# math

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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

• math -

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.

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