<PRE><B>Question 82665</B>
A person invested $6,700 for one year, part at 8%, part at 10% and the remainder at 12%.  The total annual income from these investments was $716.  The amount of money invested at 12% was $300 more than the amount invested at 8% and 10% combined.  Find the amount invested at each rate.
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Let the amt at 8% be x; at 10% be y; at 12% be z:
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Then: 
x     +  y    + z     = 6700
0.08x + 0.10y + 0.12z = 716
x     +  y    - z     = -300
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Solution: 
x= $1200
y= $2000
z= $3500
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Cheers,
Stan H.</PRE>