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? Need the amount at 4% and then the amount at 2.5%. Please and Thanks
If $x at 4%, then
.04x + .025(14000-x) = 458
To solve this problem, we can set up a system of equations based on the information given.
Let's assume that the amount invested in the account earning 4% annual interest is x dollars, and the amount invested in the account earning 2.5% annual interest is y dollars.
According to the problem, the total investment amount is $14,000, so we can write the equation: x + y = 14,000.
The interest earned from the account earning 4% interest can be calculated as 0.04x, and the interest earned from the account earning 2.5% interest can be calculated as 0.025y.
Since the total interest earned for one year is $458, we can write another equation: 0.04x + 0.025y = 458.
Now we have a system of two equations with two unknowns:
x + y = 14,000
0.04x + 0.025y = 458
We can solve this system of equations to find the values of x and y.
Multiply the first equation by 0.025 to match the coefficient of y in the second equation:
0.025x + 0.025y = 0.025 * 14,000
This simplifies to:
0.025x + 0.025y = 350
Now we can subtract this equation from the second equation:
0.04x + 0.025y - (0.025x + 0.025y) = 458 - 350
Simplifying this equation gives us:
0.015x = 108
Divide both sides of the equation by 0.015 to solve for x:
x = 108 / 0.015
Calculating this gives us:
x = 7200
Now we can substitute the value of x back into the first equation to solve for y:
7200 + y = 14,000
Subtracting 7200 from both sides gives us:
y = 14,000 - 7200
Calculating this gives us:
y = 6800
Therefore, $7,200 was invested in the 4% account, and $6,800 was invested in the 2.5% account.