document.write( "Question 231674: A janitor invested $11,000 for one year, part at 8%, and the rest at 11%. The 11% investment returns $450 more than the 8% investment. How much was invested at 11%? \n" ); document.write( "
Algebra.Com's Answer #171400 by stanbon(75887)\"\" \"About 
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A janitor invested $11,000 for one year, part at 8%, and the rest at 11%. The 11% investment returns $450 more than the 8% investment. How much was invested at 11%?
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\n" ); document.write( "Let amount invested at 8% be \"x\".
\n" ); document.write( "Then amount invested at 11% is \"11,000-x\"
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\n" ); document.write( "Equation:
\n" ); document.write( "0.11(11000-x)-0.08x = 450
\n" ); document.write( "11(11000-x) - 8x = 45000
\n" ); document.write( "121000 - 11x - 8x = 45000
\n" ); document.write( "-19x = -76000
\n" ); document.write( "x = $4000 (amt invested at 8%
\n" ); document.write( "11000-x = $7000 (amt invested at 11%)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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