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%
|
|
|