$13,055 is invested, part at 14% and the rest at 9%. if the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 9% by $770.62, how much is invested at each rate?

Let
x> the amount invested at 14%
y> the amount invested at 9%

we know that
x+y=$13,055> x=13,055-y> equation 1
x*[1.14]=y*[1.09]+770.62> equation 2

substitute equation 1 in equation 2

1.14*[13,055-y]=1.09y+770.62
14882.7-1.14y=1.09y+770.62
1.14y+1.09y=14882.7-770.62
2.23y=14112.08
y=6,328.29
x=13,055-6,328.29> x=6,726.71

the answer is
the amount invested at 14% is $6,726.71
the amount invested at 9% is $6,328.29