An executive invests $22,000, some at 8% and some at 6% annual interest. If he receives an annual return of $1,680, how much is invested at each rate?
If x is at 8%, the remainder (22000-x) is at 6% so,
.08x + .06(22000-x) = 1680
...
To solve this problem, we need to set up a system of equations based on the given information. Let's assume the amount invested at 8% is x dollars, and the amount invested at 6% is y dollars.
From the given information, we know that the total amount invested is $22,000, so we can express this as:
x + y = 22,000 -- Equation (1)
We also know that the annual return from the investments is $1,680, which is the sum of the returns from the two investments:
0.08x + 0.06y = 1,680 -- 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 of equations to find the values of x and y.
One way to solve the system of equations is by substitution. From Equation (1), we can express x in terms of y as follows:
x = 22,000 - y
Now, substitute this expression for x into Equation (2):
0.08(22,000 - y) + 0.06y = 1,680
Simplify the equation:
1,760 - 0.08y + 0.06y = 1,680
Combine like terms:
-0.02y = -80
Divide both sides by -0.02:
y = -80 / -0.02
y = 4,000
Now, substitute this value of y back into Equation (1) to find x:
x + 4,000 = 22,000
x = 22,000 - 4,000
x = 18,000
Therefore, $18,000 is invested at 8% and $4,000 is invested at 6%.