document.write( "Question 630993: A part of $10,000 is invested at 4% and the remaining amount at 3 1/2%.The annual income both is $380.Find the amount of each investment? \n" ); document.write( "
Algebra.Com's Answer #397271 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A part of $10,000 is invested at 4% and the remaining amount at 3 1/2%.The annual income both is $380.Find the amount of each investment? \n" ); document.write( "------ \n" ); document.write( "Equations: \n" ); document.write( "value: x + y = 10,000 \n" ); document.write( "interest: 0.04x + 0.035y = 360 \n" ); document.write( "----- \n" ); document.write( "Modify: \n" ); document.write( "40x + 40y = 40\"10,000 \n" ); document.write( "40x + 35y = 360,000 \n" ); document.write( "---- \n" ); document.write( "5y = 40,000 \n" ); document.write( "y = 8000 (amt. invested at 3.5% \n" ); document.write( "x = 2000 (amt. invested at 4% \n" ); document.write( "----------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |