How do you solve #2/3y-6=12#?

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Answer 1 · 2016-10-27T14:53:41

#y = 27#

First add #6# to each side to give:

#(2/3)y = 18#

Then multiple each side by #3/2# to give:

#y = 27#

Answer 2 · 2016-10-27T14:59:49

Work backwards to isolate y ( PE SADM)

Do parenthesis and then use the order of operations backwards.

SA comes first so do the subtraction and addition first. The opposite of of -6 is + 6 so add 6 to both sides.

# 2/3xx y - 6 + 6 = 12 + 6# This results in

# 2/3 xx y = 18 #

Next comes DM so divide and multiply by the opposite.

The opposite or inverse of # 2/3# is # 3/2# so multiply both sides by # 3/2#

# 2/3 xx 3/2 xx y = 18 xx 3/2# : # 2/3 xx 3/2 = 1# so

y  =  27