...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 12% per year. If the interest earned in the first year is $730, how much did she invest in each account?
To solve this problem, we can set up a system of equations based on the information given.
Let's represent the amount of money invested in the account that pays 5% per year as x (in dollars), and the amount of money invested in the account that pays 12% per year as y (in dollars).
According to the problem, the total amount of money invested is $9,000, so we have the equation:
x + y = 9,000 -------------- (Equation 1)
Now, let's consider the interest earned in the first year. The interest earned in the account that pays 5% per year is calculated as 5% of x, which is 0.05x. Similarly, the interest earned in the account that pays 12% per year is calculated as 12% of y, which is 0.12y. So, the total interest earned in the first year is $730, which gives us the equation:
0.05x + 0.12y = 730 -------------- (Equation 2)
Now, we have a system of equations (Equation 1 and Equation 2) that we can solve to find the values of x and y.
There are various methods to solve this system of equations, such as substitution, elimination, or using matrix methods. Let's solve it using the substitution method:
From Equation 1, we can rearrange it to get:
x = 9,000 - y
Substituting this value of x in Equation 2, we have:
0.05(9,000 - y) + 0.12y = 730
Now, we can proceed to solve for y:
450 - 0.05y + 0.12y = 730
0.07y = 730 - 450
0.07y = 280
Dividing both sides by 0.07, we get:
y = 4,000
Now, substitute this value of y back into Equation 1 to find x:
x + 4,000 = 9,000
x = 9,000 - 4,000
x = 5,000
Therefore, the woman invested $5,000 in the account that pays 5% per year, and $4,000 in the account that pays 12% per year.