SOLUTION: John has two investments that produce a $150 income each month. If $1000 more is invested at 9% than at 10% per year, how much was invested at each percent? There were 12 months

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: John has two investments that produce a $150 income each month. If $1000 more is invested at 9% than at 10% per year, how much was invested at each percent? There were 12 months       Log On


   

Question 396765: John has two investments that produce a $150 income each month. If $1000 more is invested at 9% than at 10% per year, how much was invested at each percent? There were 12 months in that year,
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

If $1000 more is invested at 9% than at 10% per year

Let x and (x+1000) represent the amount at 10% and 9% respectively

question states*** $150/mo (interst over 12months)

 .10x + .09(x+1000) = $1800

solving for x

   .19x + 90 = 1800

       .19x = 1710

          x = $9000, amount invested at 10%.  10,000 at 9%



CHECKING our Answer***

   $900 + $900 = $1800