An investment advisor invested a total of $12,000, part at 2.5% annual simple interest and part at 4% annual simple interest. The amount of interest earned for 1 year was $426. How much was invested at the 2.5% rate and how much was invested at the 4% rate?
let the amount invested at 2.5% be x
then the amount invested at 4% is 12000-x
solve the equation
.025x + .04(12000-x) = 426
I figured it out. Thanks for your help!
To solve this problem, we can use a system of equations. Let's call the amount invested at 2.5% rate "x" and the amount invested at the 4% rate "y".
From the given information, we can create two equations:
Equation 1: x + y = $12,000 (since the total amount invested is $12,000)
Equation 2: 0.025x + 0.04y = $426 (since the interest earned is $426, and the interest for each investment is calculated as the principal amount multiplied by the interest rate)
Now we can solve this system of equations. I'll use the substitution method.
From Equation 1, we can isolate x to get:
x = $12,000 - y
Substituting this value of x into Equation 2, we get:
0.025($12,000 - y) + 0.04y = $426
Simplifying the equation, we have:
$300 - 0.025y + 0.04y = $426
Combining like terms, we get:
0.015y = $126
To isolate y, we divide both sides by 0.015:
y = $126 / 0.015
y = $8,400
Now we can substitute this value of y back into Equation 1 to find x:
x + $8,400 = $12,000
Subtracting $8,400 from both sides, we have:
x = $12,000 - $8,400
x = $3,600
So $3,600 was invested at the 2.5% rate, and $8,400 was invested at the 4% rate.