$12,400 is invested, part at 6% and the rest at 5%. if the interest earned from the amount invested at 6% exceeds the interest earned from the amount invested at 5% by $577.00, how much is invested at each rate? (round to two decimal places if necessary.) define variables x and y and set up a system of two linear equations that represents the information given in the problem.

 $10881.81 is invested at 6% and $1518.18 is invested at 5%.

Step-by-step explanation:

Since we have given that

Amount invested = $12400

Rate of interest for first part = 6%

Rate of interest for second part = 5%

Let the amount invested for 6% be 'x'.

Let the amount invested for 5% be 'y'

According to question, we get that

So, it becomes,

x=12400-y=12400-1518.18=$10881.81

Hence, $10881.81 is invested at 6% and $1518.18 is invested at 5%.

the answer would be 484

step-by-step explanation:

just multiply 0.484 times 1,000 (formula)