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

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Answer 1 · 2016-06-06T17:14:18

#x=-35#

#x-1# is common to both the terms in the left side of the equation.

Take #x-1# and write the remaining terms inside brackets:

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

Now, solve the fractions inside the brackets:

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

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

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

Next, multiply both sides by #-6#:

#-(x-1)/6color(red)(xx-6)=6color(red)(xx-6)#

#x-1=-36#

Add #1# to both sides:

#x-1color(red)(+1)=-36color(red)(+1)#

#x=-35#

Let us check our solution

Plug in the value of #x=-35# in the given equation and solve:

Left side of the equation is #(x-1)/3-(x-1)/2#

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

#=(-36)/3-(-36/2)#

#=-12-(-18)#

#=-12+18#

#=6=# right side of the equation.

Hence, our solution is correct.