Suppose $6,500 is divided into two bank accounts. One account pays 10% simple interest per year and the other pays 4.6%. After two years there is a total of $1000 in interest between the two accounts. How much was invested into the bank account that pays 4.6% simple interest (rounded to the nearest cent)?
You want x where
0.046x + 0.10(6500-x) = 1000
To solve this problem, we can set up a system of equations. Let's say the amount of money invested in the bank account that pays 10% simple interest is "x", and the amount invested in the bank account that pays 4.6% simple interest is "6500 - x" (since the total amount invested is $6,500).
The interest earned in the first account after 2 years is calculated as: 0.10 * x * 2 (as it pays 10% simple interest per year, and it has been 2 years).
The interest earned in the second account after 2 years is calculated as: 0.046 * (6500 - x) * 2 (as it pays 4.6% simple interest per year, and it has been 2 years).
According to the problem, the total interest earned between the two accounts is $1000. So we can write the equation:
0.10 * x * 2 + 0.046 * (6500 - x) * 2 = 1000
To solve for "x", we can simplify the equation:
0.20 * x + 0.092 * (6500 - x) = 1000
0.20x + 598 - 0.092x = 1000
0.20x - 0.092x = 1000 - 598
0.108x = 402
x = 402 / 0.108
Now, we can calculate the value of "x" to find out how much was invested in the bank account that pays 4.6% interest.