My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree3Thanks
  • 1 Post By greg1313
  • 1 Post By greg1313
  • 1 Post By aurel5
Reply
 
LinkBack Thread Tools Display Modes
October 10th, 2014, 05:43 AM   #1
Senior Member
 
Joined: Aug 2014
From: Mars

Posts: 101
Thanks: 9

LCD and prime numbers(with binomials)

So I was given this problem.............

Find the LCD=
$\displaystyle \frac{3}{2x-6}, \frac{4}{x^2-9}, \frac{18}{6x+18}$

The break down (do I call these the prime numbers?
$\displaystyle 2(x-3), (x-3)(x+3), 6(x+3))$


My answer:
$\displaystyle 6(x+3) \cdot 2(x-3)$


Their answer:
$\displaystyle 6(x^2-9)$????? what?

It seems they disregarded the 2, but why? Help me understand please.

Last edited by skipjack; October 11th, 2014 at 04:04 PM.
Opposite is offline  
 
October 10th, 2014, 06:33 AM   #2
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,950
Thanks: 1141

Math Focus: Elementary mathematics and beyond
$\displaystyle 2(x-3)\cdot3(x+3)=6(x^2-9)$ (The least common multiple of 2 and 6 is 6).

They are not necessarily prime numbers.
Thanks from Opposite

Last edited by greg1313; October 10th, 2014 at 06:35 AM.
greg1313 is offline  
October 10th, 2014, 08:21 PM   #3
Senior Member
 
Joined: Aug 2014
From: Mars

Posts: 101
Thanks: 9

Quote:
Originally Posted by greg1313 View Post
$\displaystyle 2(x-3)\cdot3(x+3)=6(x^2-9)$ (The least common multiple of 2 and 6 is 6).

They are not necessarily prime numbers.


On that same note if I had $\displaystyle 4(x-3), 6(x+3)$ would I some how choose $\displaystyle 12(x^2-9)$ I don't quite understand this.
Opposite is offline  
October 10th, 2014, 08:43 PM   #4
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,950
Thanks: 1141

Math Focus: Elementary mathematics and beyond
That's correct. You're looking for the least common denominator - the least common multiple
of 4 and 6 is 12. So if you had

$\displaystyle \frac{3}{4(x-3)}+\frac{3}{6(x+3)}$

you'd write it as

$\displaystyle \frac{3}{4(x-3)}\cdot\frac{3(x+3)}{3(x+3)}+\frac{3}{6(x+3)} \cdot\frac{2(x-3)}{2(x-3)}$

$\displaystyle =\frac{9(x+3)+6(x-3)}{12(x-3)(x+3)}$

$\displaystyle =\frac{15x+9}{12(x-3)(x+3)}$

$\displaystyle \frac{3(5x+3)}{12(x-3)(x+3)}$

$\displaystyle =\frac{5x+3}{4(x-3)(x+3)}$
Thanks from Opposite

Last edited by greg1313; October 10th, 2014 at 08:50 PM.
greg1313 is offline  
October 10th, 2014, 09:59 PM   #5
Senior Member
 
aurel5's Avatar
 
Joined: Apr 2014
From: Europa

Posts: 584
Thanks: 177

Quote:
Originally Posted by Opposite View Post

Their answer:
$\displaystyle 6(x^2-9)$????? what?
$\displaystyle \color{green}{2x-6=2(x-3)
\\\;\\
x^2-9=x^2-3^2=(x-3)(x+3)
\\\;\\
6x+18=6(x+3)=2\cdot3(x+3)
\\\;\\
LCD
\\\;\\
2\cdot3(x+3)(x-3)=6(x^2-9)}$
Thanks from Opposite

Last edited by skipjack; October 11th, 2014 at 04:05 PM.
aurel5 is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
binomials, lcd, numberswith, prime



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Prime numbers Shadowdust New Users 1 February 18th, 2014 09:59 PM
Prime Numbers gabrielcj Algebra 7 July 19th, 2013 08:16 PM
The paradox between prime numbers and natural numbers. Eureka Number Theory 4 November 3rd, 2012 03:51 AM
prime numbers dedede1 Algebra 1 October 24th, 2008 11:03 AM
Prime Numbers johnny Number Theory 2 August 20th, 2007 07:16 AM





Copyright © 2019 My Math Forum. All rights reserved.