document.write( "Question 2371: Question: Suppose you invested $10,000, part at 6% annual intrest and the rest at 9% annual intrest. If you received $684 in intrest after on year, how much did you invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #997 by longjonsilver(2297)\"\" \"About 
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First define...\r
\n" ); document.write( "\n" ); document.write( "Let x = amount invested at 6% rate
\n" ); document.write( "Let y = amount invested at 9% rate\r
\n" ); document.write( "\n" ); document.write( "so, after 1 year, 0.06x + 0.09y = 684 --eqn1\r
\n" ); document.write( "\n" ); document.write( "Also, we know that x+y=10000 --eqn2\r
\n" ); document.write( "\n" ); document.write( "Eqn2 is re-written as x=10000-y, and then sub this into eqn1, to give \r
\n" ); document.write( "\n" ); document.write( "0.06(10000-y) + 0.09y = 684
\n" ); document.write( "600 - 0.06y + 0.09y = 684
\n" ); document.write( "0.03y = 84
\n" ); document.write( "y = 2,800\r
\n" ); document.write( "\n" ); document.write( "so, x= 7200\r
\n" ); document.write( "\n" ); document.write( "jon.
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