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)![]() ![]() You can put this solution on YOUR website! x=amount at 3%; y=amount at 5% \n" ); document.write( ". \n" ); document.write( "x+y=$100000 \n" ); document.write( "y=$100000-x \n" ); document.write( ". \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%. \n" ); document.write( ". \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 \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |