Answer: The amount invested at 6% was $1393.57, the amount invested at 9% was $306.13, and the amount invested at 15% was $1393.57 + $306.13 + $100 = $1800.70.
Explanation:
Let's denote the amount invested at 6% as "x". Then, the amount invested at 9% would be "y", and the amount invested at 15% would be "z".
According to the problem, we have the following information:
x + y + z = $7500 (the total amount invested)
0.06x + 0.09y + 0.15z = $867 (the total annual income)
And,
z = x + y + $100 (the amount invested at 15% was $100 more than the amounts invested at 6% and 9% combined)
We can substitute the third equation into the first equation to get:
x + y + x + y + $100 = $7500
Simplifying:
2x + 2y + $100 = $7500
2x + 2y = $7400
Next, we can substitute "z = x + y + $100" into the second equation:
0.06x + 0.09y + 0.15(x + y + $100) = $867
Expanding:
0.06x + 0.09y + 0.15x + 0.15y + 0.15 * $100 = $867
0.21x + 0.24y + $15 = $867
0.21x + 0.24y = $852
Finally, we can solve for x and y using the equations "2x + 2y = $7400" and "0.21x + 0.24y = $852".
We can use elimination method to find x:
2x + 2y = $7400
0.21x + 0.24y = $852
Multiplying the first equation by 0.21:
0.42x + 0.42y = $1557.8
Subtracting the second equation from the first:
0.42x + 0.42y - 0.21x - 0.24y = $1557.8 - $852
0.21x - 0.03y = $705.8
Dividing both sides by 0.21:
x = $705.8 / 0.21 + 0.03y / 0.21
x = $3357.14 / 0.24 + y
Finally, substituting x back into the equation "2x + 2y = $7400":
2($3357.14 / 0.24 + y) + 2y = $7400
Expanding:
$6714.29 + 2.24y = $7400
Solving for y:
2.24y = $7400 - $6714.29
y = $685.71 / 2.24
y = $306.13
And finally, substituting y back into the equation "x = $3357.14 / 0.24 + y":
x = $3357.14 / 0.24 + $306.13
x = $1393.57
Therefore, the amount invested at 6% was $1393.57, the amount invested at 9% was $306.13, and the amount invested at 15% was $1393.57 + $306.13 + $100 = $1800.70.