How do you solve the inequality #3( d + 2 )> 6#?

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Answer 1 · 2018-07-16T10:22:48

#d > 0#

As explained below:

#3(d+2)>6#

Open the brackets:

#3d + 6 > 6#

Subtract #6# both sides:

#3d+6-6 > 6 -6#

#3d+cancel(6-6) > 0 #

#3d > 0#

Divide by #3# both sides:

#(3d)/3 > 0/3#

#d > 0#

Check the answer:

Let #d = 1#

#3(1+2) > 6#

#3(3) > 6#

#9 > 6# ---> Right!