Nested Loop Time Complexity vs Successive Loops
I think that the first algorithms running time is actually $\mathcal{O}(x^4)$. 


The outer for loop executes exactly $x$ times,
the inner for loop executes at most $x$ times during every one of these $x$ iterations of the outer loop


In fact, the inner loop executes exactly $\frac{x(x+1)}{2}$ times, and $\frac{x(x+1)}{2} = \frac{x^2 + x}{2} = \mathcal{O}(x^2)$, since our asymptotic notation allows us to essentially ignore the constant factors.

So already we have a growing complexity of $\mathcal{O}(x^2)$. If we now consider the contents of the inner for loop, which you say is a quadratic time calculation of $x$. I read this as follows: the complexity of the inner loops body is a quadratic function of x $\rightarrow \mathcal{O}(x^2)$. My answer is based largely upon this assumption. 

If my assumption is correct, then a time complexity of $\mathcal{O}(x^4)$ easily follows from this reasoning: the outer and inner for loop collectively produce $\mathcal{O}(x^2)$ iterations of some algorithm that takes $\mathcal{O}(x^2)$ steps to complete itself. An extremely informal (read: not quite right, but right enough to demonstrate a point) mathematical argument for this might be: 

($x$ outer loop iterations) $\cdot$ ($x$ inner loop iterations) $\cdot$ ($x^2$ loop body steps) $= x^4$ total steps. 

With the second algorithm, the same reasoning can be used. Again, we have some procedure, which takes $\mathcal{O}(x^2)$ steps to run, which is being executed $x$ times. This gives us $\mathcal{O}(x^3)$ steps in total. Since we have two loops of this form, we can conclude that algorithm 2 has a time complexity of: 

$\mathcal{O}(x^3) + \mathcal{O}(x^3) = \mathcal{O}(x^3)$

Note that in this case we have two consecutive loops that, asymptotically speaking, do essentially the same amount of work. In reality it might be that our first loop which calculates $y$ performs some constant factor amount more or less steps than our second, which calculates $z$, but here that doesn't matter, since we're focusing on the rate of growth of our running time. One way of thinking about it is as follows:

$\mathcal{O}(x^3) + \mathcal{O}(x^3) = \mathcal{O}(x^3 + x^3) = \mathcal{O}(2 \cdot (x^3)) = \mathcal{O}(x^3)$. 

Here we can drop the 2, since its a constant factor.