This question is fairly specific in the manner of steps taken to solve the problem.
Given $T(n)=2T(2n/3)+O(n)$ prove that $T(n)=O(n^2)$.
So the steps were as follows. We want to prove that $T(n) \le cn^2$.
$$\begin{align*} T(n)&=2T(2n/3)+O(n) \\ &\leq 2c(2n/3)^2+an\\ &\leq (8/9)(cn^2)+an \end{align*}$$ and then my prof went on to do:
$$T(n) \leq cn^2+(an-(1/9)cn^2)\,,$$ which comes out to:
$$T(n)\leq cn^2 \text{ for any }c >= 9a\,.$$
My question is, how were they able to switch from 8/9 to 1/9 while introducing a new term? Is this allowed? She never explained, this was just in her solutions.