Larry Mitchell invested part of his $10,000 advance at 2% annual simple interest and the rest at 6% annual simple interest.

Question

 If his total yearly interest from both accounts was $480​, find the amount invested at each rate.

Answer 1

.06x +.02(10-x)= .48 (10, x and .48 in units of $1000)

.04x = .48-.2=.28

x=28/4 = 7 = $7,000 invested at 6%

and $3,000 invested at $2%

check the answers

.06(7)+.02(3)=.42+.06=.48 =$480

480/10000= 4.8% average interest

6-4.8=1.2

4.8-2=2.8

1.2/2.8 =3/7 = $3,000/$7,000