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Chapter 7
Rational
Expressions
7.1
7.2
7.3
7.4
7.5
7.6
7.7
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7.1
Rational Functions and
Multiplying and Dividing
Rational Expressions
Objectives:
Find the domain of a rational function
Simplify rational expressions
Multiply and divide rational expressions
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Rational expression is an expression that can be
written as the quotient P of two polynomials P and Q as
Q
long as Q is not 0.
Examples of Rational Expressions
3x  2 x  4
4x  5
2
4x  3y
2
2
2 x  3xy  4 y
3x
4
A rational expression is undefined if the denominator is 0. If a
variable in a rational expression is replaced with a number that
makes the denominator 0, we say that the rational expression is
undefined for this value of the variable.
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2
Example 1
Find the domain of each rational expression.
4x2  5
g ( x) 
x2
solve the equation “denominator ≠ 0”:
x–2 ≠0
x≠2
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9 x3  3x 2  27
f ( x) 
5
solve the equation “denominator ≠ 0”:
Since the denominator will never = 0
The domain is all real numbers
Example 2
Find the domain of the rational expression.
9x 1
h( x )  2
x  x6
Set the denominator ≠ 0.
x2 + x – 6 ≠ 0
(x – 2) (x + 3) ≠ 0
x–2≠0
or
x+3≠0
x≠2
or
x ≠ -3
The domain of h is all real numbers except 2 and 3.
x ≠ 2, x ≠ -3
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Simplifying or Writing a Rational Expression
in Lowest Terms
1.
Completely factor the numerator and
denominator of the rational expression.
2.
Divide out factors common to the numerator and
denominator. (This is the same thing as
“removing the factor of 1.”)
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Example 3
Simplify each rational expression.
2
3x
12 x3  3x 2
3x 2
3 x 2 1
 2
3
2
2 x  3x
3x (4 x  1)
1

4x 1
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x 2  3x  4
x 2  x  20
( x  1)( x  4)

( x  5)( x  4)
x 1

1
x5
x 1

x 5
8 x
x 8
x8

1
x8
Example 4
Simplify each rational expression.
48  3 x 2
x 2  3x  4
3(16  x 2 )

( x  1)( x  4)
3(4  x)(4  x)

( x  1)( x  4)
3(4  x)  1( x  4)

( x  1)( x  4)
3(4  x)

x 1
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The terms in the numerator differ by the
sign of the terms in the denominator. The
polynomials are opposites of each other.
Factor out a 1 from the numerator.
x3  8
2 x
( x  2)( x 2  2 x  4)

x2
 x2  2 x  4
Example 5
Multiply each.
2 3 x  x 5 x
1
6 x 2 5x

 
3
25 x  x  x  2 23
4
10 x 12
(m  n)(m  n)  m
( m  n) 2
m
 2

(m  n)  m(m  n)
m  n m  mn
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mn

mn
Example 6
Perform the indicted operation.
( x  3) 2 5 x  15 ( x  3) 2
25



5
25
5
5 x  15
( x  3)( x  3)  5  5

5  5( x  3)
 x3
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7.1 Summary
Objectives:
Find the domain of a rational function
Simplify rational expressions
Multiply and divide rational expressions
Do the following now:
7.1 #5, 6, 10, 15, 18
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7.2
Adding and Subtracting
Rational Expressions
Objectives:
Add/subtract rational expressions with common denominators
Identify the least common denominator of two or more rational expressions
Add/subtract rational expressions with unlike denominators
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Example 1
Add or subtract.
a. a  3a
6 6
2
x
36
b.

x6 x6
c. 4 p  3 3 p  8 4 p  3  3 p  8 7 p  5



2p 7 2p 7
2p 7
2p 7
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Example 2
Subtract.
8y
16

y2 y2
8 y  16
8( y  2)


8
y2
y2
3y  6
3( y  2)
3y
6

 2
 2
2
y  3 y  10 ( y  5)( y  2)
y  3 y  10 y  3 y  10
3

y5
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To add or subtract rational expressions with unlike denominators,
first write the rational expressions as equivalent rational expressions
with common denominators.
The least common denominator (LCD) is usually the easiest
common denominator to work with.
Finding the Least Common Denominator (LCD)
1. Factor each denominator completely,
2. The LCD is the product of all unique factors each raised to a
power equal to the greatest number of times that the factor
appears in any factored denominator.
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Example 3
Find the LCD of the rational expressions.
1
3x
,
6 y 4 y  12
4
4x  2
, 2
2
x  4 x  3 x  10 x  21
Factor each denominator:
1st denom factored: 6y
2nd denom factored: 4(y + 3)
LCD: 12y(y + 3)
x 2  4 x  3  ( x  3)( x  1)
x 2  10 x  21  ( x  3)( x  7)
The LCD is ( x  3)( x  1)( x  7).
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Example 4
Find the LCD of the rational expressions.
3x
4x2
Factor each denominator:
, 2
2
5x  5 x  2 x  1
5x 2  5  5( x 2  1)  5( x  1)( x  1)
x 2  2 x  1  ( x  1) 2
The LCD is 5( x  1)( x  1)2 .
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Adding or Subtracting Rational Expressions with
Unlike Denominators
1. Find the LCD of the rational expressions.
2. Write each rational expression as an equivalent
rational expression whose denominator is the LCD
found in Step 1.
3. Add or subtract numerators, and write the result
over the common denominator.
4. Simplify resulting rational expression.
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Example 5
Add.
73
146
15 8
15  6 8  7
90
56







