|3y+6|=|3y+3|

CPhill is correct - He's rarely wrong.  I just want to look at it myself.

Think about this for a moment

$$|a|=|t|$$

This would be true if x=t and it would also be true if x=-t 

The same thing applies to your question:

$$\begin{array}{rlll} |3y+6|&=&|3y+3|\\ (3y+6)&=&\pm (3y+3)\quad &\mbox{now factorise both sides}\\ 3(y+2)&=&\pm 3(y+1)\quad &\mbox{next divide both sides by 3}\\ y+2&=&\pm (y+1) \end{array}$$

Now we seperate this into 2 equations

$$\begin{array}{rlllrl} y+2&=&+(y+1)\quad &\mbox{and}\qquad y+2&=&-(y+1)\\ y+2&=&y+1\quad &\mbox{and}\qquad y+2&=&-y-1\\ y&=&y-1\quad &\mbox{and}\qquad 2y&=&-3\\ y&=&\mbox{no soln}\quad &\mbox{and}\qquad y&=&\frac{-3}{2}\\ \end{array}$$

So the answer is y=-1.5

I'm going to change the "y's"  to "x's"

So we have

l3x+6l = l3x+3l

Factor a 3 out of each and we get

lx+2l = lx+1l

Take both things out of the absolute value bars. So we have

x+2 = x+1    and there's no solution to this

Now set x+2 = -(x+1)   .......So we have

x+2 = -x -1        Add x to both sides

2x + 2 = -1          Subtract 2 from both sides

2x = -3                Divide by 2 on both sides

x = -3/2  = -1.5

Here's a graph of the solution:

P.S.   ...   You can back-substitute "y" for "x" if you'd like....

The leftmost function on the graph is l3x+6l and the rightmost is l3x+3l

You can back-substitute "y" for "x" if you'd like......