document.write( "Question 990896: You invested $7000 in two accounts paying 6% and 7% annual interest, respectively. If the total interest earned for the year was $471, how much was invested at each rate?\r
\n" ); document.write( "\n" ); document.write( "(Part 1: Let x denote the amount of money invested at 6%. Find an expression in terms of x for the amount invested at 7%.)????????????\r
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Algebra.Com's Answer #610875 by rothauserc(4718)\"\" \"About 
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let x be the amount of money invested at 6% and y be amount of money invested at 7%, then
\n" ); document.write( "x + y = 7000
\n" ); document.write( ".06x +.07y = 471
\n" ); document.write( "solve first equation for x
\n" ); document.write( "x = 7000 - y
\n" ); document.write( "now substitute for x in second equation
\n" ); document.write( ".06(7000 - y) + .07y = 471
\n" ); document.write( "420 - .06y + .07y = 471
\n" ); document.write( ".01y = 51
\n" ); document.write( "y = 5100 and
\n" ); document.write( "x = 7000 - 5100 = 1900
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\n" ); document.write( "$1900 was invested at 6% and $5100 was invested at 7%
\n" ); document.write( "solve equation 1 for y
\n" ); document.write( "y = 7000 - x
\n" ); document.write( " \n" ); document.write( "
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