exponent.

$\displaystyle 10=2^1\cdot5^1$
$\displaystyle 12=2^2\cdot3^1$
$\displaystyle 18=2^1\cdot3^2$

$\displaystyle LCD \ \ = \ \ 2^2\cdot3^2\cdot5=180$

Works with variables, too.