I need help with this math problem?

Question

The problem is:

A part of $5,000 was invested at 7% annual interest and the remaining at 8% annual interest. At the end of the year, the total amount received was $5398. How much money was invested at each rate?

....I know the procedure on how to do this problem, but i just need someone to validate me with my answer. Here is my work:

*Amount invested at 7% = X
*Amount invested at 8% = (5,000-X)

So, Investment= Rate X Amount = Interest

*7% investment= 0.07 rate (X) = 0.07
*% investment = 0.08 rate (5,000-X) = 0.08(5,000-X)
*Total investment = 0.07(X) + 0.08(5,000-X) = 5,398
So,: 0.07x + 400 - 0.08x = 5,398
(0.07-0.08)x = 5,398-400
(-0.01x/-0.01) = (4,998/-0.01)
(x = -498,800 @ 7%)

.....Is my answer correct? If not, would you please be so kind to tell me where my mistake is?

Answer 1

you made one small but important error.
The total investment at 7 percent is 1.07X at the end of the year. The total investiment at 8 percent is 1.08(5000-x). Remember, you didn't lose the original amount.
If you make this minor adjustment to your math, then you will get to the right answer.

I'd work it out, but it looks like you have a pretty good handle on the rest.

Answer 2

Your answer is incorrect.
The source of the mistake is:
*Total investment = 0.07(X) + 0.08(5,000-X) = 5,398

which should be:
*Total amount received = 1.07(X) + 1.08(5,000-X) = 5,398

Answer 3

*Amount invested at 7% = X
*Amount invested at 8% = (5,000-X)

Interest_Rate X Amount = Investment

*7% investment @ 1.07 rate (X) = 1.07(X)
*8% investment @ 1.08 rate (5,000-X) = 1.08(5,000-X)
*Total investment = 1.07(X) + 1.08(5,000-X) = 5,398
So: 1.07X + 5,400 - 1.08X = 5,398
(1.07-1.08)X = 5,398-5,400
-0.01X = -2
Multiply by -100
x = 200 @ 7% = 14
Therefore 4,800 @ 8% = 384
Total interest = 398