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Transcript of Section 7.1 Introduction to Rational Expressions Copyright © 2013, 2009, and 2005 Pearson...
Section 7.1
Introduction to Rational
Expressions
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Basic Concepts
Rational expressions can be written as quotients (fractions) of two polynomials.Examples include:
2
3
2
, , 5
4
6
3
4 1
4 8
x x
x
x
xx
Example
If possible, evaluate each expression for the given value of the variable.a. b. c.
Solutiona. b. c.
1; 3
3x
x
2
; 43 4
ww
w
4; 5
4
ww
w
1; 3
3x
x
2
;3 4
4w
ww
4;
45
ww
w
1
3
1
3 6
2( )
3
4
( 44)
16
12 4
16
28
4 ( )
5)
5
( 4
91
9
Example
Find all values of the variable for which each expression is undefined.a. b. c.
Solutiona. b. c.
2
1
x
2
4
w
w 2
6
4w
2
1
x
2
4
w
w 2
6
4w
Undefined when x2 = 0 or when x = 0.
Undefined when w – 4 = 0 or when w = 4.
Undefined when w2 – 4 = 0 or when w = 2.
Example
Simplify each fraction by applying the basic principle of fractions.a. b. c.
Solutiona. The GCF of 9 and 15 is 3.
b. The GCF of 20 and 28 is 4.
c. The GCF of 45 and 135 is 45.
9
15
20
28
45
135
9
15 5
3
3
3
3
5
20
28 7
4
4
5
5
7
45
135
145
5 34
1
3
Example
Simplify each expression.a. b. c.
Solutiona. b. c.
2
16
4
y
y
3 12
4 16
x
x
2
2
25
2 7 15
x
x x
2
16
4
y
y
3 12
4 16
x
x
44
4
y
y y
4
y
3( )
4(
4
)4
x
x
3
4
2
2
25
2 7 15
x
x x
( )( 5)
(2
5
3)( )5
x
x
x
x
5
2 3
x
x
Example
Simplify each expression.a. b.
Solutiona. b.
7
2 14
y
y
8
8
x
x
7
2 14
y
y
1( )
2(
7
7)
y
y
1
2
8
8
x
x
(8 )
8
x
x
8
8
x
x
81
8
x
x
Example
Suppose that n balls, numbered 1 to n, are placed in a container and two balls have the winning number.a. What is the probability of drawing the winning ball at random?b. Calculate this probability for n = 100, 1000 and 10,000.c. What happens to the probability of drawing the winning ball as the number of balls increases?Solutiona. There are 2 chances of drawing the winning ball. 2
n
Example (cont)
b. Calculate this probability for n = 100, 1000 and 10,000.
c. What happens to the probability of drawing the winning ball as the number of balls increases?
The probability decreases.
2 1
100 50
2 1
1000 500
2 1
10,000 5000
Section 7.2
Multiplication and Division of Rational
Expressions
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Review of Multiplication and Division of Fractions
• Multiplication of Rational Expressions
• Division of Rational Expressions
Example
Multiply and simplify your answers to lowest terms. a. b. c.
Solutiona.
b.
c.
4 5
9 7
415
5 2 5
7 8
4 5 20
9 7 63
4 15 4 6015 12
5 1 5 5
2 5 5 2 5 1 5
7 8 7 8 7 4 28
Example
Divide and simplify your answers to lowest terms. a. b. c.
Solutiona. b.
c.
1 3
6 5
618
7 4 11
5 15
1 3
6 5
618
7
4 11
5 15
1 5
6 3
5
18
1 6
7 18
6 1
7 18
1
21
60
55
4 15
5 11
12
11
12 5
11 5
Example
Multiply and simplify to lowest terms. Leave your answers in factored form. a. b.
Solutiona. b.
