Add 5 to both sides

#|3x+2|<5 #

Note that #|3x+2|# is always positive so there is a limit to the negative value of #3x+2#

Suppose #3x+2=5 => x= 1 #

Then the maximum value attributable to #x# is such that #x < 1#

Suppose #3x+2=-5 => x=-7/3#

So the minimum value of #x# is such that # -7/3 < x#

So #" " -7/3 < x < 1 #

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#color(blue)("In summery:")#

solve for #3x+2=|+-5|# and make the appropriate value of #x# the #color(magenta)(ul("not inclusive"))# upper and lower bounds. #" " -7/3color(magenta)( <) xcolor(magenta)( <) 1 #

'.............................................................................................

#color(brown)("Just for information:")#

If they had been #color(blue)(ul("inclusive"))# then we would have #-7/3 <= x <= 1#