document.write( "Question 928658: Bradley has $25,000 that he wants to invest. He invests it in accounts paying 12%, 7%, and 6% simple interest. The account paying 12% is higher-risk account, so he wants the amount in that account to be half of the amount he has in the account paying 6% simple interest. If his annual interest is $1945, how much is invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #563770 by ewatrrr(24785)\"\" \"About 
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.12x + .07(25000 - 3x) + .06(2x) = 1945
\n" ); document.write( ".12x - (.21x) + .12x = 1945 - 1750
\n" ); document.write( " .03x= 195
\n" ); document.write( "x = $6500, amount at 12%
\n" ); document.write( "2x = $13,000, amount at 6%
\n" ); document.write( "\"25000-3%2A6500%29\" = $5,500, amount at 7%
\n" ); document.write( "And..checking
\n" ); document.write( "Sum = $25,000
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