An investment adviser invested $14,000 in two accounts. One investment earned 4% annual simple interest, and the other investment earned 2.5% annual simple interest. The amount of interest earned for 1 year was $458. How much was invested in each account
If $x at 4%, then
.04x + .025(14000-x) = 458
To solve this problem, we can use a system of equations. Let's assume that the amount invested at 4% interest is x dollars, and the amount invested at 2.5% interest is y dollars.
Since the total investment amount is $14,000, we have the equation:
x + y = 14,000 -- Equation (1)
The interest earned from the investment at 4% is calculated as:
interest_4% = x * (4/100)
The interest earned from the investment at 2.5% is calculated as:
interest_2.5% = y * (2.5/100)
The total interest earned for one year is given as $458:
interest_4% + interest_2.5% = 458 -- Equation (2)
Now we can solve this system of equations to find the values of x and y.
First, we can simplify Equation (2):
0.04x + 0.025y = 458
To get rid of the decimals, we can multiply this equation by 100:
4x + 2.5y = 45,800 -- Equation (3)
Now we have a system of equations with Equation (1) and Equation (3). We can solve it using any method, such as substitution or elimination.
Let's use the elimination method by multiplying Equation (1) by 2.5:
2.5x + 2.5y = 35,000 -- Equation (4)
Now subtract Equation (4) from Equation (3):
4x + 2.5y - (2.5x + 2.5y) = 45,800 - 35,000
4x + 2.5y - 2.5x - 2.5y = 10,800
1.5x = 10,800
x = 10,800 / 1.5
x = 7,200
Substituting the value of x in Equation (1):
7,200 + y = 14,000
y = 14,000 - 7,200
y = 6,800
So, the adviser invested $7,200 in the account earning 4% annual interest and $6,800 in the account earning 2.5% annual interest.