How do you solve #3y + 1/2 = 4 + 2/3 y#?

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Answer 1 · 2015-05-24T21:50:01

First subtract #1/2# from both sides to get:

#3y = 7/2 + 2/3y#

Subtract #2/3y# from both sides to get:

#7/3y = 7/2#

Multiply both sides by #3/7# to get:

#y = 3/2#

Check by substituting this value back into the original equation:

LHS #= 3y+1/2 = 3(3/2)+1/2 = 9/2+1/2 = 10/2 = 5#

RHS #= 4 + 2/3y = 4+(2/3*3/2) = 4+1 = 5#