2
6 5
10 12
x
x
3 4
2 1 3 9
x x
x x
2
6 5
10 12
x
x
210 12
6 5x
x
21
0
0
3
2
x
x
1
4x
3 4
2 1 3 9
x x
x x
( 3)( 4)
(2 1) 3 )9(x x
x x
( )( 4)
3(2 1)( 3)
3 x
x
x
x
4
3(2 1)
x
x
Example
Multiply and simplify to lowest terms. Leave your answer in factored form.
Solution
2
2
16 3
9 4
x x
x x
2
2
( 3)
(
( )
4)
6
( 9)
1
x
x
x
x
2
2
16 3
9 4
x x
x x
( 3)( 4)( 4)
( 3)( 3 4))(x
xx x
x x
( 4)( )( )
( 3)
3
( 3 )
4
4)(
x
x
x x
x x
( 4)
( 3)
x
x
Example
Divide and simplify to lowest terms.a. b.
Solutiona. b.
3 2 1
6
x
x x
2
2
16 4
2 8 2
x x
x x x
3 2 1
6
x
x x
3 6
2 1
x
x x
18
(2 1)
x
x x
18
2 1x
2
2
16
2 8
4
2
x
x x
x
x
2
2
16
42 8
2x
x x
x
x
( 4)( 4) 2
( 2)( 4) 4
x x x
x x x
( )( )( )
( )( )(
4
2 4
24
)4
x
xx
x
x
x
1
Section 7.3
Addition and Subtraction with
Like Denominators
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Review of Addition and Subtraction of Fractions
• Rational Expressions Having Like Denominators
Example
Simplify each expression to lowest terms. a. b.
Solution
a.
b.
4 1
7 7
1 5
9 9
4 1
7 7 4 1
7
5
7
1 5
9 9 1 5
9
6
9
2
3
Example
Simplify each expression to lowest terms. a. b.
Solution
a.
b.
13 7
18 18
15 11
30 30
13 7
18 18
13 7
18
6
18
1
3
15 11
30 30
15 11
30
2
15
4
30
To add two rational expressions having like denominators, add their numerators. Keep the same denominator.
C is nonzero
SUMS OF RATIONAL EXPRESSIONS
A B A B
C C C
Example
Add and simplify to lowest terms.a. b.
Solutiona.
b.
4 1 2
3 3
x x
x x
2 2
5
7 10 7 10
x
x x x x
3
4
3
1 2
x x
x x
4 1 2
3
x x
x
3
5 1x
x
2 2
5
7 10 7 10
x
x x x x
2
5
7 10
x
x x
5
5 2
x
x x
1
2x
Example
Add and simplify to lowest terms.a. b.
Solutiona.
b.
7 4
ab ab
2 2 2 2
w y
w y w y
7 4
ab ab
7 4
ab
11
ab
2 2 2 2
w y
w y w y
2 2
w y
w y
( )( )
w y
w y w y
1
w y
To subtract two rational expressions having like denominators, subtract their numerators. Keep the same denominator.
C is nonzero
DIFFERENCES OF RATIONAL EXPRESSIONS
A B A B
C C C
Example
Subtract and simplify to lowest terms.a. b.
Solutiona.
b.
2 2
6 6x
x x
2 2
2 3 4
1 1
x x
x x
2 2
6 6x
x x
2
6 6x
x
2
6 6x
x
2
x
x
1
x
2 2
2 3 4
1 1
x x
x x
2
2 3 4
1
x x
x
1
1 1
x
x x
2
1
1
x
x
1
1x
Example
Subtract and simplify to lowest terms.
Solution
7 2
2 2
a a
a a
7 2
2 2
a a
a a
7 ( 2)
2
a a
a
6 2
2
a
a
Section 7.4
Addition and Subtraction with
Unlike Denominators
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Finding Least Common Multiples
• Review of Fractions Having Unlike Denominators
• Rational Expressions Having Unlike Denominators
The least common multiple (LCM) of two or more polynomials can be found as follows.
Step 1: Factor each polynomial completely.
Step 2: List each factor the greatest number of times that it occurs in either
factorization.
Step 3: Find the product of this list of factors. The result is the LCM.
FINDING THE LEAST COMMON MULTIPLE
Example
Find the least common multiple of each pair of expressions.a. 6x, 9x4 b. x2 + 7x + 12, x2 + 8x + 16 SolutionStep 1: Factor each polynomial completely.
6x = 3 ∙ 2 ∙ x 9x4 = 3 ∙ 3 ∙ x ∙ x ∙ x ∙ x
Step 2: List each factor the greatest number of times.3 ∙ 3 ∙ 2 ∙ x ∙ x ∙ x ∙ x
Step 3: The LCM is 18x4.
Example (cont)
b. x2 + 7x + 12, x2 + 8x + 16
Step 1: Factor each polynomial completely.x2 + 7x + 12 = (x + 3)(x + 4)x2 + 8x + 16 = (x+ 4)(x + 4)
Step 2: List each factor the greatest number of times.(x + 3), (x + 4), and (x + 4)
Step 3: The LCM is (x + 3)(x + 4)2.
Example
Simplify each expression.a. b.
Solutiona. The LCD is the LCM, 42.
b. The LCD is 60.
4 1
7 6
5 11
12 30
4 1
7 6
4 6 1 7
7 6 6 7
24 7
42 42
31
42
5 11
12 30
5 5 11 2
12 5 30 2
25 22
60 60
3
60
1
20
Example
Find each sum and leave your answer in factored form.a. b.
Solutiona. The LCD is x2.
b.
2
2 5
x x
4 3
1 1x x
2
2 5
x x
2
2 5x
x x x
2 2
2 5x
x x
2
2 5x
x
4 3
1 1x x
4 3 1
1 1 1x x
4 3
1 1x x
1
1x
Example
Simply the expression. Write your answer in lowest terms and leave it in factored form. SolutionThe LCD is x(x + 7).
3 5
7
x
x x
3 5
7
x
x x
3 7 5
7 7
x x x
x x x x
3 7 5
7 7
x x x
x x x x
3 7 5
7
x x x
x x
2 4 21 5
7
x x x
x x
2 21
7
x x
x x
Example
Simplify the expression. Write your answer in lowest terms and leave it in factored form.Solution
2 2
6 5
6 9 9x x x
6 5
3 3 3 3x x x x
2 2
6 5
6 9 9x x x
6 5
3 3 3 33
33
3
x
x x xx x
x
x
6 3 5 3
3 3 3 3 3 3
x x
x x x x x x
6 18 5 15
3 3 3
x x
x x x
33
3 3 3
x
x x x
Example
Add and then find the reciprocal of the result.
SolutionThe LCD is RS.
1 1,
R S
1 1 1 1
R S R
R
S
S
S R
S R
RS RS
S R
RS
The reciprocal is
.RS
S R
Basic Concepts
A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both.
Complex fractions involve division of fractions. When dividing two fractions, we can multiply the numerator by the reciprocal of the denominator.
, 0 and 1,xf x a a a For any real numbers a, b, c, and d,
where b, c, and d are nonzero.
SIMPLIFYING BASIC COMPLEX FRACTIONS
,
aa db
c b cd
Simplifying Complex Fractions (Method 1)
To simplify a complex fraction, perform the following steps.STEP 1: Write the numerator as a single fraction; write the denominator as a single fraction.
STEP 2: Divide the denominator into the numerator by multiplying the numerator and the reciprocal of the denominator.
STEP 3: Simplify the result to lowest terms.
Example
Simplify each complex fraction.a. b.
Solutiona.
b.
56
23
2
10
15
xy
xy
56
23
35
6 2
5
4
2
10
15
xy
xy
215
10
x y
y x
215
10
xy
xy 3
2
y
Example
Simplify. Write your answer in lowest terms.
Solution
13
13
3
3n
n
n
n
13
13
3
3n
n
n
n
2
2
9 13 3
9 13 3
nn n
nn n
2
2
9 13
9 13
nn
nn
2
2
9 1 3
3 9 1
n n
n n
2
2
9 1
9 1
n
n
Example
Simplify. Write your answer in lowest terms.1 1
2 2
2 24 4
x x
x x
For the numerator, the LCD is (x + 2)(x – 2).
For the denominator, the LCD is (x + 4)(x – 4).
2 22 2 2 2
2 4 2 4
4 4 4 4
x xx x x x
x x
x x x x
42 2
164 4
x x
x x
4 44
2 2 16
x x
x x
4 4
4 2 2
x x
x x
Simplifying Complex Fractions (Method II)
To simplify a complex fraction, perform the following steps.STEP 1: Find the LCD of all fractions within the complex fraction.
STEP 2: Multiply the numerator and the denominator of the complex fraction by the LCD.
STEP 3: Simplify the result to lowest terms.
Example
Simplify.
Solution
641
811
x x
x x
641
811
x x
x x
641
811
1
1x x
x x
x x
x x
4 1 6 111 8 1
1
x x x xx xx x x xx x
4 6 1
8 1
x x
x x
4 6 6
8 8
x x
x x
2 6
9 8
x
x
Example
Simplify.
Solution
3 3
3 3
1 1
11x y
x y
3 3
3 3
1 1
11x y
x y
3 3
3 3
3 31 1
3 311
x y
x y
x y
x y
3 3
3 3 1
y x
x y
Section 7.6
Rational Equations and
Formulas
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Solving Rational Equations
• Rational Expressions and Equations
• Graphical and Numerical Solutions
• Solving a Formula for a Variable
• Applications
Rational Equations
If an equation contains one or more rational expressions, it is called a rational equation.
Example
Solve each equation.a. b.
Solutiona. b.
9 4
7 x
6
2 1x
x
9 4
7 x
9 28x
28
9x
6
2 1x
x
6
2 1 1
x
x
2 1 6(1)x x
22 6x x 22 6 0x x
2 3 2 0x x
2 3 0x
3
2x 2x
The solutions are 3
2 and .2
The solution is 28
.9
2 0x
Example
Determine whether you are given an expression or an equation. If it is an expression, simplify it and then evaluate it for x = 4. If it is an equation, solve it.a. b.
Solutiona. There is an equal sign, so it is an equation.
24
3 3
x x
x x
2 3 10
2 2
x x
x x
23 4 3 3
3 3
x xx x x
x x
2 4 12x x x
4x 3 12x The answer checks.
The solution is −4.
Example (cont)
b. There is no equals sign, so it is an expression. The common denominator is x – 2, so we can add the numerators.
2 3 10
2
x x
x
2 3 10
2 2
x x
x x
2 5
2
x x
x
5x
When x = 4, the expression evaluates 4 + 5 = 9.
2 3 10
2 2
x x
x x
Example
Solve graphically and numerically.
SolutionGraph and
2
1x
x
1
2
1y
x
2 .y x
The solutions are −2 and 1.
x −3 −2 −1 0 1 2 3
−1 −2 −− 2 1 2
3
1
2(−2, −2)
(1, 1)
Example (cont)
Solve graphically and numerically.
SolutionNumerical Solution
2
1x
x
The solutions are −2 and 1.
x −3 −2 −1 0 1 2 3
−1 −2 −− 2 1
−3 −2 −1 0 1 2 3
1
2
1y
x
2
31
2
2y x
Example
Solve the equation for the specified variable.
Solution
2 for C r r
2 for C r r
2
2 2
rC
2
Cr
Example
Solve the equation for the specified variable.
Solution
2 for
A
h AB b
2h
b
A
B
( ) 2h B b A
( )
2
h B bA
Example
Solve the equation for the specified variable.
