Question 396765
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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