A person has invested $6000 . Part of the money is invested at 3% and the remaining amout at 4% . The annual income from the two investment is $225.How much is invested at each rate?
amount invested at 3% ---- x
amount invested at 4% --- 6000-x
solve for x
.03x + .04(6000-x) = 225
I would multiply each of the 3 terms by 100 to get
3x + 4(6000-x) = 22500
I will let you finish it