document.write( "Question 992142: Mr. Garcia invested $100,000 part at 5% and part at 3%. If the annual income is $4,200, find the amount of invested at each rate. \n" ); document.write( "
Algebra.Com's Answer #611861 by macston(5194)\"\" \"About 
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x=amount at 3%; y=amount at 5%
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\n" ); document.write( "x+y=$100000
\n" ); document.write( "y=$100000-x
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\n" ); document.write( "0.03x+0.05y=$4200
\n" ); document.write( "0.03x+0.05($100000-x)=$4200
\n" ); document.write( "0.03x+$5000-0.05x=$4200
\n" ); document.write( "-0.02x=-$800
\n" ); document.write( "x=$40000 ANSWER 1:$40000 was invested at 3%.
\n" ); document.write( "y=$100000-x=$100000-$40000=$60000
\n" ); document.write( "ANSWER 2: $60000 was invested at 5%.
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\n" ); document.write( "CHECK:
\n" ); document.write( "0.03x+0.05y=$4200
\n" ); document.write( "0.03($40000)+0.05($60000)=$4200
\n" ); document.write( "$1200+$3000=$4200
\n" ); document.write( "$4200=$4200\r
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