Given :
Roberto invested a total of $10,000, part at 6% and part at 9%.
To Find :
How much did he invest in the higher interest account if the total interest earned in one year was $810 .
Solution :
Let , x amount is invested in 6% interest .
Therefore , 10000-x amount is invested in 9% interest .
Now , total interest is $810 .
We know , simple interest is given by :
![I=(PRt)/(100)](https://img.qammunity.org/2021/formulas/mathematics/high-school/n95xazm6xddqyb7gsj1ms9yensneuyizn6.png)
So ,
![(x(1)(6))/(100)+((10000-x)(1)(9))/(100)=810\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/dy4roko1epksx13oqe1d2juc158jmmi9il.png)
Solving above equation , we get :
.
Therefore , investment on higher i.e 9 % is 10000-3000 = $7000 .
Hence , this is the required solution .