How do you find the least common denominator of #2/(7x-14# and #x/(3x-6#?

2 Answers
May 29, 2018

L C M of #(7x - 14) , (3x - 6)# is #color(crimson)(=> 21 (x-2)#

Explanation:

To find L C M of #2 / (7x - 14) , x / (3x - 6)#

# 7x - 14 = 7 * color(green)((x - 2)#

#3x - 6 = 3 * color(green)((x - 2)#

L C M of #(7x - 14) , (3x - 6)# is #7 * 3 * (x-2)#

#color(crimson)(=> 21 (x-2)#

May 29, 2018

#21(x-2)#

Explanation:

Given: #2/(7x-14) and x/(3x-6)#

Notice that both 7 from #7x# and 3 from #3x# are prime numbers. As these are the #x# terms we will need to focus on those and just see what happens with the constants

I imagine that that the school would expect you to do it like this:

#color(white)("dddddddddddddddd") ubrace(3xx7)#
Multiples of 7 #->color(green)(7,14,color(red)(21),28," etc")#

Multiples of 3 #->color(green)(3,6,9,12,15,18,color(red)(21),24," etc")#
#color(white)("ddddddddddddddddddddddddd")obrace(7xx3)#

Multiply by 1 and you do not change the value. However, 1 comes in many forms.

#color(green)([2/(7x-14)color(red)(xx1) ]and [x/(3x-6)color(red)(xx1)])#

#color(green)([2/(7x-14)color(red)(xx3/3) ]and [x/(3x-6)color(red)(xx7/7)])#

#color(white)("d")color(green)([6/(21x-42)]color(white)("dd")and color(white)("dd")[(7x)/(21x-42)] #

Consider just the denominator #21x-42#

Notice that #2xx21=42# so we can write it as #21(x-2)#