document.write( "Question 1006953: A total of $2000 was invested: part at 8% and the rest at 12% per year. If after one year, both parts earned equal interest, how much was invested at each rate, and how much interest was earned in total?
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #628860 by macston(5194)\"\" \"About 
You can put this solution on YOUR website!
.
\n" ); document.write( "E=amount at 8%; T=amount at 12%
\n" ); document.write( ".
\n" ); document.write( "E+T=2000
\n" ); document.write( "E=2000-T
\n" ); document.write( ".
\n" ); document.write( "0.08E=0.12T
\n" ); document.write( "0.08(2000-T)=0.12T
\n" ); document.write( "160-0.08T=0.12T
\n" ); document.write( "160=0.20T
\n" ); document.write( "800=T
\n" ); document.write( "ANSWER 1: $800 was invested at 12%.
\n" ); document.write( ".
\n" ); document.write( "E=$2000-$800=$1200
\n" ); document.write( "ANSWER 2: $1200 was invested at 8%.
\n" ); document.write( ".
\n" ); document.write( "Total interest=0.08($1200)+0.12($800)=$96+$96=$192
\n" ); document.write( "ANSWER 3: Total interest was $192.
\n" ); document.write( "
\n" ); document.write( "
\n" );