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)![]() ![]() You can put this solution on YOUR website! 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. \n" ); document.write( " \n" ); document.write( " |