Idon’t know how to set this up.. i need on it !

  $37,600 was invested at 3%

Step-by-step explanation:

The problem statement shown seems to ask only for the amount invested at 3%, so we'll let that amount be represented by x. The x/2 is the amount invested at 9%, and the total annual interest is ...

  0.03·x + 0.09·(x/2) = 2820

  0.075x = 2820 . . . . . . . . . . . . . simplify

  2820/0.075 = x = 37,600 . . . . divide by the coefficient of x

Jolene invested $37,600 at 3%.

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The amount invested at 9% is half that, $18,800.