you invested $4000 in two accounts paying 2% and 9% annual interest, respectively if the total interest earned for the year was $150, how much was invested at each rate.
$_was invested at 2% and _$ was invested at 9%
$3000 was invested at 2% and $1000 was invested at 9%
you invest $5000 in two accounts paying 5% and 8% annual interest, respectively the total interest earned was $331.
How much was invested at each rate
26566.6
To solve this problem, we can set up a system of equations.
Let's assume the amount invested at 2% is x dollars, and the amount invested at 9% is y dollars.
According to the problem, the total amount invested is $4000, so we have the equation:
x + y = 4000 [Equation 1]
We are also given that the total interest earned for the year was $150.
The interest earned on the amount invested at 2% would be 0.02x, and the interest earned on the amount invested at 9% would be 0.09y. Therefore, we have the equation:
0.02x + 0.09y = 150 [Equation 2]
Now, we can solve this system of equations using substitution or elimination method.
Let's solve it using the substitution method. Rearrange Equation 1 to solve for x:
x = 4000 - y [Equation 3]
Substitute Equation 3 into Equation 2:
0.02(4000 - y) + 0.09y = 150
80 - 0.02y + 0.09y = 150
0.07y = 70
y = 1000
Substitute the value of y back into Equation 3 to find x:
x = 4000 - 1000
x = 3000
Therefore, $3000 was invested at 2% and $1000 was invested at 9%.