SOLUTION: I am completely stuck; please help me. Thank you! Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in in

Question 151268: I am completely stuck; please help me. Thank you!
Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?

Answer by mducky2(62) (Show Source): You can put this solution on YOUR website!
Since part of the $6000 went into one account and the rest went into the other account, the variables can be set up like this:
amount invested in the 9% account: x
amount invested in the 11%account: 6000-x

Now we can set up the entire equation in these terms. 9% of the amount in the first account plus 11% of the amount in the second account will equal the interest.
0.09x + 0.11(6000-x) = 624

Now its just a matter of solving for x and 6000-x.
0.09x + 0.11(6000-x) = 624
0.09x + 0.11(6000) - 0.11x = 624
-0.02x + 660 = 624
-0.02x + 0.02x + 660 = 624 + 0.02x
660 = 624 + 0.02x
660 - 624 = 624 - 624 + 0.02x
36 = 0.02x
x = 1800
6000 - x = 4200

That means that $1800 was invested into the first account at 9% interest and $4200 was invested into the second account at 11% interest.
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