How do you solve #(x-1) /3 - (x-1)/2 = 6#?
1 Answer
Oct 16, 2016
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
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())#