Final answer:
To solve the problem, a system of equations was set up and solved using substitution and simplification to find the amount invested at each interest rate: $2200 at 8%, $1400 at 10%, and $4400 at 13%.
Step-by-step explanation:
Let x be the amount invested at 8%, y be the amount at 10%, and z be the amount at 13%. From the problem, we have the following system of equations:
- x + y + z = 8000 (total investment)
- 0.08x + 0.10y + 0.13z = 910 (total annual income)
- z = x + y + 800 (amount at 13% is $800 more than the sum invested at 8% and 10%)
Substitute the third equation into the first equation to get a simplified system:
- x + y + (x + y + 800) = 8000
- 0.08x + 0.10y + 0.13(x + y + 800) = 910
Simplify each equation:
- 2x + 2y = 7200
- 0.08x + 0.10y + 0.13x + 0.13y + 104 = 910
Simplify further and solve the linear system for x and y:
- x + y = 3600
- 0.21x + 0.23y = 806
From the first simplified equation, we can express y in terms of x: y = 3600 - x. Substitute this into the second simplified equation and solve for x to get x = 2200. Then y = 3600 - 2200 = 1400. Finally, using the initial relationships, z = 8000 - x - y, which gives us z = 8000 - 2200 - 1400 = 4400.
The amount invested at each rate is then: $2200 at 8%, $1400 at 10%, and $4400 at 13%.