document.write( "Question 549420: A total of $15,000 is invested in two accounts. One of the two accounts pays 8% per year, and the other account pays 12% per year. If the total interest paid in the first year is $1,600, how much was invested in each account?\r
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document.write( "Two answers are needed amount for 8% account and money made for 12% account.
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document.write( "Please help! \n" );
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Algebra.Com's Answer #357742 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Let x = amt invested at 8% \n" ); document.write( "Let y = amt invested at 12% \n" ); document.write( ": \n" ); document.write( "Write an equation for each statement: \n" ); document.write( ": \n" ); document.write( "A total of $15,000 is invested in two accounts. \n" ); document.write( "x + y = 15000 \n" ); document.write( "or \n" ); document.write( "x = (15000-y), we can use this form for substitution \n" ); document.write( ": \n" ); document.write( " One of the two accounts pays 8% per year, and the other account pays 12% per year. \n" ); document.write( " If the total interest paid in the first year is $1,600, \n" ); document.write( ".08x + .12y = 1600 \n" ); document.write( "substitute (15000-y) for x, find y \n" ); document.write( ".08(15000-y) + .12y = 1600 \n" ); document.write( "1200 - .08y + .12y = 1600 \n" ); document.write( "-.08y + .12y = 1600 - 1200 \n" ); document.write( ".04y = 400 \n" ); document.write( "y = 400/.04 \n" ); document.write( "y = $10,000 invested at 12% \n" ); document.write( "then \n" ); document.write( "15000 - 10000 = $5,000 invested at 8% \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "You can check this out \n" ); document.write( ".08(5000) + .12(1000) =\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |