How do you solve #3x + 3< x + 6#?

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Answer 1 · 2018-04-26T14:55:38

#x<3/2# or #x<1.5#

Given: #3x+3<x+6#.

Subtract #x# from both sides.

#3x+3-x<color(red)cancelcolor(black)x+6-color(red)cancelcolor(black)x#

#2x+3<6#

Subtract #3# from both sides.

#2x+color(red)cancelcolor(black)3-color(red)cancelcolor(black)3<6-3#

#2x<3#

Divide by #2# to get #x#.

#(color(red)cancelcolor(black)2x)/(color(red)cancelcolor(black)2)<3/2#

#x<3/2=1.5#