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Algebra Word Problem

$6300 is invested, part of it at 10% and part of it at 9%. For a certian year, the total yeild is $598.00 How much was invested at each rate?
 
How much was invested at 10%____
 
How much was invested at 9%_____

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Francisco E. | Francisco; Civil Engineering, Math., Science, Spanish, Computers.Francisco; Civil Engineering, Math., Sci...
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The equations are:
x+y=6300
0.1x + 0.09Y=598
multiplying the second equation by -10 we obtain
-x - 0.9y =-5980
adding equations 1 and 2 modified we have:
0.1y = 320
Y=3200 and x=6300-3200= 3100
3200X0.09+3100X0.1=598
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