Solution
2 for S B sl s
2 for S B ls s 2 S B B B ls
2 S B ls 2
2 2
S B
l
sl
l
2
S Bs
l
Example
A pump can fill a swimming pool ¾ full in 6 hours, another can fill the pool ¾ full in 9 hours. How long would it take the pumps to fill the pool ¾ full, working together?Solution
16 9
t t
36 36(1)6 9
t t 6 4 36t t
6 33 3
10 5t
The two pumps can fill the pool ¾ full in hours.3
35
Proportions
A proportion is a statement (equation) that two ratios (fractions) are equal.
The following property is a convenient way to solve proportions:
a
b d
c is equivalent to ,ad bc
provided b ≠ 0 and d ≠ 0.
Example
On an elliptical machine, Francis can burn 370 calories in 25 minutes. If he increases his work time to 30 minutes, how many calories will he burn?SolutionLet x be the equivalent amount of calories.
25 30
370 x
Minutes Minutes=
Calories Calories
25 11,100x
444x Thus, in 30 minutes, Francis will burn 444 calories.
Example
A 6-foot tall person casts a shadow that is 8-foot long. If a nearby tree casts a 32-foot long shadow, estimate the height of the tree.SolutionThe triangles are similar because the measures of its corresponding angles are equal. Therefore corresponding sides are proportional.
6 ft h
8 ft 32 ft
32
6 8
h
Height Shadow length=
Height Shadow length8 192h
24h The tree is 24 feet tall.
Example
Let y be directly proportional to x, or vary directly with x. Suppose y = 9 when x = 6. Find y when x = 13.SolutionStep 1 The general equation is y = kx. Step 2 Substitute 9 for y and 6 for x in y = kx. Solve for k.Step 3 Replace k with 9/6 in the equation y = 9x/6.Step 4 To find y, let x = 13.
9 6
9
6
y kx
k
k
9
69
(13)619.5
y x
y
y
Example
The table lists the amount of pay for various hours worked.
a. Find the constant of proportionality. b. Predict the pay for 19 hours of work.
Hours Pay
6 $138
11 $253
15 $345
23 $529
31 $713
6, 138
11, 253
15, 345
23, 529
31, 713
0
100
200
300
400
500
600
700
800
0 10 20 30 40P
ay
(do
llars
)
Hours
Example (cont)
The slope of the line equals the proportionality, k. If we use the first and last data points (6, 138) and (31, 713), the slope is
The amount of pay per hour is $23. The graph of the line y = 23x, models the given graph.To find the pay for 19 hours, substitute 19 for x.
713 138
31 6k
23
y = 23x,y = 23(19)y = 437
19 hours of work would pay $437.00
Example
Let y be inversely proportional to x, or vary inversely with x. Suppose y = 6 when x = 4. Find y when x = 8.SolutionStep 1 The general equation is y = k/x. Step 2 Substitute 6 for y and 4 for x in Solve for k.Step 3 Replace k with 24 in the equation y = k/x.Step 4 To find y, let x = 8.
64
24
ky
xk
k
24
83
ky
x
y
y
Example
Determine whether the data in each table represent direct variation, inverse variation, or neither. For direct and inverse variation, find the equation.a.
b.
c.
x 3 7 9 12
y 12 28 32 48
x 5 10 12 15
y 12 6 5 4
x 8 11 14 21
y 48 66 84 126
Neither the product xy nor the ratio y/x are constant in the data in the table. Therefore there is neither direct variation nor indirect variation in this table.
As x increases, y decreases. Because xy = 60 for each data point, the equation y = 60/x models the data. This represents an inverse variation.
The equation y = 6x models the data. The data represents direct variation.
, 0 and 1,xf x a a a
Let x, y, and z denote three quantities. Then z varies jointly with x and y if there is a nonzero number k such that
JOINT VARIATION
.z kxy
Example
The strength S of a rectangular beam varies jointly as its width w and the square of its thickness t. If a beam 5 inches wide and 2 inches thick supports 280 pounds, how much can a similar beam 4 inches wide and 3 inches thick support?SolutionThe strength of the beam is modeled by S = kwt2.
2280 5 2k
280
5 4k
14k