A woman has a total of $10,000 to invest. She invests part of the money in an account that pays 6% per year and the rest in an account that pays 9% per year. If the interest earned in the first year is $750, how much did she invest in each account?
To solve this problem, let's start by assigning variables to represent the unknowns. Let's say she invests x dollars at 6% and (10,000 - x) dollars at 9%.
The interest earned from the first account, at a rate of 6%, can be calculated using the formula: interest = principal * rate. In this case, the interest for the first account is 0.06x.
Similarly, the interest earned from the second account, at a rate of 9%, is: interest = principal * rate. In this case, the interest for the second account is 0.09(10,000 - x).
According to the problem, the total interest earned in the first year is $750. Therefore, we can write the equation:
0.06x + 0.09(10,000 - x) = 750
Now, let's solve the equation to find the value of x:
0.06x + 900 - 0.09x = 750
-0.03x = 750 - 900
-0.03x = -150
x = (-150) / (-0.03)
x = 5,000
So, she invests $5,000 at 6% and $10,000 - $5,000 = $5,000 at 9%.
Therefore, she invested $5,000 in the account that pays 6% per year and $5,000 in the account that pays 9% per year.
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But, just to get you started, if there are $x at 6%, the rest (1000-x) is at 9%. So, add up the interests earned:
.06x + .09(10000-x) = 750.00
Now just crank it out.