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%
 
 
  | 
 
  
 
 |   
 
 |