document.write( "Question 900481: Investment of $9,000. Part at 3%, part at 4% (simple interest over 1 year). How much was invested in each account if the interest earned in the 3% account was $32 greater than the interest earned in the 4% account?\r
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Algebra.Com's Answer #546401 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! x+y=9000, \n" ); document.write( "0.03*x-0.04*y=32 \n" ); document.write( "x=9000-y \n" ); document.write( "0.03*(9000-y)-0.04*y=32 \n" ); document.write( "270-0.03y-0.04*y=32 \n" ); document.write( "-0.07*y=-238 \n" ); document.write( "y=3400 at 4% \n" ); document.write( "x=9000-y \n" ); document.write( "x=5600 at 3% \n" ); document.write( "check \n" ); document.write( "0.03*5600-0.04*3400=32 \n" ); document.write( "168-136=32 \n" ); document.write( "32=32 \n" ); document.write( "ok \n" ); document.write( " \n" ); document.write( " |