document.write( "Question 948949: Phyllis invested $9,500, a portion earning a simple interest rate of
\n" ); document.write( "8 1/5% per year and the rest earning a rate of 8% per year. After one year the total interest earned on these investments was $777.00. How much money did she invest at each rate?
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Algebra.Com's Answer #579245 by macston(5194)\"\" \"About 
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X=invested amount at 8.2%; Y=invested amount at 8%
\n" ); document.write( "X+Y=$9500
\n" ); document.write( "X=$9500-Y
\n" ); document.write( "0.082(X)+0.08((Y)=$777.00 Substitute for X.
\n" ); document.write( "0.082($9500-Y)+0.08Y=$777.00
\n" ); document.write( "$779-0.082Y+0.08Y=$777.00 Subtract $779 from each side.
\n" ); document.write( "-0.002Y=-$2.00 Divide each side by -0.002
\n" ); document.write( "Y=$1000 ANSWER 1: $1000 was invested at 8%.
\n" ); document.write( "X=$9500-Y=$9500-$100=$8500 ANSWER 2: $8500 was invested as 8.2%
\n" ); document.write( "CHECK
\n" ); document.write( "0.082(X)+0.08((Y)=$777.00
\n" ); document.write( "0.082($8500)+0.08($1000)=$777.00
\n" ); document.write( "$697+$80=$777.00
\n" ); document.write( "$777.00=$777.00\r
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