Alg_Complete

# Lets do that 2007 paul dawkins 48

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Unformatted text preview: ed operation. 4 1 5 (a) - 5 + 3 [Solution] 2 6 x 3x 2 x 2 z -1 (b) [Solution] z +1 z + 2 y 2 3 (c) 2 + [Solution] y - 2 y +1 y -1 y + 2 2x 1 2 (d) 2 [Solution] x -9 x +3 x -3 4 1 (e) - + 1 [Solution] y+2 y Solution (a) 4 1 5 - 5+ 3 2 6 x 3x 2 x For this problem there are coefficients on each term in the denominator so we’ll first need the least common denominator for the coefficients. This is 6. Now, x (by itself with a power of 1) is the only factor that occurs in any of the denominators. So, the least common denominator for this part is x with the largest power that occurs on all the x’s in the problem, which is 5. So, the least common denominator for this set of rational expression is lcd : 6 x5 So, we simply need to multiply each term by an appropriate quantity to get this in the denominator and then do the addition and subtraction. Let’s do that. © 2007 Paul Dawkins 48 http://tutorial.math.lamar.edu/terms.aspx College Algebra 4 ( x3 ) 5 ( 3x 2 ) 1( 2 ) 4 1 5 + = + 6 x 2 3x 5 2 x3 6 x 2 ( x 3 ) 3x 5 ( 2 ) 2 x3 ( 3x 2 ) 4 x3 2 15 x 2 = 5- 5+ 5 6x 6x 6x 3 4...
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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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