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| Question 113012:  Roberto invested some money at 7%, and then invested $2000 more than twice this amount at 11%. His total annual income from the two investments was $3990. How much was invested at 11%?
 Answer by SHUgrad05(58)
      (Show Source): 
You can put this solution on YOUR website! .07x+.11y=3990 Let x=amount invested at 7%
 Let y=amount invested at 11%; y=2x+2000
 .07x+(2x+2000).11=3990
 .07x+.22x+220=3990
 .07x+.22x=3770
 .29x=3770
 x=$13,000-> plug back into original equation
 .07(13,000)+.11y=3990
 910+.11y=3990
 .11y=3080
 y=$28,000
 $28,000 invested at 11%
 
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