Alg_Complete

# Alg_Complete

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Unformatted text preview: the least common denominator. y ( y - 1) 2 - 2 3 + y -1 y + 2 So, there are two factors in the denominators a y-1 and a y+2. So we will write both of those down and then take the highest power for each. That means a 2 for the y-1 and a 1 for the y+2. Here is the least common denominator for this rational expression. lcd : ( y + 2 )( y - 1) 2 Now determine what’s missing in the denominator for each term, multiply the numerator and denominator by that and then finally do the subtraction and addition. © 2007 Paul Dawkins 49 http://tutorial.math.lamar.edu/terms.aspx College Algebra y ( y + 2) 2 ( y - 1)( y + 2 ) 3 ( y - 1) y 2 3 + = + y 2 - 2 y + 1 y - 1 y + 2 ( y - 1) 2 ( y + 2 ) ( y - 1)( y - 1)( y + 2 ) ( y - 1)2 ( y + 2 ) 2 y ( y + 2 ) - 2 ( y - 1)( y + 2 ) + 3 ( y - 1) = 2 ( y - 1) ( y + 2 ) 2 Okay now let’s multiply the numerator out and simplify. In the last term recall that we need to do the multiplication prior to distributing the 3 through the parenthesis. Here is the simplification work for this part. y 2 + 2 y - 2 ( y 2 + y - 2 ) + 3 ( y 2 - 2 y + 1) y 2 3 +...
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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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