Eliminate one recurrence at a time, by plugging in.
For instance, you can use the equation $P(i)=5 M(i-1)$ to eliminate every instance of $P(\cdot)$ in each of the other recurrences. Plugging in, we get
$$M(i) = R(i-1) + 10 M(i-2)$$
$$R(i) = R(i-1) + 15 M(i-2).$$
Now you have two inter-related recurrence relations, instead of three; you've eliminated one.
Do it one more time, say to remove each instance of $M(\cdot)$. You get something like
$$R(i) = R(i-1) + 15 R(i-3) + 150 M(i-4).$$
Do it again, to remove the $M(\cdot)$:
$$R(i) = R(i-1) + 15 R(i-3) + 150 R(i-5) + 1500 M(i-7).$$
Keep doing this, and you'll get a summation, say something like:
$$R(i) = R(i-1) + 15 R(i-3) + 150 R(i-5) + \cdots + 15 \times 10^{(i-3)/2} \times M(0)$$
(if $i$ is odd.)
Now you're down to a single recurrence relation. Solve it. Then, use the equations above to derive the solutions for $P(\cdot)$ and $M(\cdot)$.