Miles invests $\$ 4500$ in two different accounts. The first account paid $14 \%$, the second account paid $4 \%$ in interest. At the end of the first year he had earned $\$ 390$ in interest. How much was in each account? \[ \begin{array}{l} \$ \square \text { at } 14 \% \\ \$ \square \text { at } 4 \% \end{array} \]

Step 1 ：Let's denote the amount of money invested at 14% as \(x\) and the amount of money invested at 4% as \(y\).

Step 2 ：From the problem, we know that the total amount of money invested is \$4500, so we have the equation \(x + y = 4500\).

Step 3 ：We also know that the total interest earned is \$390, and the interest is calculated as 14% of \(x\) and 4% of \(y\), so we have the equation \(0.14x + 0.04y = 390\).

Step 4 ：We can solve the first equation for \(x\), which gives us \(x = 4500 - y\).

Step 5 ：Substitute \(x = 4500 - y\) into the second equation, we get \(0.14(4500 - y) + 0.04y = 390\), which simplifies to \(630 - 0.1y = 390\).

Step 6 ：Solving this equation for \(y\), we get \(y = 2400\).

Step 7 ：Substitute \(y = 2400\) back into the first equation, we get \(x = 4500 - 2400 = 2100\).

Step 8 ：So, Miles invested \(\boxed{\$2100}\) at 14% and \(\boxed{\$2400}\) at 4%.