Almost correct , $\displaystyle x \equiv 4 (\bmod.5)$ means that either $\displaystyle x\equiv 4 (\bmod.10)$ or $\displaystyle x\equiv 9 (\bmod.10)$

Note that you have a

*degree of freedom* in what $\displaystyle x$ is in module 2 . This happnes because fixing the values of $\displaystyle x(\bmod. b_1)$ and $\displaystyle x (\bmod.b_2)$ determines uniquely $\displaystyle x (\bmod.b_1\cdot b_2)$ and viceversa, when $\displaystyle (b_1,b_2)=1$. (this is a particular case of the

Chinese Remainder Theorem )