You invested $14,000 in two accounts paying 5% and 9% annual interest, respectively. If the total interest earned for the year was $820, how much was invested at each rate?
If you have x at 5%, then you have (14000-x) at 9%. Now just calculate the interest on each and add it up:
.05x + .09(14000-x) = 820
x = 11000
so, you have $11,000 at 5% and $3000 at 9%.
Let's assume that the amount invested at 5% annual interest is x dollars.
According to the given information, the amount invested at 9% annual interest would be $(14,000 - x) since the total investment is $14,000.
We can now calculate the interest earned on each account.
The interest earned on the first account (at 5% interest) would be (x * 0.05), which can also be written as 0.05x.
The interest earned on the second account (at 9% interest) would be ((14,000 - x) * 0.09), which can also be written as (0.09 * (14,000 - x)).
The total interest earned for the year is given as $820, so we can write the equation:
0.05x + 0.09(14,000 - x) = 820
Now, let's solve the equation to find the value of x and then calculate the amount invested at each rate.
To solve this problem, we can use a system of equations. Let's assume that the amount invested at 5% interest is denoted by "x", and the amount invested at 9% interest is denoted by "y".
We are given that the total amount invested is $14,000, so we have the equation:
x + y = 14000 (Equation 1)
We are also given that the total interest earned for the year is $820. The interest earned on the amount invested at 5% is 5% of x, which is 0.05x, and the interest earned on the amount invested at 9% is 9% of y, which is 0.09y. So, we have the equation:
0.05x + 0.09y = 820 (Equation 2)
Now we can solve this system of equations.
1. Solve equation 1 for x in terms of y:
x = 14000 - y
2. Substitute this expression for x into equation 2:
0.05(14000 - y) + 0.09y = 820
3. Simplify and solve for y:
700 - 0.05y + 0.09y = 820
0.04y = 120
y = 3000
4. Substitute the value of y back into equation 1 to solve for x:
x + 3000 = 14000
x = 11000
Therefore, $11,000 was invested at 5% interest, and $3,000 was invested at 9% interest.