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.