The task is to solve

3y+6=y-2

We first try to group terms containing the unknown ("y" in this case) to one side of the equation and those that do not contain the unknown to the other side.

With that in mind, we have to abide by some rules pertaining to algebraic manipulation.  Simply put, the rule can be sum up as "what ever operation you do on one side, do it equally on the other"

First, we want to be able to move the y on the RHS (Right hand side) to the left.  So we minus y on both side:

3y +6 -y = y -2 -y

We then get

2y +6 = -2

Now, we want to get rid of the 6 on the LHS (left hand side).  We minus 6 on both sides:

2y +6 -6 = -2 -6

Which yield

2y = -8

Now the final step.  We want to find y.  So we divide both sides by 2:

2y /2 =  -8 /2

and we get

y = -4

The required answer is y=-4

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Counter checking

We can put this newly found value of y back into the original equation to see if it works out:

LHS= 3(-4)+6 = -6

RHS= (-4)-2 = -6

LHS=RHS.  Hence the answer found was correct.

I think the question is to solve the equation for the value of y.

Now you have given 3y + 6 = y - 2

subtract y from both the sides of the equation

=> 3y - y + 6 = y - y -2

=> 2y + 6 = -2

subtract 6 from both the sides of the equation

=> 2y + 6 - 6 = -2 - 6

=> 2y = -8

divide both the sides by 2

=> 2y / 2 = -8 / 2

=> y = -4

Therefore the value of y that satisfies the equation is -4.