How do you solve #(3x-5.7)/6 = 4.1#?

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Answer 1 · 2016-11-04T16:39:13

#x = 10.7#

You need to isolate #x# which is in the numerator.
Get rid of the denominator first: # xx 6# on both sides

#color(blue)(cancel6xx)((3x-5.7))/cancel6 = 4.1color(blue)(xx6)#

#3x-5.7 = 26.4" "larr# add 5.7 to both sides

#3x-5.7color(red)(+5.7) = 26.4color(red)(+5.7)#

#3x = 32.1" "larr div 3# on both sides

#(3x)/3 = 32.1/3#

#x = 10.7#