How do you solve #6= 2( y + 2) #?

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Answer 1 · 2017-08-13T02:48:57

#y=1#

Begin by distributing #2# to #(y+2)#

#6=(2*y)+(2*2)#

#6=2y+4#

Subtract #y# from both sides:

#6-4=2y+cancel(4-4)#

#2=2y#

Divide #2# from both sides to solve for #y#

#2/2=cancel2/cancel2y#

#1=y#