Mitchell invested part of his $34,000 advance at 2% annual simple interest and the rest at 9% annual simple interest. if his total yearly interest from both accounts was $2,570, find the amount invested at each rate. invested at 2%: $ invested at 9%: $

Let x = Amount invested at 2%
      y = Amount invested at 9%

x + y = 34000            ×(-0.02)→ -0.02x - 0.02y = -680
0.02x + 0.09y = 2570             →  0.02x + 0.09y = 2570
                      add the equations               0.07y = 1890
           divide by 0.07 both sides                      y = 27000
x = 34000 - 27000 = 7000
ANSWER:
Invested at 2%: $7000
Invested at 9%: $27000