I'll offer up a more general claim and a proof, you can apply it to your scenario as needed.
The Uneven Split Theorem
Let $c$ and $k$ be positive constants.
Then let $\{a_1, a_2, \dots a_k\}$ be positive constants such that $\sum_1^k a_i < 1$.
We also must have a recurrence of the form:
$$\begin{align}
T(n) & \leq c & 0 < n < max\{a_1^{-1}, a_2^{-1}, \dots a_k^{-1}\}\\
T(n) & \leq cn + T(a_1 n) + T(a_2 n) + \dots T(a_k n) & n \geq max\{a_1^{-1}, a_2^{-1}, \dots a_k^{-1}\}
\end{align}$$
Claim
Then I claim $T(n) \leq bn$ where:
$$b = \frac{c}{1 - \left(\sum_1^k a_i\right)}$$
Proof by Induction
Basis: $n < max\{a_1^{-1}, a_2^{-1}, \dots a_k^{-1}\} \implies T(n) \leq c < b < bn$
Induction: Assume true for any $n' < n$, we then have
$$\begin{align}
T(n) & \leq cn + T(\lfloor a_1 n \rfloor) + T(\lfloor a_2 n \rfloor) + \dots + T(\lfloor a_k n \rfloor)\\
& \leq cn + b \lfloor a_1 n \rfloor + b \lfloor a_2 n \rfloor + \dots + b \lfloor a_k n \rfloor\\
& \leq cn + b a_1 n + b a_2 n + \dots + b a_k n\\
& = cn + bn \sum_1^k a_i\\[0.5em]
& = \frac{cn - cn \sum_1^k a_i }{1 - \left(\sum_1^k a_i\right)} + \frac{cn \sum_1^k a_i}{1 - \left(\sum_1^k a_i\right)}\\[0.5em]
& = \frac{cn}{1 - \left(\sum_1^k a_i\right)}\\
& = bn & \square
\end{align}$$
Then we have $T(n) < bn \implies T(n) = O(n)$.
We also know $T(n) = \Omega(n)$ by the recurrence definition, therefore $T(n) = \Theta(n)$.
Next apply it to your recurrence of the form:
$$T(n)=cn + T\left(\frac{n}{5}\right) + T\left(\frac{n}{5}\right) +T\left(\frac{n}{10}\right) + T\left(\frac{n}{10}\right) + T\left(\frac{n}{10}\right)$$
I'll leave this for you to figure out the $a_k$'s and $b$.
A pretty quick rule of thumb if the work done per recursion is linear $(cn)$:
If the constants in the recursion calls sum to less than $1$, it's $O(n)$.
If the constants sum to exactly $1$, it's $O(n \log n)$.
