Solve the problem. Roberto invested some money at 7%, and then invested $2000 more than twice this amount at 11%. His total annual income from the two investments was $3990. How much was invested at 11%?
Set up an algebra equation
X*.07 + (2000+2*X)*.11 = 3990.
Solve for X and then for 2000+2X
To solve this problem, we can set up a system of equations.
Let's say Roberto invested some amount of money at 7%, which we'll call "x".
According to the problem, he then invested $2000 more than twice this amount at 11%. This means that the amount invested at 11% is $2000 + 2x.
We know that his total annual income from the two investments was $3990.
The interest earned from the investment at 7% can be calculated as 0.07*x, and the interest earned from the investment at 11% can be calculated as 0.11*(2000 + 2x).
Therefore, we can set up the equation:
0.07x + 0.11(2000 + 2x) = 3990
To solve for x, we can simplify the equation and then solve:
0.07x + 0.22x + 220 = 3990
0.29x = 3770
x ≈ 13,000
So, Roberto invested approximately $13,000 at 7%.
Now, to find how much was invested at 11%, we can substitute this value back into our expression: $2000 + 2x.
2*13,000 + 2000 = $28,000
Therefore, Roberto invested approximately $28,000 at 11%.