You invested $\$ 8000$ between two accounts paying $7 \%$ and $8 \%$ annual interest, respectively. If the total interest earned for the year was $\$ 610$, how much was invested at each rate? $\$ \square$ was invested at $7 \%$ and $\$ \square$ was invested at $8 \%$.

Step 1 ：Let's denote the amount invested at 7% as x and the amount invested at 8% as y.

Step 2 ：We know that the total amount invested is $8000, so we can write this as an equation: \(x + y = 8000\).

Step 3 ：We also know that the total interest earned is $610, so we can write this as another equation: \(0.07x + 0.08y = 610\).

Step 4 ：We can solve this system of equations to find the values of x and y.

Step 5 ：The solution to the system of equations is \(x = 3000\) and \(y = 5000\).

Step 6 ：Final Answer: \(\boxed{\$ 3000}\) was invested at 7% and \(\boxed{\$ 5000}\) was invested at 8%.