Problema Solution
A man invests $2,200 in three accounts that pay 6%, 8% and 9% in annual interest, respectively. He has three time as much invested at 9% as does at 6%. If the total interest for the year is $178, how much is invested at each rate?
Answer provided by our tutors
let
x = the money invested at 6%
y = the money invested at 8%
z = the money invested at 9%
a man invests $2,200 in three accounts
x + y + z = 2200
he has three times as much invested at 9% as does at 6%
z = 3x
the total interest for the year is $178
0.06x + 0.08y + 0.09z = 178
by solving the system of equations
x + y + z = 2200
z = 3x
0.06x + 0.08y + 0.09z = 178
we find
x = $200
y = $1400
z = $600
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