Two investments totaling $23,500 produce an annual income of $795. One investment yields 2% per year, while the other yields 7% per year. How much is invested at each rate?
if x is at 2%, then the rest (23500-x) is at 7%. So, add up the interest:
.02x + .07(23500-x) = 795
Two investments totaling $23,500 produce an annual income of $795. One investment yields 2% per year, while the other yields 7% per year. How much is invested at each rate?
To determine how much is invested at each rate, we can set up a system of equations. Let's assume the amount invested at 2% is x dollars and the amount invested at 7% is y dollars. We can use the following equations to represent the given information:
Equation 1: x + y = $23,500 (total amount invested)
Equation 2: 0.02x + 0.07y = $795 (total annual income)
To solve this system of equations, we can use the substitution method. Let's solve Equation 1 for x and substitute it into Equation 2:
x = $23,500 - y
0.02($23,500 - y) + 0.07y = $795
Now, we can distribute 0.02:
$470 - 0.02y + 0.07y = $795
Combine like terms:
0.05y = $325
Divide both sides by 0.05:
y = $6,500
Now, substitute the value of y back into Equation 1 to find x:
x + $6,500 = $23,500
x = $23,500 - $6,500
x = $17,000
Therefore, $17,000 is invested at a rate of 2% per year, and $6,500 is invested at a rate of 7% per year.