document.write( "Question 190386This question is from textbook algebra for college students
\n" ); document.write( ": Bob invested some money at 5% simple interest and some at 9% simple interest. The amount invested at the higher rate was twice the amount invested at the lower rate. If the total interst on the investments for 1 year was $920, then how much did he invest at each rate?\r
\n" ); document.write( "\n" ); document.write( "I know what the answer is but I have tried every forumla and can't figure out where the numbers come from. The anser is $4000 @ 5% and $8000 @ 9%. Your help is greatly appreciated!
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Algebra.Com's Answer #142882 by jojo14344(1513)\"\" \"About 
You can put this solution on YOUR website!

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\n" ); document.write( "Let x = amount inv. @ 5%
\n" ); document.write( "And, 2x = amount for 9% (twice that of lower rate 5%)\r
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\n" ); document.write( "\n" ); document.write( "Then it follows,
\n" ); document.write( "Interest on 5% + Interest on 9% = $920
\n" ); document.write( "\"0.05x%2B0.09%282x%29=920\"
\n" ); document.write( "\"0.05x%2B0.18x=920\"
\n" ); document.write( "\"0.23x=920\" ----> \"cross%280.23%29x%2Fcross%280.23%29=cross%28920%294000%2Fcross%280.23%29\"
\n" ); document.write( "x = $4,000.00 , amount for 5%\r
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\n" ); document.write( "\n" ); document.write( "Also, 2*($4,000) = $8,000.00 , amount for 9%\r
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\n" ); document.write( "\n" ); document.write( "Let's check,
\n" ); document.write( "\"0.05%284000%29%2B0.09%288000%29=920\"
\n" ); document.write( "\"200%2B720=920\"
\n" ); document.write( "\"920=920\"\r
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\n" ); document.write( "\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo
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