How do you solve #1/3x + 2 = 1/6x + 11/2#?

3 Answers
Sep 6, 2016

Answer:

#x=21#

Explanation:

#1/3x+2=1/6x+11/2#

or

#1/3x-1/6x=11/2-2#

or

#1/6x=7/2#

or

#x=7/2times6#

or

#x=21#

Sep 6, 2016

Answer:

The same thing but in disguise

#x=21#

Explanation:

Lets get rid of the fractions.

Make all the denominators 6

#2/6x+12/6=1/6x+33/6#

Now totally disregard the denominators

#2x+12=x+33#

#2x-x=33-12#

#x=21#

Sep 6, 2016

Answer:

#x= 21#

Explanation:

If you have an equation which has fractions, you can get rid of the fractions by

"multiply through by the LCD"

This allows you to cancel all the denominators, which makes the entire equation simpler.

#color(white)(xxxxxxx)1/3x +2 = 1/6x+11/2 color(white)(xxxxxxxxx)larr LCD = 6#

#(color(red)(cancel6^2xx)1)/cancel3x +color(red)(6xx)2 = (color(red)(cancel6xx)1)/cancel6x+(color(red)(cancel6^3xx)11)/cancel2#

#color(white)(xxx.xxx)2x +12 = x +33#

#color(white)(xxxxxxxxxxx)x= 21#