A person invested $\mathrm{\$}6100$ for 1 year, part at $5\mathrm{\%}$, part at $11\mathrm{\%}$, and the remainder at $13\mathrm{\%}$. The total annual income from these investments was $\mathrm{\$}665$. The amount of money invested at $13\mathrm{\%}$ was $\mathrm{\$}500$ more than the amounts invested at $5\mathrm{\%}$ and $11\mathrm{\%}$ combined. Find the amount invested at each rate.

The person invested $\mathrm{\$}◻$ at $5\mathrm{\%},\mathrm{\$}◻$ at $11\mathrm{\%}$, and $\mathrm{\$}◻$ at $13\mathrm{\%}$.

Step 1 ：Let's denote the amount invested at 5% as $x$, the amount invested at 11% as $y$, and the amount invested at 13% as $z$.

Step 2 ：From the problem, we know that the total amount invested is $6100, so we have the equation $x+y+z=6100$.

Step 3 ：We also know that the total annual income from these investments was $665, which gives us the equation $0.05x+0.11y+0.13z=665$.

Step 4 ：Finally, we know that the amount of money invested at 13% was $500 more than the amounts invested at 5% and 11% combined, so we have the equation $z=x+y+500$.

Step 5 ：Solving this system of equations, we find that $x=1200$, $y=1600$, and $z=3300$.

Step 6 ：So, the final answer is: The person invested $\boxed{{\displaystyle \mathrm{\$}1200}}$ at 5%, $\boxed{{\displaystyle \mathrm{\$}1600}}$ at 11%, and $\boxed{{\displaystyle \mathrm{\$}3300}}$ at 13%.