Suppose $500 is divided into two bank accounts. One account pays 15% simple interest per year and the other pays 10%. After three years there is a total of $200 in interest between the two accounts. How much was invested into the bank account that pays 10% simple interest (rounded to the nearest cent)?

x = amount invested in the 15% account

x*.15*3 + (500-x)*.1*3 = 200
x = 333.33

so, 500-333.33 = 166.67 at 10%

Let's assume that the amount invested into the bank account that pays 10% simple interest is "x" dollars.

The amount invested into the bank account that pays 15% simple interest would then be (500 - x) dollars.

The interest earned on the account paying 10% simple interest can be calculated using the formula I1 = P1 * R1 * T, where:
- I1 is the interest earned on the account paying 10% simple interest,
- P1 is the principal (amount invested into the account) which is "x" dollars,
- R1 is the rate of interest which is 0.10 (10% written as a decimal), and
- T is the time in years which is 3 years.

So, we have I1 = x * 0.10 * 3 = 0.3x dollars.

Similarly, the interest earned on the account paying 15% simple interest can be calculated using the same formula but with different values. Let's call this interest I2.

So, I2 = (500 - x) * 0.15 * 3 = 0.45(500 - x) dollars.

According to the given information, the total interest earned from both accounts is $200. Therefore, we can add the interest from both accounts and set it equal to $200.

So, I1 + I2 = 0.3x + 0.45(500 - x) = 200.

Now, let's solve this equation to find the value of x.

0.3x + 0.45(500 - x) = 200
0.3x + 225 - 0.45x = 200
-0.15x = -25
x = (-25) / (-0.15)
x ≈ 166.67 (rounded to the nearest cent)

Therefore, approximately $166.67 was invested into the bank account that pays 10% simple interest.

To find out how much was invested in the bank account that pays 10% simple interest, we can set up a system of equations.

Let's assume that the amount invested in the bank account paying 15% simple interest is x dollars. Since the total amount invested is $500, the amount invested in the bank account paying 10% simple interest would be ($500 - x) dollars.

The interest earned on the first account would be (x * 0.15) dollars, and the interest earned on the second account would be (($500 - x) * 0.1) dollars.

According to the problem, the total interest earned after three years is $200, so we can set up the equation:

(x * 0.15) + (($500 - x) * 0.1) = $200

Now, we can solve this equation to find the value of x, which represents the amount invested in the bank account paying 15% simple interest.

Multiply the terms:
0.15x + 0.1 * ($500 - x) = $200

Distribute the terms inside the parentheses:
0.15x + 0.1 * $500 - 0.1x = $200

Simplify:
0.15x + $50 - 0.1x = $200

Combine like terms:
0.05x + $50 = $200

Subtract $50 from both sides of the equation:
0.05x = $150

Divide both sides of the equation by 0.05 to solve for x:
x = $150 / 0.05
x = $3000

Therefore, the amount invested in the bank account that pays 10% simple interest is ($500 - $3000) = -$2500.

However, this result does not make sense as you cannot invest a negative amount of money. It means that there is no solution to this problem, and we cannot determine the amount invested in the bank account that pays 10% simple interest.