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Inside Collection (Textbook):

Textbook by: Wade Ellis, Denny Burzynski. E-mail the authors

Rational Equations

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary:

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr.

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.

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.

Objectives of this module: be able to identify rational equations, understand and be able to use the method of solving rational expressions, be able to recognize extraneous solutions.

Overview

• Rational Equations
• The Logic Behind The Process
• The Process
• Extraneous Solutions

Rational Equations

Rational Equations

When one rational expression is set equal to another rational expression, a rational equation results.

Some examples of rational equations are the following (except for number 5):

Example 1

3x 4 = 15 2 3x 4 = 15 2

Example 2

x+1 x2 = x7 x3 x+1 x2 = x7 x3

Example 3

5a 2 =10 5a 2 =10

Example 4

3 x + x3 x+1 = 6 5x 3 x + x3 x+1 = 6 5x

Example 5

x6 x+1 x6 x+1 is a rational expression, not a rational equation.

The Logic Behind The Process

It seems most reasonable that an equation without any fractions would be easier to solve than an equation with fractions. Our goal, then, is to convert any rational equation to an equation that contains no fractions. This is easily done.

To develop this method, let’s consider the rational equation

1 6 + x 4 = 17 12 1 6 + x 4 = 17 12

The LCD is 12. We know that we can multiply both sides of an equation by the same nonzero quantity, so we’ll multiply both sides by the LCD, 12.

12( 1 6 + x 4 )=12· 17 12 12( 1 6 + x 4 )=12· 17 12

Now distribute 12 to each term on the left side using the distributive property.

12· 1 6 +12· x 4 =12· 17 12 12· 1 6 +12· x 4 =12· 17 12

Now divide to eliminate all denominators.

2·1+3·x = 17 2+3x = 17 2·1+3·x = 17 2+3x = 17

Now there are no more fractions, and we can solve this equation using our previous techniques to obtain 5 as the solution.

The Process

We have cleared the equation of fractions by multiplying both sides by the LCD. This development generates the following rule.

Clearing an Equation of Fractions

To clear an equation of fractions, multiply both sides of the equation by the LCD.

When multiplying both sides of the equation by the LCD, we use the distributive property to distribute the LCD to each term. This means we can simplify the above rule.

Clearing an Equation of Fractions

To clear an equation of fractions, multiply every term on both sides of the equation by the LCD.

The complete method for solving a rational equation is

1. Determine all the values that must be excluded from consideration by finding the values that will produce zero in the denominator (and thus, division by zero). These excluded values are not in the domain of the equation and are called nondomain values.

2. Clear the equation of fractions by multiplying every term by the LCD.

3. Solve this nonfractional equation for the variable. Check to see if any of these potential solutions are excluded values.

4. Check the solution by substitution.

Extraneous Solutions

Extraneous Solutions

Potential solutions that have been excluded because they make an expression undefined (or produce a false statement for an equation) are called extraneous solutions. Extraneous solutions are discarded. If there are no other potential solutions, the equation has no solution.

Sample Set A

Solve the following rational equations.

Example 6

3x 4 = 15 2 . Since the denominators are constants, there are no excluded values.  No values must be excluded. The LCD is 4. Multiply each term by 4. 4· 3x 4 = 4· 15 2 4 · 3x 4 = 4 2 · 15 2 3x = 2·15 3x = 30 x = 10 10 is not an excluded value. Check it as a solution. Check: 3x 4 = 15 2 3( 10 ) 4 = 15 2 Is this correct? 30 4 = 15 2 Is this correct? 15 2 = 15 2 Yes, this is correct. 10 is the solution. 3x 4 = 15 2 . Since the denominators are constants, there are no excluded values.  No values must be excluded. The LCD is 4. Multiply each term by 4. 4· 3x 4 = 4· 15 2 4 · 3x 4 = 4 2 · 15 2 3x = 2·15 3x = 30 x = 10 10 is not an excluded value. Check it as a solution. Check: 3x 4 = 15 2 3( 10 ) 4 = 15 2 Is this correct? 30 4 = 15 2 Is this correct? 15 2 = 15 2 Yes, this is correct. 10 is the solution.

