Jesus invests some money at 6%, and then invests $5,000 more than twice this

amount at 9%. His total income from the two investments is $3,330. How much

money did he invest at 9%? I can figure out a regular simple interest problem,

but get lost on problems like this. Help?!?! Kathy

>>...Jesus invests some money at 6%...<<

Say, he invests $x at 6%, getting $.06x in interest

>>...and then invests $5,000 more than twice this amount at 9%.

So he invests %5,000 more than $2x at 9%.  That is.

So he invests $(2x + 5000) at 9% getting $.09(3x + 5000)

>>...His total income from the two investments is $3,330...<<

$.06x plus $.09(2x + 5000) must equal $3330

             .06x + .09(3x + 5000) = 3330

Can you solve that? Answer x = $12000 at 6%, then

since he invests $(2x + 5000) at 9%, that's $(2·12000 + 5000) or

                                            $(24000 + 5000) or

                                            $29000 

Edwin

 

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