Problema Solution
Larry Mitchell invested part of his $34,000 advance at 6% annual simple interest and the rest at (% annual simple interest. If his total yearly interest from both accounts was $2,220, find the amount of invested at each rate.
Answer provided by our tutors
let he invest in 6% = $x
in 9% = $(34000-x)
=> 6x/100 + 9(34000 - x)/100 = 2220
=> -3x = 22200 - 306000
=> x = 28000
in 6% = $28000
in 9% = $6000