Answer:
It was invested $18,000 at a 3% annual interest and $3,000 at a 9% annual interest.
Step-by-step explanation:
To determine how much was invested at each rate, we can create a system of equations.
Let x be the amount invested at 3%
Let y be the amount invested at 9%
If the total invested was $21,000 and the interest earned was $810
[tex]\begin{gathered} \text{0}.03x+0.09y=810 \\ \text{ Then, if} \\ y=21,000-x \end{gathered}[/tex]Substitute y into the first equation, and solve for x.
[tex]\begin{gathered} 0.03x+0.09(21000-x)=810 \\ 0.03x+1890-0.09x=810 \\ \text{0}.09x-0.03x=1890-810 \\ 0.06x=1080 \\ x=\frac{1,080}{0.06} \\ x=\text{ \$18,000} \end{gathered}[/tex]Now, if x=18,000, substitute it into the y-equation:
[tex]\begin{gathered} y=21000-18000 \\ y=\text{ \$3,000} \end{gathered}[/tex]It was invested $18,000 at a 3% annual interest and $3,000 at a 9% annual interest.
