A woman has a total of $9,000 to invest. She invests part of the money in an account that pays 5% per year and the rest in an account that pays 10% per year. If the interest earned in the first year is $650, how much did she invest in each account?
(9,000 -x)(0.05) + x (0.1) = 650
To solve this problem, let's assume the woman invests x dollars in the account that pays 5% per year. Therefore, she invests (9000 - x) dollars in the account that pays 10% per year.
The formula to calculate simple interest is: Interest = Principal * Rate * Time.
In this scenario, the interest earned in the first year ($650) is equal to the sum of the interest earned from both accounts.
For the first account, we have:
Principal = x dollars,
Rate = 5% (0.05),
Time = 1 year.
So, the interest earned from the first account is: x * 0.05 * 1 = 0.05x.
For the second account, we have:
Principal = (9000 - x) dollars,
Rate = 10% (0.1),
Time = 1 year.
Therefore, the interest earned from the second account is: (9000 - x) * 0.1 * 1 = 0.1(9000 - x).
According to the problem, the total interest earned from both accounts is $650.
So, we can write the equation: 0.05x + 0.1(9000 - x) = 650.
Now, let's solve this equation to find the value of x.
0.05x + 0.1(9000 - x) = 650
0.05x + 900 - 0.1x = 650
-0.05x + 900 = 650
-0.05x = 650 - 900
-0.05x = -250
x = (-250) / (-0.05)
x = 5000
Therefore, the woman invested $5000 in the account that pays 5% per year, and she invested $9000 - $5000 = $4000 in the account that pays 10% per year.