Answer:
11,500 at 2% and $6,000 at 12%.
Step-by-step explanation:
Let x represent amount invested at 2% and y represent amount invested at 12%.
We have been given that Theodore invests a total of $17,500 in two accounts. We can represent this information in an equation as:


We are also told that he invested $17,500 in two accounts paying 2% and 12% annual interest, respectively. After one year, the total interest was $950.00.
Interest earned at 2% in one year would be
.
Interest earned at 12% in one year would be 

Upon substituting equation (1) in equation (2), we will get:







Therefore, Theodore invested $6,000 at 12%.
Upon substituting
in equation (1), we will get:


Therefore, Theodore invested $11,500 at 2%.