document.write( "Question 957696: Nancy invested a sum of money at 6%. She invested a second sum, $500 more than the first at 8%. The total interest earned for the year was $180. HOW MUCH DID NANCY INVEST AT EACH RATE? \n" ); document.write( "
Algebra.Com's Answer #585307 by macston(5194)![]() ![]() You can put this solution on YOUR website! S=amount at 6%; E=amount at 8%=S+$500 \n" ); document.write( "0.06S+0.08E=$180 Substitute for E \n" ); document.write( "0.06S+0.08(S+$500)=$180 \n" ); document.write( "0.06S+0.08S+$40=$180 Subtract $40 from each side. \n" ); document.write( "0.14S=$140 Divide each side by 0.14 \n" ); document.write( "S=1000 ANSWER 1: The amount invested at 6% was $1000. \n" ); document.write( "E=S+$500=$1000+$500=$1500 ANSWER 2: The amount invested at 8% was $1500. \n" ); document.write( "CHECK: \n" ); document.write( "0.06S+0.08E=$180 \n" ); document.write( "0.06($1000)+0.08($1500)=$180 \n" ); document.write( "$60+$120=$180 \n" ); document.write( "$180=$180 \n" ); document.write( " |