How do you solve #(x-1) /3 - (x-1)/2 = 6#?

1 Answer
Oct 16, 2016

Answer:

#x = -35#

Explanation:

Start by making sure that you're working with fractions that have equal denominators.

You start with

#(x-1)/3 - (x-1)/2 = 6/1#

The common denominator here is #6#, so multiply the first fraction by #1 = 2/2#, the second fraction by #1 = 3/3#, and the third fraction by #1 = 6/6#.

This will get you

#(x-1)/3 * 2/2 - (x-1)/2 * 3/3 = 6/1 * 6/6#

#(2(x-1))/6 - (3(x-1))/6 = 36/6#

At this point, drop the denominators and focus exclusively on the numerators.

#2(x-1) - 3(x-1) = 36#

This will get you

#-(x-1) = 36#

#-x + 1 = 36#

#x = - 35#

Do a quick check to make sure that the calculations are correct

#(-35 - 1)/3 - (-35 - 1)/2 = 6#

#-36/3 + 36/2 = 6#

#-12 + 18 = 6 " "color(green)(sqrt())#