How do you solve #2( t + 3) < 3( t + 2)#?

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Answer 1 · 2018-05-14T09:40:26

#t > 0# graph{2 ( x + 3 ) < 3 ( x + 2 ) [-10, 10, -5, 5]}

#2t+6 < 3t + 6#
#-t < 0#
#t > 0#

Answer 2 · 2018-05-14T09:40:55

#t>0#

First lets simplify the inequality:

#2(t+3) < 3(t+2)#
#2t + 6 < 3t + 6#-----> opening the brackets

#2t-3t < 6-6#

#-t<0#

When multiplying or dividing by a negative number, the inequality sign changes, from #<# to #>#.

Hence we get:

#t>0#

Check the answer:
Take #t = 2#

#2(2+3) < 3(2+2)##

#(2xx5) < (3 xx 4)#

#10 < 12#