document.write( "Question 949868: Jody invested $4700 less in an account paying 4% simple interest than she did in an account paying 3% simple interest. At the end of the first year, the total interest from both accounts was $659. Find the amount listed in each account.\r
\n" ); document.write( "\n" ); document.write( "The amount Jody invested in the 4% account is:\r
\n" ); document.write( "\n" ); document.write( "The amount Jody invested in the 3% account is:
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Algebra.Com's Answer #579928 by macston(5194)\"\" \"About 
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x=amount in 4% account; y=amount in 3% account
\n" ); document.write( "x=y-$4700
\n" ); document.write( "0.04x+0.03y=$659 Substitute for x.
\n" ); document.write( "0.04(y-$4700)+0.03y=$659
\n" ); document.write( "0.04y-$188+0.03y=$659 Add $188 to each side.
\n" ); document.write( "0.07y=$847
\n" ); document.write( "y=$12100 ANSWER 1: $12100 was in the 3% account
\n" ); document.write( "x=y-$4700=$12100-$4700=$7400 ANSWER 2:$7400 was in the 4% account
\n" ); document.write( "CHECK:
\n" ); document.write( "0.04($7400)+0.03($12100)=$659
\n" ); document.write( "$296+$363=$659
\n" ); document.write( "$659=$659
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