Big O Subtraction: $g(x) = f(x) + O(x) \rightarrow f(x) = g(x) + O(x)$

Is this true, that:

$$g(x) = f(x) + O(x) \rightarrow f(x) = g(x) + O(x)$$

In other words, is the order relation symmetric? How can I prove this?

I think this is true, since:

$$g(x) = f(x) + O(x) \rightarrow g(x) - f(x) = O(x)$$

So in other words:

$$\frac{|g(x) - f(x)|}{x} = K$$

For some value $$K$$ as $$x \rightarrow \infty$$. Similarly, we see that:

$$f(x) = g(x) + O(x) \rightarrow f(x) - g(x) = O(x)$$

So in other words:

$$\frac{|f(x) - g(x)|}{x} = M$$

For some other value $$M$$ as $$x \rightarrow \infty$$. But since $$|g(x) - f(x)|$$ = $$|f(x) - g(x)|$$, we have that $$K = M$$. So the relation holds.

Is this the way to prove this?

The idea is correct, but the problem with these proofs is that what we write as $$O(x)$$ is actually just some element from the set $$O(x)$$. So the most rigorous way would probably be to pick a representative element. Something like: let $$h\in O(x)$$ such that $$f(x) =g(x) +h(x)$$ then $$g(x) =f(x) - h(x)$$ and since $$-h \in O(x)$$ our hypotesis holds. If you find it tricky, try to find the rigorous definition of $$O(f)$$.