math
if $9,000 is to be invested, part at 13% and the rest at 8% simple interest , how much should be invested at each rate so that the total annual return will be the same as $9,000 invested at 9% set up as a system of linear equation
asked by
jota

It's a linear partition problem, since both investments are paid simple interest.
Amount invested at 8%
=$9000*(139)/(138)
=$7200
Amount invested at 13%
=$9000*(98)/(138)
=$1800
Alternatively, using algebra,
Let x=amount invested at 8%,
Then (9000x)=amount invested at 13%
and the interest obtained should be the same as $9000 invested at 9%:
x*(8/100)+(9000x)*(13/100)=9000*(9/100)
Isolate x,
x((138)/100)=9000(139)/100
x=9000*(139)/(138)=7200 as before.posted by MathMate
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