# If $8700 is invested at 3% annual simple interest, how much should be invested at 6% annual simple interest so that the total yearly income for both investments is $393?

first off, see how much 8700 as principal, yields at 3% APR that is [latex]\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$8700\\ r=rate\to 3\%\to \frac{3}{100}\to &0.03\\ t=years\to &1 \end{cases}[/latex] it will yield some amount subtract that amount from 393 the difference is how much the yield will be on the 6% investment so [latex]\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\quad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$8700\\ r=rate\to 3\%\to \frac{3}{100}\to &0.03\\ t=years\to &1 \end{cases} \\\\\\ \implies \boxed{?}\\\\ -----------------------------\\\\ \textit{how much to invest at 6\%?} \\\\\\ [/latex] [latex]\bf \qquad \textit{Simple Interest Earned}\\\\ (393-\boxed{?}) = Prt\quad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\to& \$\\ r=rate\to 6\%\to \frac{6}{100}\to &0.06\\ t=years\to &1 \end{cases} \\\\\\ \textit{solve for "P", to see how much should the Principal be}\\\\ \textit{keep in mind that }P+\boxed{?}=393\leftarrow \textit{both yields added}[/latex]