you invested $8000 between two accounts paying 7% and 9% annual interest, respectively. If the total interest earned for the year was $600 how much was invested at each rate?
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Let's assume x dollars were invested at 7% interest rate.
Therefore, the amount invested at 9% interest rate would be $8000 - x.
To find the interest earned at each rate, we can set up the following equation:
0.07x + 0.09(8000 - x) = 600
Simplifying the equation, we have:
0.07x + 720 - 0.09x = 600
Combining like terms, we have:
-0.02x = -120
Dividing both sides by -0.02, we get:
x = 6000
So, $6000 was invested at 7% interest rate, and $2000 was invested at 9% interest rate.
To solve this problem, we can use a system of equations. Let's assume you invested "x" dollars in the account with a 7% interest rate, and you invested "8000 - x" dollars in the account with a 9% interest rate.
The interest earned from the account with a 7% interest rate can be calculated by multiplying the amount invested by the interest rate: 0.07x. The interest earned from the account with a 9% interest rate can be calculated by multiplying the amount invested by the interest rate: 0.09(8000 - x).
According to the problem, the total interest earned is $600. So we set up an equation: 0.07x + 0.09(8000 - x) = 600.
Now, let's solve the equation to find the value of "x."
0.07x + 0.09(8000 - x) = 600
0.07x + 720 - 0.09x = 600
-0.02x = 600 - 720
-0.02x = -120
Divide both sides by -0.02:
x = (-120) / (-0.02)
x = 6000
Therefore, $6000 was invested at a 7% interest rate (x), and $2000 (8000 - x) was invested at a 9% interest rate.
if there is $x at 7%, the rest (8000-x) is at 9%. So, add up the interest:
.07x + .09(8000-x) = 600
Now just find x and 8000-x