$27,797 is invested, part at 12% and the rest at 11%. if the interest earned from the amount invested at 12% exceeds the interest earned from the amount invested at 11% by $211.32, how much is invested at each rate? (round to two decimal places if necessary.)

Let
X is the amount invested at 0.12
Y is the amount invested at 0.11
x+y=27727First equation
0.12x-0.11y=211.32...second equation
multiply both sides of the first equation by 0.11 to get
0.11x+0.11y=3,049.97
0.12x-0.11y=211.32
add these equations together we will get
0.23x=3,261.29
Divide both sides by 0.23 to get
X=14,179.52 this is the amount invested at 0.12

Now the amount invested at 0.11 is
27727-14179.52=13,547.48

Check your answer using the second equation
14,179.52×0.12−13,547.48×0.11
=211.32. Correct

Ibelieve it would be letter c (: