The amounts invested in the first, second, and third parts of the investment are $72,000, $48,000, and $10,000, respectively.
We can solve this problem using a system of equations. Let x, y, and z be the amounts invested in the first, second, and third parts of the investment, respectively. Then we have:
x + y + z = 130,000 (the total amount invested is $130,000)
0.08x + 0.06y + 0.09z = 9,540 (the total interest earned is $9,540)
0.08x = 2(0.06y) (the interest earned on the first investment is twice the interest earned on the second)
Simplifying the third equation, we get:
0.08x = 0.12y
x = 1.5y
Substituting x = 1.5y into the first equation, we get:
1.5y + y + z = 130,000
2.5y + z = 130,000
Substituting x = 1.5y into the second equation, we get:
0.08(1.5y) + 0.06y + 0.09z = 9,540
0.12y + 0.06y + 0.09z = 9,540
0.18y + 0.09z = 9,540
Now we have two equations with two variables:
2.5y + z = 130,000
0.18y + 0.09z = 9,540
Solving for z in the first equation, we get:
z = 130,000 - 2.5y
Substituting z = 130,000 - 2.5y into the second equation, we get:
0.18y + 0.09(130,000 - 2.5y) = 9,540
0.18y + 11,700 - 0.225y = 9,540
-0.045y = -2,160
y = 48,000
Substituting y = 48,000 into x = 1.5y, we get:
x = 1.5(48,000) = 72,000
Substituting y = 48,000 into z = 130,000 - 2.5y, we get:
z = 130,000 - 2.5(48,000) = 10,000
Therefore, the amounts invested in the first, second, and third parts of the investment are $72,000, $48,000, and $10,000, respectively.