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

Answer 1

It's a linear partition problem, since both investments are paid simple interest.
Amount invested at 8%
=$9000*(13-9)/(13-8)
=$7200
Amount invested at 13%
=$9000*(9-8)/(13-8)
=$1800

Alternatively, using algebra,
Let x=amount invested at 8%,
Then (9000-x)=amount invested at 13%
and the interest obtained should be the same as $9000 invested at 9%:
x*(8/100)+(9000-x)*(13/100)=9000*(9/100)
Isolate x,
x((13-8)/100)=9000(13-9)/100
x=9000*(13-9)/(13-8)=7200 as before.