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%?

see my answer to your re-post

To solve this problem, let's break it down step by step:

Step 1: Let's assume that Roberto initially invested x dollars at 7%.
So, the amount invested at 7% will earn an annual interest of 0.07x.

Step 2: According to the problem, Roberto then invested $2000 more than twice the amount he invested at 7%.
Twice the amount invested at 7% is 2x, and with an additional $2000, the total amount invested at 11% becomes 2x + $2000.

Step 3: The amount invested at 11% will earn an annual interest of 0.11(2x + $2000).

Step 4: The total annual income from the two investments is given as $3990.
Therefore, we can write the equation: 0.07x + 0.11(2x + $2000) = $3990.

Step 5: Simplify and solve the equation:
0.07x + 0.22x + $220 = $3990.
0.29x + $220 = $3990.
0.29x = $3990 - $220.
0.29x = $3770.
x = $3770 / 0.29.
x ≈ $13,000.

Step 6: We have found that Roberto initially invested approximately $13,000 at 7%.
Therefore, the amount invested at 11% is 2x + $2000 = 2($13,000) + $2000 = $28,000.

So, the amount invested at 11% is approximately $28,000.