document.write( "Question 902461: Phyllis invested $12,500, a portion earning a simple interest rate of 3.5%
\n" ); document.write( " per year and the rest earning a rate of 3% per year. After one year the total interest earned on these investments was $420.00. How much money did she invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #547300 by richwmiller(17219)\"\" \"About 
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We know the total amount of money invested. $12500
\n" ); document.write( "x+y=12500,
\n" ); document.write( "We know that the total interest for the year for the two accounts is $420
\n" ); document.write( "0.035*x+0.03*y=420
\n" ); document.write( "x=12500-y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.035*(12500-y)+0.03*y=420
\n" ); document.write( "We multiply out
\n" ); document.write( "437.5-0.035y+0.03*y=420
\n" ); document.write( "We combine like terms.
\n" ); document.write( "-0.005*y=-17.5
\n" ); document.write( "Isolate y
\n" ); document.write( "y=3500 at 3%
\n" ); document.write( "x=12500-y
\n" ); document.write( "Calculate x
\n" ); document.write( "x=9000 at 3.5%
\n" ); document.write( "We check
\n" ); document.write( "0.035*9000+0.03*3500=420
\n" ); document.write( "315.0+105=420
\n" ); document.write( "420.0=420
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