document.write( "Question 82665This question is from textbook Introductory & Intermediate Algebra for College Students
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document.write( ": A person invested $6,700 for one year, part at 8%, part at 10% and the remainder at 12%. The total annual income from these investments was $716. The amount of money invested at 12% was $300 more than the amount invested at 8% and 10% combined. Find the amount invested at each rate. \n" );
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Algebra.Com's Answer #59261 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A person invested $6,700 for one year, part at 8%, part at 10% and the remainder at 12%. The total annual income from these investments was $716. The amount of money invested at 12% was $300 more than the amount invested at 8% and 10% combined. Find the amount invested at each rate. \n" ); document.write( "------------- \n" ); document.write( "Let the amt at 8% be x; at 10% be y; at 12% be z: \n" ); document.write( "----------- \n" ); document.write( "Then: \n" ); document.write( "x + y + z = 6700 \n" ); document.write( "0.08x + 0.10y + 0.12z = 716 \n" ); document.write( "x + y - z = -300 \n" ); document.write( "-------------- \n" ); document.write( "Solution: \n" ); document.write( "x= $1200 \n" ); document.write( "y= $2000 \n" ); document.write( "z= $3500 \n" ); document.write( "============= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |