If we factor a minus out of the numerator we can do

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Unformatted text preview: = 2 y2 - 2 y +1 y -1 y + 2 ( y - 1) ( y + 2 ) = = y2 + 2 y - 2 y2 - 2 y + 4 + 3 y2 - 6 y + 3 ( y - 1) ( y + 2 ) 2 2 y2 - 6 y + 7 ( y - 1) ( y + 2 ) 2 [Return to Problems] (d) 2x 1 2 x -9 x +3 x -3 2 Again, factor the denominators and get the least common denominator. 2x 1 2 ( x - 3)( x + 3) x + 3 x - 3 The least common denominator is, lcd : ( x - 3) ( x + 3) Notice that the first rational expression already contains this in its denominator, but that is okay. In fact, because of that the work will be slightly easier in this case. Here is the subtraction for this problem. 1( x - 3) 2 ( x + 3) 2x 1 2 2x = x - 9 x + 3 x - 3 ( x - 3)( x + 3) ( x + 3)( x - 3) ( x - 3)( x + 3) 2 = 2 x - ( x - 3) - 2 ( x + 3) ( x - 3)( x + 3) = 2x - x + 3 - 2x - 6 ( x - 3)( x + 3) = -x - 3 ( x - 3)( x + 3) Notice that we can actually go one step further here. If we factor a minus out of the numerator we can do some canceling. © 2007 Paul Dawkins 50 http://tutorial.math.lamar.edu/terms.aspx College Algebra - ( x + 3) 2x 1 2 -1 = = x - 9 x + 3 x - 3 ( x - 3)( x + 3) x - 3 2 Sometimes this kind of canceling will happen after the addition/...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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