For any a, b, c, if a = b and b = c, then a = c, right? The transitive property of equality.
For any a, b, x, if a = b, then x + a = x + b. The substitution property.
By combining both of these properties, for any a, b, x, y, if a = b and y = b + x, it follows that b + x = a + x and y = a + x.

In our example, a is x' (notice the ') and b is 0.999… (by definition). y is 10x' and x is 9. Let’s fill in the values.

If x' = 0.9999… (true by definition) and 10x = 0.999… + 9 (true by algebraic manipulation), then 0.999… + 9 = x' + 9 and 10x' = x' + 9.

if you are rearranging algebra you have to do the exact same thing on both sides

If you actually change any of the sides. Since, after substitution, the numeric value doesn’t change (literally the definition of equality), I don’t have to do anything – as I’m not rearranging. I’m merely presenting the same value in an equivalent manner. By contrast, when multiplying both sides by 10, since multiplication by 10 changes the concrete numeric value, I have to do it on both sides to maintain the equality relation (ditto for subtracting x'). But substitution never changes a numeric value – only rearranges what we already know.

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(Edit)

Take the following simple system of equations.

5y = 3
x + y = 6

How would you solve it? Here’s how I would:

\begin{gather*} %% Ignore the LaTeX boilerplate, just so I could render it
\begin{cases}
y = \frac{3}{5} \\ % Isolate y by dividing both sides by 5
x = 6 - y % Subtract y from both sides
\end{cases} \\
x = 6 - \frac{3}{5} \\ % SUBSTITUTE 3/5 for y
x = 5.4 \\
(x, y) = (5.4, 0.6)
\end{gather*}

Here’s how Microsoft Math Solver would do it.