SOLUTION: Sam invested $2100, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $28 more than twice the income from the 6% investment. How much

Question 1083955: Sam invested $2100, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $28 more than twice the income from the 6% investment. How much did he invest at each rate?
Found 2 solutions by Fombitz, MathTherapy:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let A be the amount at 6%, B the amount at 8%.
1.
.
.
.

2.
Substituting from 2 into 1,

Solve for B, then use either equation to solve for A.

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

Sam invested $2100, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $28 more than twice the income from the 6% investment. How much did he invest at each rate?

Let amount invested in the 8% fund be E
Then amount invested in the 6% fund = 2,100 - E
We then get the following INTEREST equation: .08E = 2(.06)(2,100 - E) + 28
Solve this for E, the amount invested in the 8% fund. You should get $1,400.
Subtract value of E from $2,100 to get amount invested in 6% fund.
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