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)\"\" \"About 
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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
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