Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $700 more invested at 8% than at 2%, find the amount invested at each rate if the total annual interest received is $380. Let x = amount invested at 8% and y = amount invested at 2%.

In the given problem, two assumptions are already provided.

Let the amount of money invested at 8% = x

Let the amount of money invested in 2% = y

Then from the above question we can find that

x = y + 700

Also

(8x/100) + (2y/100) = 380

8x + 2y = 380 * 100

8x + 2y = 38000

Now replacing the value of x from the first equation we get

8 (y + 700) + 2y = 38000

8y + 5600 + 2y = 38000

10y = 38000 - 5600

10 y = 32400

y = 32400/10

= 3240

Putting the value of y in the first equation, we get

x = y + 700

= 3240 + 700

= 3940

So the amount of money invested at 8% is $3940 and the amount of money invested at 2% is $3240.