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? looking for the amount at 4% and also the amt. at 2.5%. all I am getting for replys is $458, just don't get it.

You have one equation. Solve for x and 14,000-x. The x will be the amount invested at 4% and 14,000-x will be the amount invested at 2.5%. Solve the equation and you'll get. If you have trouble with the equation post (respond here and not with a different post) exactly what you problem is and show your work. We can help you through it but you need to learn to do if you don't know how.

To solve this problem, we can use a system of equations. Let's assume the amount invested at 4% is represented by 'x', and the amount invested at 2.5% is represented by 'y'.

Since the total amount invested is $14,000, we know that:
x + y = 14,000 (Equation 1)

The total interest earned after 1 year is given as $458, and we can calculate the interest earned from each investment:
0.04x (4% interest on x) + 0.025y (2.5% interest on y) = 458 (Equation 2)

Now, we have a system of two equations:

x + y = 14,000
0.04x + 0.025y = 458

There are multiple methods to solve this system, but I will use the substitution method for this explanation.

Start by solving Equation 1 for 'x':
x = 14,000 - y

Substitute this value of 'x' into Equation 2:
0.04(14,000 - y) + 0.025y = 458

Simplify and solve for 'y':
560 - 0.04y + 0.025y = 458
0.015y = 458 - 560
0.015y = -102
y = -102 / 0.015
y ≈ $6,800

Now, substitute the value of 'y' back into Equation 1 to find 'x':
x + 6,800 = 14,000
x = 14,000 - 6,800
x ≈ $7,200

Therefore, the amount invested at 4% is approximately $7,200, and the amount invested at 2.5% is approximately $6,800.