Section 6: Complex Fractions

Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: Section 7.6 Complex Fractions 695 Version: Fall 2007 7.6 Complex Fractions In this section we learn how to simplify what are called complex fractions , an example of which follows. 1 2 + 1 3 1 4 + 2 3 (1) Note that both the numerator and denominator are fraction problems in their own right, lending credence to why we refer to such a structure as a “complex fraction.” There are two very different techniques we can use to simplify the complex fraction ( 1 ). The first technique is a “natural” choice. Simplifying Complex Fractions — First Technique. To simplify a complex fraction, proceed as follows: 1. Simplify the numerator. 2. Simplify the denominator. 3. Simplify the division problem that remains. Let’s follow this outline to simplify the complex fraction ( 1 ). First, add the fractions in the numerator as follows. 1 2 + 1 3 = 3 6 + 2 6 = 5 6 (2) Secondly, add the fractions in the denominator as follows. 1 4 + 2 3 = 3 12 + 8 12 = 11 12 (3) Substitute the results from ( 2 ) and ( 3 ) into the numerator and denominator of ( 1 ), respectively. 1 2 + 1 3 1 4 + 2 3 = 5 6 11 12 (4) The right-hand side of ( 4 ) is equivalent to 5 6 ÷ 11 12 . This is a division problem, so invert and multiply, factor, then cancel common factors. Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/ 1 696 Chapter 7 Rational Functions Version: Fall 2007 1 2 + 1 3 1 4 + 2 3 = 5 6 · 12 11 = 5 2 · 3 · 2 · 2 · 3 11 = 5 2 · 3 · 2 · 2 · 3 11 = 10 11 Here is an arrangement of the work, from start to finish, presented without comment. This is a good template to emulate when doing your homework. 1 2 + 1 3 1 4 + 2 3 = 3 6 + 2 6 3 12 + 8 12 = 5 6 11 12 = 5 6 · 12 11 = 5 2 · 3 · 2 · 2 · 3 11 = 5 2 · 3 · 2 · 2 · 3 11 = 10 11 Now, let’s look at a second approach to the problem. We saw that simplifying the numerator in ( 2 ) required a common denominator of 6. Simplifying the denominator in ( 3 ) required a common denominator of 12. So, let’s choose another common denomina- tor, this one a common denominator for both numerator and denominator, namely, 12. Now, multiply top and bottom (numerator and denominator) of the complex fraction ( 1 ) by 12, as follows. 1 2 + 1 3 1 4 + 2 3 = 1 2 + 1 3 12 1 4 + 2 3 12 (5) Distribute the 12 in both numerator and denominator and simplify. Section 7.6 Complex Fractions 697 Version: Fall 2007 1 2 + 1 3 12 1 4 + 2 3 12 = 1 2 12 + 1 3 12 1 4 12 + 2 3 12 = 6 + 4 3 + 8 = 10 11 Let’s summarize this second technique....
View Full Document

This document was uploaded on 01/31/2008.

Page1 / 8

Section 6: Complex Fractions - Section 7.6 Complex...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online