Answer
Claim: If a is any odd integer and b is any even integer, then 2a + 3b is even.
Proof: Suppose a and b are a particular but arbitrarily chosen integers such that a is odd and b is even.
Since a is an odd integer, then a = 2r + 1.
Since b is an even integer, then b = 2s.
By substitution, 2a + 3b = 2(2r + 1) + 3(2s).
Using the Distributive law and algebra, 2(2r+1) + 3(2s) = 2(2r+1) + 2(3s) = 2(2r+1 + 3s).
Let t = (2r+1 + 3s), then 2a + 3b = 2t.
By definition of even, 2t = 2a + 3b is even.
Work Step by Step
Define a as an odd integer such that a = 2r + 1. Define b as an even integer such that b = 2s. Substitute these values into the equation 2a + 3b. Use the Distributive law to determine 2a + 3b = 2(2r+1 + 3s). Let t = (2r+1 + 3s), then 2a + 3b = 2t. 2t is an even integer by definition of even.
