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?
F*.05+(9000-F).12=730
solve for F, the amount of money invested at five percent. then 9000-F is the amount at 12 percetn.
To solve this problem, we can use a system of equations. Let's say the amount invested in the 5% account is x dollars, and the amount invested in the 12% account is y dollars.
We are given two pieces of information:
1. The total amount invested is $9,000:
x + y = 9,000
2. The interest earned in the first year is $730:
0.05x + 0.12y = 730
Now, we can solve this system of equations to find the values of x and y.
Method 1: Substitution method
Solve one equation for one variable and substitute it into the other equation.
From the first equation, we can express x in terms of y:
x = 9,000 - y
Substitute this value of x into the second equation:
0.05(9,000 - y) + 0.12y = 730
Simplify the equation:
450 - 0.05y + 0.12y = 730
0.07y = 280
y = 4,000
Now, substitute the value of y back into the first equation:
x + 4,000 = 9,000
x = 5,000
So, she invested $5,000 in the account that pays 5% per year and $4,000 in the account that pays 12% per year.
Method 2: Elimination method
Multiply the first equation by -0.05 to get -0.05x - 0.05y = -450. Now, add this equation to the second equation to eliminate x:
(-0.05x - 0.05y) + (0.05x + 0.12y) = -450 + 730
0.07y = 280
y = 4,000
Substitute this value of y back into the first equation:
x + 4,000 = 9,000
x = 5,000
Again, we find that she invested $5,000 in the account that pays 5% per year and $4,000 in the account that pays 12% per year.
So, she invested $5,000 in the 5% account and $4,000 in the 12% account.