Problema Solution
a person invested $8300 for 1 year, part at 6%, part at 10% and the remainder at 13%. the total annual income from these investments was $902. the amount of money invested at 13% was $500 more than the amounts invested at 6% and 10% combined. find the amounts invested at each rate
Answer provided by our tutors
let
x = the money invested at 6% rate
y = the money invested at 10% rate
z = the money invested at 13% rate
a person invested $8300 for 1 year
x + y + z = 8300
the total annual income from these investments was $902
0.06x + 0.10*y + 0.13x = 902
the amount of money invested at 13% was $500 more than the amounts invested at 6% and 10% combined
z = 500 + (x + y)
by solving the system of equations
x + y + z = 8300
0.06x + 0.10*y + 0.13x = 902
z = 500 + (x + y)
we find
x = $5,688.89
y = - $1,788.89
z = $4,400.00
click here to see the step by step solution of the system of equations
we got negative value for y and we need y to be non-negative number thus the problem has no solutions.