document.write( "Question 79731: You invested $10,000 in two stocks paying 9% and 12% annual interest,
\n" ); document.write( "respectively. At the end of the year, the total interest from these investments
\n" ); document.write( "was $930. How much was invested at each rate?
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Algebra.Com's Answer #57254 by checkley75(3666)\"\" \"About 
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.12x+.09(10000-x)=930
\n" ); document.write( ".12x+900-.09x=930
\n" ); document.write( ".03x=30
\n" ); document.write( "x=30/.03
\n" ); document.write( "x=1000 invested @ 12%
\n" ); document.write( "10000-1000=9000 invested @ 9%
\n" ); document.write( "proof
\n" ); document.write( ".12*1000+.09*9000=930
\n" ); document.write( "120+810=930
\n" ); document.write( "930=930
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