How do you solve 3x - 1 < x + 12 ?

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Answer 1 · 2018-02-26T23:33:57

The solution is #x<13/2#

Isolate the variable #x# and then simplify.

#3x-1 < x+12#

#3x-1 color(red)(-x) < x+12color(red)(-x) #

#3x-x-1 < color(red)cancel(color(black)(x-x))+12 #

#2x-1 < 12#

#2x -1 color(red)(+1) < 12color(red)(+1) #

#2x color(red)cancel(color(black)(-1+1)) < 12 + 1#

#2x < 12 + 1#

#2x < 13#

#(2x)/2<13/2#

#(color(red)cancel(color(black)2)x)/(color(red)cancel(color(black)2)) < 13/2#

#x<13/2#

or

#x<6.5#