<PRE><B>Question 675272</B>
(3y^2 - 2)/(3y^2 + 10y - 8) 
- (y + 10)/(3y^2 + 10y - 8) 
- (y^2 - 6y)/(3y^2 + 10y - 8)
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The denominators are all the same, so combine the numerators
over the common denominator::
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= [3y^2-2-y-10-y^2+6y]/(3y^2+10y-8)
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= (2y^2 +5y -12)/(3y^2+10y-8)
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Factor:
= (2y^2+8y-3y-12)/(3y^2+12y-2y-8)
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= (2y(y+4)-3(y+4))/(3y(y+4)-2(y+4))
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= [(y+4)(2y-3)]/[(y+4)(3y-2)]
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= (2y-3)/(3y-2)
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cheers,
Stan H.</PRE>