Problema Solution
conner invested a part of $6,000 at 10% and the rest at 8%. his annual income from these investments is$556. how much did he invest at each rate of interest?
Answer provided by our tutors
let
x = the part invested at 10%
y = the part invested at 8%
Conner invested a part of $6,000 means
x + y = 6000
his annual income from these investments is $556
0.10x + 0.08y = 556
by solving the system of equations
x + y = 6000
0.10x + 0.08y = 556
we find
x = $3,800
y = $2,200
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Conner invested $3,800 at 10% and $2,200 at 8%.