$22,102 is invested, part at 11% and the rest at 7%. if the interest earned from the amount invested at 11% exceeds the interest earned from the amount invested at 7% by $913.46, how much is invested at each rate? (round to two decimal places if necessary.)

X + y = 22,102
.11x = .07y + 913.46  =      x= (7/11)y + 8304.18

((7/11)y + 8304.18) + y = 22,102

(18/11)y = 13797.82

y= 13797.82/(18/11)

y=8432.00 

8,432.00 dollars are invested at 7%
13,670.00 dollars are invested at 11%

Check
.07 x 8432=  590.24
.11 x 13,670 = 1503.70

1503.70 - 590.24 = 913.46

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