# Algebra: Simultaneous Recurrence Relations

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Solve the simultaneous recurrence relations given below.

a(n) = 3a(n-1) + 2b(n-1)

b(n) = a(n-1) + 2b(n-1)

a(0) = 1

b(0) = 2

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##### Solution Summary

The expert solves a system of two recurrence sequences.

##### Solution Preview

We have the following system:

Firstly, we'll subtract the second equation from the first one, and get:

(1)

Now, if we put n-1 instead of n, we'll get:

(2)

Then, from the first equation, we'll subtract from both sides, and get:

- = 2( - ) (3).

And now we'll substitute (2) in (3) and get:

- = 4 , which is same as ...

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