Wow, nasty question. Too many old memories. I'll do my best.
To solve this we want to have an equation in the form of x=(a number)
So we need to start moving things around. To get x on its own.
Let's start with that pesky 15. We just subtract it 15-15=0 and so the term no longer exists (or rather, is just equal to 0 and not worth writing) on the LEFT hand side of the equation.
But this is an equation, the two sides have been stated to be equal and we can't just subtract numbers from one side all willy nilly, or else you could say that 7=7, therefore 7-4=7, therefore 3=7! Whatever we do to the left, we must do to the whole thing. So in the above example we would say that 7=7, 7-4=7-4 (we have to do the same thing to both sides) and therefore 3=3. Hey, that works!
So let's do it, 5x+15-15=0-15
Then we have that 15-15 on the left so
5x= - 15
Now to get rid of the 5 on the 5x. And get x on its own (solve the equation).
If we have 5 times a number and want that number on its own, we divide by 5, and once again we have to do it to both sides, or else 6=6 would be the same as 6/3=6, 3=6 which obviously isn't right.,
So we divide both sides by 5
5x/5= -15/5
x= -15/5 (and since 15=5 times 3 and we have -15)
x= -3
Hope that helps!