document.write( "Question 82665This question is from textbook Introductory & Intermediate Algebra for College Students
\n" ); 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" ); document.write( "
Algebra.Com's Answer #59261 by stanbon(75887)\"\" \"About 
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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.
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\n" ); document.write( "Let the amt at 8% be x; at 10% be y; at 12% be z:
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\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
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\n" ); document.write( "Solution:
\n" ); document.write( "x= $1200
\n" ); document.write( "y= $2000
\n" ); document.write( "z= $3500
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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