document.write( "Question 949910: $740 is split into two investments, one paying 6% and the other paying %9 interest. if the total amount of money earned from the two investments is $52.50, how much money is invested at each rate
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Algebra.Com's Answer #579971 by macston(5194)\"\" \"About 
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S=invested at 6%; N=invested at 9%
\n" ); document.write( "S+N=$740
\n" ); document.write( "S=$740-N
\n" ); document.write( "0.06S+0.09N=$52.50 Substitute for S
\n" ); document.write( "0.06($740-N)+0.09N=$52.50
\n" ); document.write( "$44.40-0.06N+0.09N=$52.50 Subtract $44.40 from each side.
\n" ); document.write( "0.03N=$8.10 Divide each side by 0.03
\n" ); document.write( "N=$270 ANSWER 1: The amount invested at 9% was $270.
\n" ); document.write( "S=$740-$270=$470 ANSWER 2: The amount invested at 6% was $470.
\n" ); document.write( "CHECK
\n" ); document.write( "0.06S+0.09N=$52.50
\n" ); document.write( "0.06($470)+0.09($270)=$52.50
\n" ); document.write( "$28.20+$24.30=$52.50
\n" ); document.write( "$52.50=$52.50
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