A total of $9000 is invested: part at 6% and the remainder at 14%. How much is invested at each rate if the annual interest is $680
To solve this problem, we can set up a system of equations.
Let's assume that the amount invested at 6% is x dollars, and the amount invested at 14% is y dollars.
According to the given information, the total investment is $9000, so our first equation is:
x + y = 9000 -- Equation 1
Next, we have the information about the annual interest earned. The interest earned from the amount invested at 6% is given by the expression 0.06x (since 6% is equivalent to 0.06 in decimal form), and the interest earned from the amount invested at 14% is given by the expression 0.14y (since 14% is equivalent to 0.14 in decimal form).
The total interest earned is $680, so we can write our second equation as:
0.06x + 0.14y = 680 -- Equation 2
Now we have a system of two equations (Equation 1 and Equation 2) with two variables (x and y). We can solve this system to find the values of x and y.
First, let's rewrite Equation 1 in terms of x:
x = 9000 - y
Next, substitute this expression for x in Equation 2:
0.06(9000 - y) + 0.14y = 680
Simplifying:
540 - 0.06y + 0.14y = 680
0.08y = 140
y = 140 / 0.08
y = 1750
Now substitute this value for y back into Equation 1:
x + 1750 = 9000
x = 9000 - 1750
x = 7250
So, $7250 is invested at 6% and $1750 is invested at 14%.