An investor invested a total of $2000 in two mutual funds. One fund earned 9% profit while the other earned a 6% profit. If the investor's total profit was $129, how much was invested in each mutual fund?

let one be $x

the other $ (2000-x)
.09x + .06(2000-x) = 129
times 100
9x + 6(2000-x) = 12900
9x + 12000 - 6x = 12900
3x = 900
x=300

$300 was invested at 9%, and 1700 was invested at 6%

check:
.09(300) + .06(1700) = 129

To solve this problem, we can use a system of equations. Let's assume that the amount invested in the mutual fund that earned 9% profit is x, and the amount invested in the mutual fund that earned 6% profit is y.

According to the problem, the total amount invested in both funds is $2000:

x + y = 2000 (Equation 1)

Now, let's calculate the profits from each fund. The profit from the first fund (9% profit) can be calculated as 0.09x, and the profit from the second fund (6% profit) can be calculated as 0.06y. The total profit is given as $129:

0.09x + 0.06y = 129 (Equation 2)

We have two equations, so we can solve this system to find the values of x and y.

One way to solve this system is by using the substitution method. Let's solve Equation 1 for x:

x = 2000 - y

Now substitute x in Equation 2:

0.09(2000 - y) + 0.06y = 129

180 - 0.09y + 0.06y = 129

Combine like terms:

0.03y = 129 - 180

0.03y = -51

Divide both sides by 0.03:

y = -51 / 0.03

y = -1700

Since the investment amount cannot be negative, the solution does not make sense in real-world terms. It seems there might be an error in the problem statement or in the calculations. Please double-check the information and try again.