an executive invests $35000, some at 9% and some at 7% annual interest. If he receives an annual return of $2910, how much is invested at each rate
.09*X + .07*(35000-x)=2910
solve for x
X is the amount invested at nine percent, 35000-X is the remaining amount.
To find out how much is invested at each interest rate, we can set up a system of equations.
Let's assume the amount invested at 9% is 'x' dollars and the amount invested at 7% is '35000 - x' dollars (since the total investment is $35,000).
The interest earned from the investment at 9% is then 0.09x.
The interest earned from the investment at 7% is 0.07(35000 - x).
Given that the total annual return is $2910, we can write the equation as:
0.09x + 0.07(35000 - x) = 2910
Now we can solve this equation to find the value of 'x', which represents the amount invested at 9% interest.
0.09x + 0.07(35000 - x) = 2910
0.09x + 2450 - 0.07x = 2910
0.02x = 2910 - 2450
0.02x = 460
x = 460 / 0.02
x = 23000
So, $23,000 is invested at 9% interest, and $12,000 (35000 - 23000) is invested at 7% interest.