A person invested $7000 for 1 year, part at 5%, part at 10%, and the remainder at 15%. The total annual income from these investments was $815. The amount of money invested at 15% was $200 more than the amount invested at 5% and 10% combined. Find the amount invested at each rate.

Let's assume the amount invested at 5% is x.

The amount invested at 10% is also x because the question says that the amount invested at 5% and 10% is the same.

Therefore, the amount invested at 15% is (x + x + $200), which simplifies to (2x + $200).

Now, let's calculate the income from each investment:

Income from the amount invested at 5% is (x * 0.05) = $0.05x.
Income from the amount invested at 10% is (x * 0.10) = $0.10x.
Income from the amount invested at 15% is ((2x + $200) * 0.15) = $0.15(2x + $200).

The total annual income from these investments is $815:

$0.05x + $0.10x + $0.15(2x + $200) = $815

Now, let's solve this equation for x:

0.05x + 0.10x + 0.15(2x + $200) = $815

0.05x + 0.10x + 0.30x + $30 = $815

0.45x + $30 = $815

0.45x = $815 - $30

0.45x = $785

x = $785 / 0.45

x ≈ $1744.44

So, the amount invested at 5% and 10% is approximately $1744.44 each.

The amount invested at 15% is (2x + $200) = (2 * $1744.44 + $200) ≈ $3488.88 + $200 ≈ $3688.88.

Therefore, the amount invested at 5% is $1744.44, the amount invested at 10% is $1744.44, and the amount invested at 15% is $3688.88.

To solve this problem, let's assign variables to the unknowns.

Let x be the amount invested at 5%.
Let y be the amount invested at 10%.
Let z be the amount invested at 15%.

Based on the given information, we can form three equations:

1. The total amount invested is $7000:
x + y + z = 7000

2. The total income from the investments is $815:
0.05x + 0.10y + 0.15z = 815

3. The amount invested at 15% is $200 more than the amount invested at 5% and 10% combined:
z = x + y + 200

Now we have a system of three equations. We can solve it using substitution or elimination method to find the values of x, y, and z. Let's use substitution:

From equation 3, we know that z = x + y + 200. We can substitute this value for z in equations 1 and 2:

Equation 1: x + y + (x + y + 200) = 7000
Simplifying, we get: 2x + 2y + 200 = 7000
Rearranging the equation: 2x + 2y = 7000 - 200
2x + 2y = 6800

Equation 2: 0.05x + 0.10y + 0.15(x + y + 200) = 815
Simplifying, we get: 0.05x + 0.10y + 0.15x + 0.15y + 30 = 815
Combining like terms: 0.20x + 0.25y = 785

Now we have a new system of two equations:

2x + 2y = 6800
0.20x + 0.25y = 785

To eliminate the decimals, we can multiply the second equation by 100:

20x + 25y = 78500

Now we have the following system of two equations:

2x + 2y = 6800
20x + 25y = 78500

We can solve this system using elimination or substitution method. Let's use elimination:

Multiply the first equation by 10:

20x + 20y = 68000

Now subtract the new equation from the second equation:

(20x + 25y) - (20x + 20y) = 78500 - 68000
5y = 10500
y = 2100

Substitute the value of y back into the first equation:

2x + 2(2100) = 6800
2x + 4200 = 6800
2x = 2600
x = 1300

Now we have the values of x and y. To find z, we can substitute the values of x and y back into equation 3:

z = x + y + 200
z = 1300 + 2100 + 200
z = 3600

Therefore, the amounts invested at each rate are:
$1300 invested at 5%
$2100 invested at 10%
$3600 invested at 15%

If x at 5%, y at 10%, then 200+x+y at 15%

x+y+200+x+y = 7000
.05x + .10y + .15(200+x+y) = 815

Now just solve for x and y, and then x+y+200