Independence does not seem physically plausible, but we are explicitly asked to assume it.
Let $X_1$ and $X_2$, respectively, be the numbers of call users and text users connected, or at least wishing to connect. Then $X_1$ and $X_2$ have binomial distribution. (In the real world there may be truncation.) Let $W=r_1X_1+r_2X_2$. Then $W$ is the random variable that we want to be $\le C$.
It would be reasonable next to use the normal approximation. This is because the distributions of the $X_i$ probably can be well approximated by normals, and then $W$ is a linear combination of independent nearly normals. The mean and variance of $W$ are not hard to find. In extreme cases (the $n_i$ large but the $p_i$ small, with $n_ip_i$ smallish) this might not work well, and we might want to look at linear combinations of Poisson random variables.
The above approach is probably what you need. But there could be problems if you want to choose $C$ so that the probability of overload is very small. In that case you would need specialized estimates for extreme tail probabilities, since the normal approximation tends to underestimate probabilities far in the tail.
Comment: The distributions involved in possible analyses do not include the geometric distribution of the title.