SOLUTION: If $3000 is invested at 6% annual simple interest, how much should be invested at 9% annual simple interest so that the total yearly income from both investments is $585?
Algebra ->
Finance
-> SOLUTION: If $3000 is invested at 6% annual simple interest, how much should be invested at 9% annual simple interest so that the total yearly income from both investments is $585?
Log On
Question 401436: If $3000 is invested at 6% annual simple interest, how much should be invested at 9% annual simple interest so that the total yearly income from both investments is $585? Answer by sofiyac(983) (Show Source):
You can put this solution on YOUR website! I=prt but you have two different amounts and percents so
I=p1r1t1+p2r2t2 both t's are 1, since it's just 1 year
585=3000*0.06*1+x*0.09*1
585=180+0.09x subtract 180 from each side
585-180=0.09x
405=0.09x divide each side by 0.09
4500=x
