Lets do that 2007 paul dawkins 48

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online