How do you solve #12-[2+6y-3(y+4)]=-3(2y-5)-[5(y-1)-6y+4]#?

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Answer 1 · 2016-09-09T12:49:11

#- 3#

We have: #12 - [2 + 6 y - 3 (y + 4)] = - 3 (2 y - 5) - [5 (y - 1) - 6 y + 4]#

Let's expand the parentheses:

#=> 12 - [2 + 6 y - 3 y - 12] = - 6 y + 15 - [5 y - 5 - 6 y + 4]#

Then, let's simplify the terms within the brackets:

#=> 12 - [3 y - 10] = - 6 y + 15 - [- y - 1]#

#=> 12 - 3 y + 10 = - 6 y + 15 + y + 1#

Now, let's move all terms containing #y# to the left-hand side of the equation:

#=> 22 - 3 y + 6 y - y = 16#

#=>22 + 2 y = 16#

Let's subtract both sides by #22#:

#=> 2 y = - 6#

Finally, to solve for #y#, let's divide both sides by #2#:

#=> y = - 3#