Problema Solution
Larry Mitchell invested part of his $35,000 advance at 6% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was $2,970, find the amount invested at each rate.
Answer provided by our tutors
let 'x' represent the amount invested at 6%, then '35000-x' represents the amount invested at 9%
interest = principal * rate * time
we take time to be equal to 1 year
2970 = 0.06x + 0.09(35000-x)
solving for 'x' we have x=6000
35000-6000 = 29000
$6,000 was invested at 6%, $29,000 was invested at 9%