The amount invested at 4% was $81, and the amount invested at 9% was $825. The amount invested at 15% was $300 more than the combined amount invested at 4% and 9%, which is $81 + $825 + $300 = $1206.
Let x be the amount of money invested at 4%, and y be the amount of money invested at 9%. We know that x + y + (x + y + 300) = 6100, so the amount invested at 15% is x + y + 300.
We also know that 0.04x + 0.09y + 0.15(x + y + 300) = 681, which is the total annual income from these investments.
Substituting the value of x + y + 300, we get:
0.04x + 0.09y + 0.15(x + y + 300) = 681
0.04x + 0.09y + 0.15x + 0.15y + 0.15*300 = 681
0.19x + 0.24y + 45 = 681
0.19x + 0.24y = 636
We can now solve this system of equations using elimination method. First, we can multiply the first equation by 4 and the second equation by 9:
4x + 4y = 6100
9x + 9y = 5676
Then, we can add the two equations to eliminate the y-term:
(4x + 4y) + (9x + 9y) = 6100 + 5676
13x + 13y = 11776
x + y = 906
Substituting this value back into either of the original equations, we can solve for x and y:
x + y = 906
x = 906 - y
4(906 - y) + y = 6100
4*906 - 4y + y = 6100
3624 - 3y = 6100
-3y = -2476
y = 825
Substituting this value back into the equation x + y = 906, we can solve for x:
x + y = 906
x + 825 = 906
x = 81
Therefore, the amount invested at 4% was $81, and the amount invested at 9% was $825. The amount invested at 15% was $300 more than the combined amount invested at 4% and 9%, which is $81 + $825 + $300 = $1206.
We can check our solution by substituting the values of x and y into the original equations to verify that they are both satisfied.