Two consecutive numbers that are odd such that three times the first is 5 more than twice the second.
Solution. Let the first odd number be 2n + 1.
Then, the next one is 2n + 3 -- because it will be 2 more.
The problem says, that is, the equation is:
3(2n + 1) = 2(2n + 3) + 5. That implies: 6n + 3 = 4n + 6 + 5 2n = 8 n = 4
Consequently, the first odd number is 2· 4 + 1 = 9.
The next one is 11.
According to the problem, that is the true solution. 3 · 9 = 2· 11 + 5.