document.write( "Question 567498: Phyllis invested $10,000, a portion earning a simple interest rate of 3 1/5%
\n" ); document.write( "per year and the rest earning a rate of 3% per year. After 1 year the total interest earned on these investments was $312. How much money did she invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #366562 by mananth(16946)\"\" \"About 
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Investment Part I 3.50% per annum ----x
\n" ); document.write( "Investment part II 3.00% per annum ----y
\n" ); document.write( "The sum of the investments is $10,000.00
\n" ); document.write( "The sum of individual interests = $312.00
\n" ); document.write( "x+y= 10000 ------------------------1
\n" ); document.write( "3.50%x+ 3.00%y =$312.00
\n" ); document.write( "Multiply by 100
\n" ); document.write( "3.5x+3y =$31,200.00 --------2
\n" ); document.write( "
\n" ); document.write( "Multiply (1) by -3.5
\n" ); document.write( "we get
\n" ); document.write( "
\n" ); document.write( "-3.5x -3.5y= -35000.00
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\n" ); document.write( "Add this to (2)
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\n" ); document.write( "0x-0.5y =-$3,800.00
\n" ); document.write( "
\n" ); document.write( "divide by-0.5
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\n" ); document.write( "y = $7,600.00 investment at 3.00%
\n" ); document.write( "Balance $2,400.00 investment at 3.50%
\n" ); document.write( "CHECK
\n" ); document.write( "$2,400.00 @ 3.50% $84.00
\n" ); document.write( "$7,600.00 @ 3.00% $228.00
\n" ); document.write( "Total -------------------- $312.00
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