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<para>This module is from<link document="col10614">Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr.</para><para>A detailed study of arithmetic operations with rational expressions is presented in this chapter, beginning with the definition of a rational expression and then proceeding immediately to a discussion of the domain. The process of reducing a rational expression and illustrations of multiplying, dividing, adding, and subtracting rational expressions are also included. Since the operations of addition and subtraction can cause the most difficulty, they are given particular attention. We have tried to make the written explanation of the examples clearer by using a "freeze frame" approach, which walks the student through the operation step by step.</para><para>The five-step method of solving applied problems is included in this chapter to show the problem-solving approach to number problems, work problems, and geometry problems. The chapter also illustrates simplification of complex rational expressions, using the combine-divide method and the LCD-multiply-divide method.</para><para>Objectives of this module: be able to distinguish between simple and complex fractions, be able to simplify complex fractions using the combine-divide and the LCD-multiply-divide method.</para>

Overview

  • Simple And Complex Fractions
  • The Combine-Divide Method
  • The LCD-Multiply-Divide Method

Simple and complex fractions

Simple fraction

In Section [link] we saw that a simple fraction was a fraction of the form P Q , where P and Q are polynomials and Q 0 .

Complex fraction

A complex fraction is a fraction in which the numerator or denominator, or both, is a fraction. The fractions 8 15 2 3 and 1 1 x 1 1 x 2 are examples of complex fractions, or more generally, complex rational expressions.

There are two methods for simplifying complex rational expressions: the combine-divide method and the LCD-multiply-divide method.

The combine-divide method

  1. If necessary, combine the terms of the numerator together.
  2. If necessary, combine the terms of the denominator together.
  3. Divide the numerator by the denominator.

Sample set a

Simplify each complex rational expression.

x 3 8 x 5 12 Steps 1 and 2 are not necessary so we proceed with step 3 . x 3 8 x 5 12 = x 3 8 · 12 x 5 = x 3 8 2 · 12 3 x 5 2 = 3 2 x 2

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1 1 x 1 1 x 2 Step 1:       Combine the terms of the numerator: LCD = x . 1 1 x = x x 1 x = x 1 x Step 2:       Combine the terms of the denominator: LCD = x 2 1 1 x 2 = x 2 x 2 1 x 2 = x 2 1 x 2 Step 3:       Divide the numerator by the denominator . x 1 x x 2 1 x 2 = x 1 x · x 2 x 2 1 = x 1 x x 2 ( x + 1 ) ( x 1 ) = x x + 1 Thus, 1 1 x 1 1 x 2 = x x + 1

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2 13 m 7 m 2 2 + 3 m + 1 m 2 Step 1:        Combine the terms of the numerator: LCD = m 2 . 2 13 m 7 m 2 = 2 m 2 m 2 13 m m 2 7 m 2 = 2 m 2 13 m 7 m 2 Step 2:        Combine the terms of the denominator: LCD = m 2 2 + 3 m + 1 m 2 = 2 m 2 m 2 + 3 m m 2 + 1 m 2 = 2 m 2 + 3 m + 1 m 2 Step 3:       Divide the numerator by the denominator . 2 m 2 13 m 7 m 2 2 m 2 + 3 m 1 m 2 = 2 m 2 13 m 7 m 2 · m 2 2 m 2 + 3 m + 1 = ( 2 m + 1 ) ( m 7 ) m 2 · m 2 ( 2 m + 1 ) ( m + 1 ) = m 7 m + 1 Thus, 2 13 m 7 m 2 2 + 3 m + 1 m 2 = m 7 m + 1

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Practice set a

Use the combine-divide method to simplify each expression.

27 x 2 6 15 x 3 8

12 5 x

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3 1 x 3 + 1 x

3 x 1 3 x + 1

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1 + x y x y 2 x

x y ( x y )

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m 3 + 2 m m 4 + 3 m

m 2 m 3

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1 + 1 x 1 1 1 x 1

x x 2

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The lcd-multiply-divide method

  1. Find the LCD of all the terms.
  2. Multiply the numerator and denominator by the LCD.
  3. Reduce if necessary.

Sample set b

Simplify each complex fraction.

