Answer:
$37,600 was invested at 3%
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.