Problema Solution
A person invested $7400 for 1 year, part at 7%, part at 9% and the remainder at 12%. The total annual income from these investments was $767. The amount of money invested at 12% was $800 more than the amounts invested at 7% and 9% combined. Find the amount invested at each rate.
Answer provided by our tutors
let
x = the amount invested at 7%
y = the amount invested at 9%
z = the amount invested at 12%
a person invested $7400 for 1 year
x + y + z = 7400
the total annual income from these investments was $767
0.07x + 0.09y + 0.12z = 767
the amount of money invested at 12% was $800 more than the amounts invested at 7% and 9% combined
z = 800 + x + y
by solving the system of equations
x + y + z = 7400
0.07x + 0.09y + 0.12z = 767
z = 800 + x + y
we find
x = $1,100
y = $2,200
z = $4,100
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$1,100 were invested at 7%, $2,200 were invested at 9% and $4,100 were invested at 12%.