A loan of 5000 can be repaid by payments of 117.38 at the end of each month for n years (12n payments), starting one month after the loan is made. At the same rate of interest, 12n monthly payments of 113.40 each accumulate to 10000 one month after the final payment. Find the equivalent effective annual rate of interest.
The wording of the question is a little confusing to me. I attempted the question, and let x = 1+i at some point, but nothing seemed to work out. Any help would be appreciated
Yikes! The teacher that gave you this sure was in a bad mood!
I've reworded your problem this way (same thing):
At annual rate of R%, a loan of 5000 is repaid with N monthly payments of 117.38,
starting one month after the loan is made.
At the same annual rate of R%, N monthly deposits of 113.40 accumulate to 10000
one month after the Nth deposit.
Find the equivalent effective annual rate of interest.
Answer is: 14.4%, 60 months.
To give you an idea of what's involved (and keeping typing to a minimum!):
i = R/12 : u = 117.38 : v = 113.40 : a = 5000 : b = 10000
Using the "loan payment" formula leads to:
(1 + i)^n = u / (u - ai) 
Using the "future value of annuity" formula leads to:
(1 + i)^n = 1 + bi / (v(1 + i)) 
u / (u - ai) = 1 + bi / (v(1 + i))
Solve for i:
i = (ub - av) / (ab + av)
Substitute back in: i = .012 ; R = .012 * 12 = .144 or 14.4%
The deposit account ends at ~9881.42 after 60th deposit; 9881.42 * 1.012 = ~10000.00