1 4 a 2 1 + 2 a Step 1:      The LCD = a 2 . Step 2:      Multiply both the numerator and denominator by  a 2 . a 2 ( 1 4 a 2 ) a 2 ( 1 + 2 a ) = a 2 · 1 a 2 · 4 a 2 a 2 · 1 + a 2 · 2 a = a 2 4 a 2 + 2 a Step 3:        Reduce . a 2 4 a 2 + 2 a = ( a + 2 ) ( a 2 ) a ( a + 2 ) = a 2 a Thus, 1 4 a 2 1 + 2 a = a 2 a

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1 5 x 6 x 2 1 + 6 x + 5 x 2 Step 1 : The LCD is x 2 . Step 2 : Multiply the numerator and denominator by x 2 . x 2 ( 1 5 x 6 x 2 ) x 2 ( 1 + 6 x + 5 x 2 ) = x 2 · 1 x 2 · 5 x x 2 · 6 x 2 x 2 · 1 + x 2 · 6 x + x 2 · 5 x 2 = x 2 5 x 6 x 2 + 6 x + 5 Step 3 : Reduce . x 2 5 x 6 x 2 + 6 x + 5 = ( x 6 ) ( x + 1 ) ( x + 5 ) ( x + 1 ) = x 6 x + 5 Thus, 1 5 x 6 x 2 1 + 6 x + 5 x 2 = x 6 x + 5

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Practice set b

The following problems are the same problems as the problems in Practice Set A. Simplify these expressions using the LCD-multiply-divide method. Compare the answers to the answers produced in Practice Set A.

27 x 2 6 15 x 3 8

12 5 x

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3 1 x 3 + 1 x

3 x 1 3 x + 1

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1 + x y x y 2 x

x y ( x y )

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m 3 + 2 m m 4 + 3 m

m 2 m 3

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1 + 1 x 1 1 1 x 1

x x 2

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Exercises

For the following problems, simplify each complex rational expression.

1 1 y 1 + 1 y

y 1 y + 1

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a b + c b a b c b

a + c a c

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1 + x x + y 1 x x + y

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2 + 5 a + 1 2 5 a + 1

2 a + 7 2 a 3

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1 1 a 1 1 + 1 a 1

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4 1 m 2 2 + 1 m

2 m 1 m

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k 1 k k + 1 k

k 1

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2 x y 2 x y y 2 x y 3

3 y 2 ( 2 x y ) 2

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1 a + b 1 a b 1 a + b + 1 a b

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5 x + 3 5 x 3 5 x + 3 + 5 x 3

3 x

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1 x 2 1 y 2 1 x + 1 y

y x x y

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1 + 5 x + 6 x 2 1 1 x 12 x 2

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1 + 1 y 2 y 2 1 + 7 y + 10 y 2

y 1 y + 5

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3 n m 2 m n 3 n m + 4 + m n

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x 4 3 x 1 1 2 x 2 3 x 1

3 x 4

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y x + y x x y x x + y + y x y

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a a 2 a a + 2 2 a a 2 + a 2 a + 2

4 a 2 + 4

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x 1 1 1 x x + 1 1 + 1 x

( x 2 ) ( x + 1 ) ( x 1 ) ( x + 2 )

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In electricity theory, when two resistors of resistance R 1 and R 2 ohms are connected in parallel, the total resistance R is

R = 1 1 R 1 + 1 R 2

Write this complex fraction as a simple fraction.

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According to Einstein’s theory of relativity, two velocities v 1 and v 2 are not added according to v = v 1 + v 2 , but rather by

v = v 1 + v 2 1 + v 1 v 2 c 2

Write this complex fraction as a simple fraction.

Einstein's formula is really only applicale for velocities near the speed of light ( c = 186 , 000 miles per second ) . At very much lower velocities, such as 500 miles per hour, the formula v = v 1 + v 2 provides an extremely good approximation.

c 2 ( V 1 + V 2 ) c 2 + V 1 V 2

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Exercises for review

( [link] ) Supply the missing word. Absolute value speaks to the question of how and not “which way.”

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( [link] ) Find the product. ( 3 x + 4 ) 2 .

9 x 2 + 24 x + 16

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( [link] ) Solve the equation 3 x 1 5 x + 3 = 0.

x = 7

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( [link] ) One inlet pipe can fill a tank in 10 minutes. Another inlet pipe can fill the same tank in 4 minutes. How long does it take both pipes working together to fill the tank?

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Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
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what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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