document.write( "Question 949329: A person invested $9000, part at 4 1/2%p.a and the rest at 5% p.a. If the total interest he earned after one year was $435, find the amount of each investment? \n" ); document.write( "
Algebra.Com's Answer #579573 by macston(5194)![]() ![]() You can put this solution on YOUR website! x=invested at 4.5%; y=invested at 5%; \n" ); document.write( "x+y=$9000 Subtract y from each side \n" ); document.write( "x=$9000-y \n" ); document.write( "0.045x+0.05y=$435 Substitute for x. \n" ); document.write( "0.045($9000-y)+0.05y=$435 \n" ); document.write( "0.045($9000)-0.045y+0.05y=$435 \n" ); document.write( "$405+0.005y=$435 subtract $405 from each side. \n" ); document.write( "0.005y=$30 Divide each side by 0.005 \n" ); document.write( "y=$6000 ANSWER 1:$6000 was invested at 5%. \n" ); document.write( "x=$9000-y=$9000-$6000=$3000 ANSWER 2: $3000 was invested at 4.5%. \n" ); document.write( "CHECK \n" ); document.write( "0.045x+0.05y=$435 \n" ); document.write( "0.045($3000)+0.05($6000)=$435 \n" ); document.write( "$135+$300=$435 \n" ); document.write( "$435=$435 \n" ); document.write( " \n" ); document.write( " |