Example 7

4 x1 = 2 x+6 . 1 and 6 are nondomain values. Exclude them from consideration.  TheLCDis( x1 )( x+6 ).Multiplyeverytermby( x1 )( x+6 ). ( x1 )( x+6 )· 4 x1 = ( x1 )( x+6 )· 2 x+6 ( x1 )( x+6 )· 4 x1 = ( x1 )( x+6 )· 2 x+6 4( x+6 ) = 2( x1 ) Solve this nonfractional equation. 4x+24 = 2x2 2x = 26 x = 13 13is not an excluded value. Check it as a solution. Check: 4 x1 = 2 x+6 4 131 = 2 13+6 Is this correct? 4 14 = 2 7 Is this correct? 2 7 = 2 7 Yes, this is correct. 13 is the solution. 4 x1 = 2 x+6 . 1 and 6 are nondomain values. Exclude them from consideration.  TheLCDis( x1 )( x+6 ).Multiplyeverytermby( x1 )( x+6 ). ( x1 )( x+6 )· 4 x1 = ( x1 )( x+6 )· 2 x+6 ( x1 )( x+6 )· 4 x1 = ( x1 )( x+6 )· 2 x+6 4( x+6 ) = 2( x1 ) Solve this nonfractional equation. 4x+24 = 2x2 2x = 26 x = 13 13is not an excluded value. Check it as a solution. Check: 4 x1 = 2 x+6 4 131 = 2 13+6 Is this correct? 4 14 = 2 7 Is this correct? 2 7 = 2 7 Yes, this is correct. 13 is the solution.

Example 8

4a a4 = 2+ 16 a4 . 4 is a nondomain value. Exclude it from consideration. The LCD is a4.Multiply every term bya4. ( a4 )· 4a a4 = 2( a4 )+( a4 )· 16 a4 ( a4 )· 4a a4 = 2( a4 )+( a4 )· 16 a4 4a = 2( a4 )+16 Solve this nonfractional equation. 4a = 2a8+16 4a = 2a+8 2a = 8 a = 4 This value, a=4,has been excluded from consideration. It is not to be considered as a solution. It is extraneous.  As there are no other potential solutions to consider, we conclude that this equation has nosolution. 4a a4 = 2+ 16 a4 . 4 is a nondomain value. Exclude it from consideration. The LCD is a4.Multiply every term bya4. ( a4 )· 4a a4 = 2( a4 )+( a4 )· 16 a4 ( a4 )· 4a a4 = 2( a4 )+( a4 )· 16 a4 4a = 2( a4 )+16 Solve this nonfractional equation. 4a = 2a8+16 4a = 2a+8 2a = 8 a = 4 This value, a=4,has been excluded from consideration. It is not to be considered as a solution. It is extraneous.  As there are no other potential solutions to consider, we conclude that this equation has nosolution.

Practice Set A

Solve the following rational equations.

Exercise 1

2x 5 = x14 6 2x 5 = x14 6

x=10 x=10

Exercise 2

3a a1 = 3a+8 a+3 3a a1 = 3a+8 a+3

a=2 a=2

Exercise 3

3 y3 +2= y y3 3 y3 +2= y y3

Solution

y=3 y=3 is extraneous, so no solution.

Sample Set B

Solve the following rational equations.

Example 9

3 x + 4x x1 = 4 x 2 +x+5 x 2 x . Factor all denominators to find any  excluded values and the LCD. 3 x + 4x x1 = 4 x 2 +x+5 x( x1 ) Nondomain values are 0 and 1.  Exclude them from consideration. The LCD is x( x1 ).Multiply each  term by x( x1 )and simplify. x ( x1 )· 3 x +x( x1 )· 4x x1 = x( x1 ) · 4 x 2 +x+5 x( x1 ) 3( x1 )+4x·x = 4 x 2 +x+5 Solve this nonfractional equation  to obtain the potential solutions. 3x3+4 x 2 = 4 x 2 +x+5 3x3 = x+5 2x = 8 x = 4 4 is not an excluded value. Check it as a solution. 3 x + 4x x1 = 4 x 2 +x+5 x 2 x . Factor all denominators to find any  excluded values and the LCD. 3 x + 4x x1 = 4 x 2 +x+5 x( x1 ) Nondomain values are 0 and 1.  Exclude them from consideration. The LCD is x( x1 ).Multiply each  term by x( x1 )and simplify. x ( x1 )· 3 x +x( x1 )· 4x x1 = x( x1 ) · 4 x 2 +x+5 x( x1 ) 3( x1 )+4x·x = 4 x 2 +x+5 Solve this nonfractional equation  to obtain the potential solutions. 3x3+4 x 2 = 4 x 2 +x+5 3x3 = x+5 2x = 8 x = 4 4 is not an excluded value. Check it as a solution. Check: 3 x + 4x x1 = 4 x 2 +x+5 x 2 x 3 4 + 4·4 41 = 4· 4 2 +4+5 164 Is this correct? 3 4 + 16 3 = 64+4+5 12 Is this correct? 9 12 + 64 12 = 73 12 Is this correct? 73 12 = 73 12 Yes, this is correct. 4 is the solution. Check: 3 x + 4x x1 = 4 x 2 +x+5 x 2 x 3 4 + 4·4 41 = 4· 4 2 +4+5 164 Is this correct? 3 4 + 16 3 = 64+4+5 12 Is this correct? 9 12 + 64 12 = 73 12 Is this correct? 73 12 = 73 12 Yes, this is correct. 4 is the solution.

The zero-factor property can be used to solve certain types of rational equations. We studied the zero-factor property in Section 7.1, and you may remember that it states that if a a and b b are real numbers and that a·b=0, a·b=0, then either or both a=0 a=0 or b=0. b=0. The zero-factor property is useful in solving the following rational equation.

Example 10

3 a 2 2 a = 1. Zero is an excluded value. The LCD is  a 2  Multiply each  term by  a 2  and simplify. a 2 · 3 a 2 a 2 · 2 a = 1· a 2 32a = a 2 Solve this nonfractional quadratic  equation. Set it equal to zero. 0 = a 2 +2a3 0 = ( a+3 )( a1 ) a = 3, a=1 Check these as solutions. 3 a 2 2 a = 1. Zero is an excluded value. The LCD is  a 2  Multiply each  term by  a 2  and simplify. a 2 · 3 a 2 a 2 · 2 a = 1· a 2 32a = a 2 Solve this nonfractional quadratic  equation. Set it equal to zero. 0 = a 2 +2a3 0 = ( a+3 )( a1 ) a = 3, a=1 Check these as solutions. Check: If a=3: 3 ( 3 ) 2 2 3 = 1 Is this correct? 3 9 + 2 3 = 1 Is this correct? 1 3 + 2 3 = 1 Is this correct? 1 = 1 Yes, this is correct. a = 3  checks and is a solution. If a=1: 3 ( 1 ) 2 2 1 = 1 Is this correct? 3 1 2 1 = 1 Is this correct? 1 = 1 Yes, this is correct. a = 1  checks and is a solution. 3 and 1 are the solutions. Check: If a=3: 3 ( 3 ) 2 2 3 = 1 Is this correct? 3 9 + 2 3 = 1 Is this correct? 1 3 + 2 3 = 1 Is this correct? 1 = 1 Yes, this is correct. a = 3  checks and is a solution. If a=1: 3 ( 1 ) 2 2 1 = 1 Is this correct? 3 1 2 1 = 1 Is this correct? 1 = 1 Yes, this is correct. a = 1  checks and is a solution. 3 and 1 are the solutions.

Practice Set B

Exercise 4

Solve the equation a+3 a2 = a+1 a1 . a+3 a2 = a+1 a1 .

a= 1 3 a= 1 3

Exercise 5

Solve the equation 1 x1 1 x+1 = 2x x 2 1 . 1 x1 1 x+1 = 2x x 2 1 .

Solution

This equation has no solution. x=1 x=1 is extraneous.

Section 7.6 Exercises

For the following problems, solve the rational equations.

Exercise 6

32 x = 16 3 32 x = 16 3

x=6 x=6

Exercise 7

54 y = 27 4 54 y = 27 4

Exercise 8

8 y = 2 3 8 y = 2 3

y=12 y=12

Exercise 9

x 28 = 3 7 x 28 = 3 7

Exercise 10

x+1 4 = x3 2 x+1 4 = x3 2

x=7 x=7

Exercise 11

a+3 6 = a1 4 a+3 6 = a1 4

Exercise 12

y3 6 = y+1 4 y3 6 = y+1 4

y=9 y=9

Exercise 13

x7 8 = x+5 6 x7 8 = x+5 6

Exercise 14

a+6 9 a1 6 =0 a+6 9 a1 6 =0

a=15 a=15

Exercise 15

y+11 4 = y+8 10 y+11 4 = y+8 10

Exercise 16

b+1 2 +6= b4 3 b+1 2 +6= b4 3

b=47 b=47

Exercise 17

m+3 2 +1= m4 5 m+3 2 +1= m4 5

Exercise 18

a6 2 +4=1 a6 2 +4=1

a=4 a=4

Exercise 19

b+11 3 +8=6 b+11 3 +8=6

Exercise 20

y1 y+2 = y+3 y2 y1 y+2 = y+3 y2

y= 1 2 y= 1 2

Exercise 21

x+2 x6 = x1 x+2 x+2 x6 = x1 x+2

Exercise 22

3m+1 2m = 4 3 3m+1 2m = 4 3

m=3 m=3

Exercise 23

2k+7 3k = 5 4 2k+7 3k = 5 4

Exercise 24

4 x+2 =1 4 x+2 =1

x=2 x=2

6 x3 =1 6 x3 =1

Exercise 26

a 3 + 10+a 4 =6 a 3 + 10+a 4 =6

a=6 a=6

Exercise 27

k+17 5 k 2 =2k k+17 5 k 2 =2k

Exercise 28

2b+1 3b5 = 1 4 2b+1 3b5 = 1 4

b= 9 5 b= 9 5

Exercise 29

3a+4 2a7 = 7 9 3a+4 2a7 = 7 9

Exercise 30

x x+3 x x2 = 10 x 2 +x6 x x+3 x x2 = 10 x 2 +x6

x=2 x=2

Exercise 31

3y y1 + 2y y6 = 5 y 2 15y+20 y 2 7y+6 3y y1 + 2y y6 = 5 y 2 15y+20 y 2 7y+6

Exercise 32

4a a+2 3a a1 = a 2 8a4 a 2 +a2 4a a+2 3a a1 = a 2 8a4 a 2 +a2

a=2 a=2

Exercise 33

3a7 a3 = 4a10 a3 3a7 a3 = 4a10 a3

Exercise 34

2x5 x6 = x+1 x6 2x5 x6 = x+1 x6

Solution

No solution; 6 is an excluded value.

Exercise 35

3 x+4 + 5 x+4 = 3 x1 3 x+4 + 5 x+4 = 3 x1

Exercise 36

2 y+2 + 8 y+2 = 9 y+3 2 y+2 + 8 y+2 = 9 y+3

y=12 y=12

Exercise 37

4 a 2 +2a = 3 a 2 +a2 4 a 2 +2a = 3 a 2 +a2

Exercise 38

2 b( b+2 ) = 3 b 2 +6b+8 2 b( b+2 ) = 3 b 2 +6b+8

b=8 b=8

Exercise 39

x x1 + 3x x4 = 4 x 2 8x+1 x 2 5x+4 x x1 + 3x x4 = 4 x 2 8x+1 x 2 5x+4

Exercise 40

4x x+2 x x+1 = 3 x 2 +4x+4 x 2 +3x+2 4x x+2 x x+1 = 3 x 2 +4x+4 x 2 +3x+2

no solution

Exercise 41

2 a5 4a2 a 2 6a+5 = 3 a1 2 a5 4a2 a 2 6a+5 = 3 a1

Exercise 42

1 x+4 2 x+1 = 4x+19 x 2 +5x+4 1 x+4 2 x+1 = 4x+19 x 2 +5x+4

Solution

No solution;  4 4 is an excluded value.

Exercise 43

2 x 2 + 1 x =1 2 x 2 + 1 x =1

Exercise 44

6 y 2 5 y =1 6 y 2 5 y =1

y=6,1 y=6,1

Exercise 45

12 a 2 4 a =1 12 a 2 4 a =1

Exercise 46

20 x 2 1 x =1 20 x 2 1 x =1

x=4,5 x=4,5

Exercise 47

12 y + 12 y 2 =3 12 y + 12 y 2 =3

Exercise 48

16 b 2 + 12 b =4 16 b 2 + 12 b =4

y=4,1 y=4,1

Exercise 49

1 x 2 =1 1 x 2 =1

Exercise 50

16 y 2 =1 16 y 2 =1

y=4,4 y=4,4

Exercise 51

25 a 2 =1 25 a 2 =1

Exercise 52

36 y 2 =1 36 y 2 =1

y=6,6 y=6,6

Exercise 53

2 x 2 + 3 x =2 2 x 2 + 3 x =2

Exercise 54

2 a 2 5 a =3 2 a 2 5 a =3

Solution

a= 1 3 ,2 a= 1 3 ,2

Exercise 55

2 x 2 + 7 x =6 2 x 2 + 7 x =6

Exercise 56

4 a 2 + 9 a =9 4 a 2 + 9 a =9

Solution

a= 1 3 , 4 3 a= 1 3 , 4 3

Exercise 57

2 x = 3 x+2 +1 2 x = 3 x+2 +1

Exercise 58

1 x = 2 x+4 3 2 1 x = 2 x+4 3 2

Solution

x= 4 3 ,2 x= 4 3 ,2

Exercise 59

4 m 5 m3 =7 4 m 5 m3 =7

Exercise 60

6 a+1 2 a2 =5 6 a+1 2 a2 =5

Solution

a= 4 5 ,1 a= 4 5 ,1

For the following problems, solve each literal equation for the designated letter.

Exercise 61

V= GMm D  forD. V= GMm D  forD.

Exercise 62

PV=nrtforn. PV=nrtforn.

Solution

n= PV rt n= PV rt

Exercise 63

E=m c 2 form. E=m c 2 form.

Exercise 64

P=2( 1+w )forw. P=2( 1+w )forw.

W= P2 2 W= P2 2

Exercise 65

A= 1 2 h( b+B )forB. A= 1 2 h( b+B )forB.

Exercise 66

A=P( 1+rt )forr. A=P( 1+rt )forr.

Solution

r= AP Pt r= AP Pt

Exercise 67

z= x x ¯ s for x ¯ . z= x x ¯ s for x ¯ .

Exercise 68

F= S x 2 S y 2 for S y 2 . F= S x 2 S y 2 for S y 2 .

Solution

S y 2 = S x 2 F S y 2 = S x 2 F

Exercise 69

1 R = 1 E + 1 F forF. 1 R = 1 E + 1 F forF.

Exercise 70

K= 1 2 h( s 1 + s 2 )for s 2 . K= 1 2 h( s 1 + s 2 )for s 2 .

Solution

S 2 = 2K h S 1  or  2Kh S 1 h S 2 = 2K h S 1  or  2Kh S 1 h

Exercise 71

Q= 2mn s+t fors. Q= 2mn s+t fors.

Exercise 72

V= 1 6 π( 3 a 2 + h 2 )for h 2 . V= 1 6 π( 3 a 2 + h 2 )for h 2 .

Solution

h 2 = 6V3π a 2 π h 2 = 6V3π a 2 π

Exercise 73

I= E R+r forR. I= E R+r forR.

Exercises For Review

Exercise 74

((Reference)) Write ( 4 x 3 y 4 ) 2 ( 4 x 3 y 4 ) 2 so that only positive exponents appear.

Solution

y 8 16 x 6 y 8 16 x 6

Exercise 75

((Reference)) Factor x 4 16. x 4 16.

Exercise 76

((Reference)) Supply the missing word. An slope of a line is a measure of the


of the line.

steepness

Exercise 77

((Reference)) Find the product. x 2 3x+2 x 2 x12 · x 2 +6x+9 x 2 +x2 · x 2 6x+8 x 2 +x6 . x 2 3x+2 x 2 x12 · x 2 +6x+9 x 2 +x2 · x 2 6x+8 x 2 +x6 .

Exercise 78

((Reference)) Find the sum. 2x x+1 + 1 x3 . 2x x+1 + 1 x3 .

Solution

2 x 2 5x+1 ( x+1 )( x3 ) 2 x 2 5x+1 ( x+1 )( x3 )

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