document.write( "Question 694590: Jack has $10,000 to invest in three accounts. He invests part of the money at 7% annual interest and twice that amount in an account paying 5.5% annual interest. The remaining money is invested at 4% annual interest. How much was invested in Each account if the total annual income from the three investments was $490?\r
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document.write( "I tried the question and got it wrong. I did:\r
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document.write( ".07x+5.5(2x)=490
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document.write( ".07x+11x=490
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document.write( "11.07x=490
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document.write( "x=44.26\r
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document.write( "4y=10000
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document.write( "y=2500\r
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document.write( "$44.26=7%
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document.write( "$88.52=5.5%
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document.write( "$2500=4% \n" );
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Algebra.Com's Answer #428080 by mananth(16946)![]() ![]() You can put this solution on YOUR website! 7.5% ------------->x \n" ); document.write( "5.5%---------------->2x\r \n" ); document.write( "\n" ); document.write( "Balance amount = (10000-(x+2x))\r \n" ); document.write( "\n" ); document.write( "(10000-(x+2x))-----------> 4%\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Sm up individual interests\r \n" ); document.write( "\n" ); document.write( "7%x+5.5%*2x+(10000-(x+2x))*4%=49000\r \n" ); document.write( "\n" ); document.write( "multiply equation by 100\r \n" ); document.write( "\n" ); document.write( "7x+11x+40000-12x=49000\r \n" ); document.write( "\n" ); document.write( "6x=9000 \n" ); document.write( "x= 9000/6 \n" ); document.write( "x= 1500 amount invested at 7% \n" ); document.write( "2x= 3000 amount at 5% \n" ); document.write( "Balance 5500 at 4%\r \n" ); document.write( "\n" ); document.write( "CHECK\r \n" ); document.write( "\n" ); document.write( "1500 7.00% 105 \n" ); document.write( "3000 5.50% 165 \n" ); document.write( "5500 4.00% 220 \n" ); document.write( " 490\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |