$\$ 17,818$ is invested, part at $11 \%$ and the rest at $6 \%$. If the interest earned from the amount invested at $11 \%$ exceeds the interest earned from the amount invested at $6 \%$ by $\$ 490.33$, how much is invested at each rate? (Round to two decimal places if necessary.)

Step 1 ：Let's denote the amount invested at 11% as x and the amount invested at 6% as y. We know that the total amount invested is $17,818, so we have the equation \(x + y = 17818\).

Step 2 ：We also know that the interest earned from the amount invested at 11% exceeds the interest earned from the amount invested at 6% by $490.33. This gives us the equation \(0.11x - 0.06y = 490.33\).

Step 3 ：We can solve this system of equations to find the values of x and y.

Step 4 ：The solution to the system of equations gives us the amounts invested at each rate. The amount invested at 11% is $9173 and the amount invested at 6% is $8645.

Step 5 ：Final Answer: The amount invested at 11% is \$9173 and the amount invested at 6% is \$8645. So, the final answer is \(\boxed{9173, 8645}\).