document.write( "Question 1129650: Amy invests $1500 in one account and $1900 in an account paying 3 % higher interest. At the end of one year she had earned $261 in interest. At what rates did she invest?\r
\n" ); document.write( "\n" ); document.write( "1500 invested at -
\n" ); document.write( "1900 invested at-
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Algebra.Com's Answer #746257 by Boreal(15235)\"\" \"About 
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interest for 1500 is x
\n" ); document.write( "interest for 1900 is x+3
\n" ); document.write( "1500*x/100=15x
\n" ); document.write( "1900*(x+3)/100=19x+57
\n" ); document.write( "that sum is 261
\n" ); document.write( "34x=204
\n" ); document.write( "x=6% for $1500=$90
\n" ); document.write( "X+3 is 9% for $1900=$171
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