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

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Answer 1 · 2016-08-12T20:50:57

#x=-35#

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

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

#(x-1)((2×1)/(2×3)-(3×1)/(3×2))=6# or

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

#(x-1)×(-1/6)=6# or

#x-1=6/(-1/6)=-6×6=-36# or

#x=-36+1=-35#

Answer 2 · 2016-08-12T20:52:03

#x = -35#

Get rid of the denominators first. # xx LCM# which is 6.

#6xx((x-1)) /3 - 6xx((x-1) )/2 = 6xx6#

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

#2x -2 -3x+3 = 36#

#-x = 35#

#x = -35#