document.write( "Question 954779: You invested 5000 between two accounts paying 3% and 8% annual interest, respectively. If the total interest earned for the year was $250, how much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #583172 by macston(5194)\"\" \"About 
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T=amount invested at 3%; E=amount invested at 8%
\n" ); document.write( "T+E=5000
\n" ); document.write( "T=5000-E
\n" ); document.write( "0.03T+0.08E=$250 Substitute for T.
\n" ); document.write( "0.03(5000-E)+0.08E=$250
\n" ); document.write( "150-0.03E+0.08E=$250 Subtract 150 from each side.
\n" ); document.write( "0.05E=$100 Divide each side by 0.05
\n" ); document.write( "E=$2000 ANSWER 1: $2000 was invested at 8%.
\n" ); document.write( "T=5000-E=5000-2000=3000 ANSWER 2: $3000 was invested at 3%.
\n" ); document.write( "CHECK:
\n" ); document.write( "0.03T+0.08E=$250
\n" ); document.write( "0.03($3000)+0.08($2000)=$250
\n" ); document.write( "$90+$160=$250
\n" ); document.write( "$250=$250
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