D=a+b+c over ac. if you under stand

You forgot to say what the question actually is ! 

But  I've seen this problem before, in the last few days, here on 
I think the problem has three parts:  1). Solve the equation for 'a'; 
2). Solve it for 'b'; and 3). Solve it for 'c'.

I'll slog through that, and I'll try to explain what I'm doing clearly enough
so that eventually, you can do it on your own ... which is really the whole
idea behind this website.

1).  Solve the equation for 'a'.  That means you have to wind up with something
that says      a = everything else.

D = (a + b + c) / c

Split the right side into 3 fractions:      D = a/c + b/c + c/c

But  c/c =1 ,  so the equation says      D = a/c + b/c + 1

Subtract  (b/c +1)  from each side:      D - b/c - 1  =  a/c

Multiply each side by 'c' :                  Dc - b - c  =  a


2).  Solve the equation for 'b'.  That means you have to wind up with something
that says      b = everything else.

D = (a + b + c) / c

Split the right side into 3 fractions:      D = a/c + b/c + c/c

But  c/c =1 ,  so the equation says      D = a/c + b/c + 1

Subtract  (a/c +1)  from each side:      D - a/c - 1  =  b/c

Multiply each side by 'c' :                  Dc - a - c  =  b


3).  Solve the equation for 'c'.  That means you have to wind up with
something that says      c = everything else.

D = (a + b + c) / c

Split the right side into 3 fractions:      D = a/c + b/c + c/c

But  c/c =1 ,  so the equation says      D = a/c + b/c + 1

Subtract  1  from each side:          D - 1 = a/c + b/c

The two fractions on the right can be added/combined:    D - 1 = (a + b) / c

Multiply each side by  'c' :          c(D - 1) = (a + b)

Divide each side by  (D - 1) :      c  =  (a + b) / (D - 1) 

Plug in the numbers

For example
A-2
B-3
C-8

Your problem would be 2+3+8 divided by 2 times 8