document.write( "Question 818417: A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is 9900 dollars. How much did he invest at each rate?
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Algebra.Com's Answer #492555 by mananth(16946)\"\" \"About 
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10% ------------------$x
\n" ); document.write( "6%--------------------$2x\r
\n" ); document.write( "\n" ); document.write( "Total interest -----------9900\r
\n" ); document.write( "\n" ); document.write( "10%x+6%*2x=9900\r
\n" ); document.write( "\n" ); document.write( "0.1x+0.06x=9900\r
\n" ); document.write( "\n" ); document.write( "0.16x=9900
\n" ); document.write( "x=61875
\n" ); document.write( "Amount invested at 10%= 61875
\n" ); document.write( "amount invested at 6% = 123,750
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