Questions on Algebra: Expressions involving variables, substitution answered by real tutors!

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Question 170451: Solve the problem. Roberto invested some money at 7%, and then invested $2,000 more than twice this amount at 11%. His total annual income from the two investments was $3,990. How much was invested at 11%?: Solve the problem. Roberto invested some money at 7%, and then invested $2,000 more than twice this amount at 11%. His total annual income from the two investments was $3,990. How much was invested at 11%?
Answer by jojo14344(1023) About Me  (Show Source):
You can put this solution on YOUR website!
Let x ---> money inv. 7%
Therefore, 2x+2000 ---> money inv. @ 11%
It follows,
0.07x+(2x+2000)0.11=3990
0.07x+0.22x+220=3990
0.29x=3990-220=3770
cross(0.290)x/cross(0.29)=cross(3770)13000/cross(0.29)
x=$13,000, amount inv @ 7%
Then,
2*13000+2000=26000+2000=$28,000, amount inv at 11% (ANSWER)
Check,
0.07*13000+28000(0.11)=3990
910+3080=3990
3990=3990
Thank you,
Jojo