document.write( "Question 151268: I am completely stuck; please help me. Thank you!\r
\n" ); document.write( "\n" ); document.write( "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? \r
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Algebra.Com's Answer #111167 by mducky2(62)\"\" \"About 
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:
\n" ); document.write( "amount invested in the 9% account: x
\n" ); document.write( "amount invested in the 11%account: 6000-x
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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.
\n" ); document.write( "0.09x + 0.11(6000-x) = 624
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Now its just a matter of solving for x and 6000-x.
\n" ); document.write( "0.09x + 0.11(6000-x) = 624
\n" ); document.write( "0.09x + 0.11(6000) - 0.11x = 624
\n" ); document.write( "-0.02x + 660 = 624
\n" ); document.write( "-0.02x + 0.02x + 660 = 624 + 0.02x
\n" ); document.write( "660 = 624 + 0.02x
\n" ); document.write( "660 - 624 = 624 - 624 + 0.02x
\n" ); document.write( "36 = 0.02x
\n" ); document.write( "x = 1800
\n" ); document.write( "6000 - x = 4200
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That means that $1800 was invested into the first account at 9% interest and $4200 was invested into the second account at 11% interest. \n" ); document.write( "
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