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

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 $\boxed{{\displaystyle \mathrm{\$}2100}}$ at 14% and $\boxed{{\displaystyle \mathrm{\$}2400}}$ at 4%.