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)\"\" \"About 
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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
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