document.write( "Question 69675: You invest 10,000, part at 6% annual interest and the rest at 9% annual interest. If you received 684 in interest after one year, how much did you invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #49610 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount at at 6%
\n" ); document.write( "Then $10000-x=amount at 9%\r
\n" ); document.write( "\n" ); document.write( "Now we are told that the amount of interest received at 6% (.06x) plus the amount of interest received at 9% (.09(10000-x)) equals $684. So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( ".06x+.09(10000-x)=684 get rid of parens\r
\n" ); document.write( "\n" ); document.write( ".06x+900-.09x=684 subtract 900 from both sides\r
\n" ); document.write( "\n" ); document.write( "-.03x=-$216\r
\n" ); document.write( "\n" ); document.write( "x=$7200---------------------amount invested at 6%\r
\n" ); document.write( "\n" ); document.write( "10000-x=10000-7200=$2800-------------------amount invested at 9%\r
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\n" ); document.write( "\n" ); document.write( "Ck\r
\n" ); document.write( "\n" ); document.write( ".06(7200)+.09(2800)=684\r
\n" ); document.write( "\n" ); document.write( "432+252=684\r
\n" ); document.write( "\n" ); document.write( "684=684\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor\r
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