Section 6: Complex Fractions

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

• Notes
• davidvictor
• 8

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

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: 1

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

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. Simplifying Complex Fractions — Second Technique. To simplify a com- plex fraction, proceed as follows: 1. Find a common denominator for both numerator and denominator.

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

This is the end of the preview. Sign up to access the rest of the document.
• '
• NoProfessor
• Fraction, Elementary arithmetic, Rational function, Greatest common divisor

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern