Posted by **jota** on Sunday, May 6, 2012 at 11:44pm.

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 -
**MathMate**, Monday, May 7, 2012 at 9:31pm
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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