document.write( "Question 1067918: $14,188 is invested, part at 10% and the rest at 6%. If the interest earned from the amount invested at 10% exceeds the interest earned from the amount invested at 6%6% by $1175.92, how much is invested at each rate? (Round to two decimal places if necessary.)
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Algebra.Com's Answer #683101 by addingup(3677) $\"\"$ $\"About$

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Solve by elimination:
\n" ); document.write( "Let x be the amount invested at 10% and y the amount invested at 0.06%
\n" ); document.write( "x+y = 14,188 multiply both sides of this equation times 0.06
\n" ); document.write( "0.06x+0.06y = 851.28 (equation 1)
\n" ); document.write( "and the problem tells us that
\n" ); document.write( "0.10x-0.06y = 1,175.92 (equation 2)
\n" ); document.write( "Add equations 1 and 2:
\n" ); document.write( "0.06x+0.06y = 851.28
\n" ); document.write( "+
\n" ); document.write( "0.10x-0.06y = 1,175.92
\n" ); document.write( "-------------------
\n" ); document.write( "0.16x-0y = 2,027.20
\n" ); document.write( "= 0.16x = 2,027.2
\n" ); document.write( "x = 12,670 this is the amount invested at 10%
\n" ); document.write( "14,188-12,670 = 1,518 this is the amount invested at 6%
\n" ); document.write( "---------------------
\n" ); document.write( "Check:
\n" ); document.write( "12,670(0.10) = 1,518(0.06)+1,175.92
\n" ); document.write( "1,267 = 1267 Correct. \n" ); document.write( "

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