7 a 6a
7a  6 6a  7 42a 42a 42a 21a
The LCD is 42a.
4
5x
4( x  3)
5 x( x  3)



x  3 x  3 ( x  3)( x  3) ( x  3)( x  3)
4 x  12  5 x 2  15 x 5 x 2  19 x  12


( x  3)( x  3)
( x  3)( x  3)
The LCD is (x + 3)(x – 3)
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Example 6
Subtract.
5
3
5
3



1(6 – 26)x)
2 x  6 6  2 x 2 x  6 –(2x
5
3(31)
8

+

2x  6 2x  6
2x  6
4

x 3
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Example 7
Add
4
x
4
x



2
2
x  x  6 x  5 x  6 ( x  3)( x  2) ( x  3)( x  2)
4( x  3)
x( x  3)


( x  3)( x  2)( x  3) ( x  3)( x  2)( x  3)
4 x  12  x 2  3x

( x  2)( x  3)( x  3)
x 2  x  12

( x  2)( x  3)( x  3)
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7.2 Summary
Objectives:
Add/subtract rational expressions with common denominators
Identify the least common denominator of two or more rational expressions
Add/subtract rational expressions with unlike denominators
Do the following now:
7.2 #2, 4, 6, 8, 10
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7.3
Simplifying Complex
Fractions
Objectives:
Simplify complex fractions by:
•Simplifying the numerator and denominator and then dividing
•Multiplying by a common denominator
Simplify expressions with negative exponents
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A rational expression whose numerator, denominator,
or both contain one or more rational expressions is
called a complex rational expression or a complex
fraction.
1
b
1
a
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x
2
3y
4x 1
8y
1
x
y
y 1
Example 1
Simplify the complex fraction.
5x2
x 1
3x
x 1
2
x
2
5x
3x
5
x
x 1




x 1 x 1
x  1 3x
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5x

3
Example 2
Simplify:
1
1
 2
2
x
y
3 3

x y
1
1
 2
2
x
y
3 3

x y
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What is the LCD for all four fractions?
LCD = x2y2
yx

3xy
Example 3
Simplify the complex fraction.
1 2

2
y
3
1 5

y 6
1 2

2
y
3
1 5

y 6
What is the LCD for all four fractions?
LCD = 6y2
6  4 y2

6 y  5 y2
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When you have a rational expression
where some of the variables have
negative exponents, rewrite the
expression using positive exponents.
Example 4
Simplify.
x 2 y 2

1
1
5x  2 y
2
2
x y

1
1
5x  2 y
What is the LCD for all three fractions?
LCD = x2y2
1
x2 y2
5 2

x y
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7.3 Summary
Objectives:
Simplify complex fractions by:
•Simplifying the numerator and denominator and then dividing
•Multiplying by a common denominator
Simplify expressions with negative exponents
Do the following now:
7.3 #1, 3, 5, 7, 9,18
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7.4
Dividing Polynomials:
Long Division and
Synthetic Division
Objectives:
Divide a polynomial by a monomial
Divide polynomials using long division and synthetic division and be able to tell
when to use which
Use the remainder theorem to evaluate polynomials
Use the factor theorem
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Example 1
Divide
12a3  36a  15 by 3a.
5
12a3  36a  15 12a3 36a 15
2
 4a  12 



a
3a
3a
3a 3a
10t 4  35t 3  5t 2 . 10t 4  35t 3  5t 2  2t 2  7t  1
5t 2
5t 2
5t 2 5t 2
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Dividing Polynomials
Dividing a polynomial by a polynomial other than a
monomial uses a “long division” technique that is
similar to the process known as long division in
dividing two numbers, which is reviewed on the next
slide.
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Example 2
168
43 7256
– 43
295
– 258
37 6
– 344
32
Divide 43 into 72.
Multiply 1 times 43.
Subtract 43 from 72.
Bring down 5.
Divide 43 into 295.
Multiply 6 times 43.
Subtract 258 from 295.
Bring down 6.
Divide 43 into 376.
Multiply 8 times 43.
Subtract 344 from 376.
Nothing to bring down.
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We then write our result as
168
32
.
43
As you can see from the previous example, there is
a pattern in the long division technique.
Divide
Multiply
Subtract
Bring down
Then repeat these steps until you can’t
bring down or divide any longer.
We will incorporate this same repeated technique
with dividing polynomials.
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Example 3
4x  5
2
7 x  3 28x  23x  15
2
28 x  12 x
 35x  15
 35x 15
Divide 7x into 28x2.
Multiply 4x times 7x+3.
Subtract 28x2 + 12x from 28x2 – 23x.
Bring down – 15.
Divide 7x into –35x.
Multiply – 5 times 7x+3.
Subtract –35x–15 from –35x–15.
Nothing to bring down.
So our answer is 4x – 5.
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Example 4
2x 10
2
2x  7 4x  6x  8
2
4 x  14x
20x  8
20x  70
78
We write our final answer as
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Divide 2x into 4x2.
Multiply 2x times 2x+7.
Subtract 4x2 + 14x from 4x2 – 6x.
Bring down 8.
Divide 2x into –20x.
Multiply -10 times 2x+7.
Subtract –20x–70 from –20x+8.
Nothing to bring down.
2x 10 
78
2x  7
Example 5
2
c
Divide:  3c  2
c 1
c 2
c  1 c 2  3c  2
2

c
 c
–


2c 2
–  2c  2 

4
c 2  3c  2  c  2  4
c 1
c 1
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1. c2 by c.
2. Multiply c by c + 1.
3. Subtract c2 + c from c2 + 3c – 2.
4. Bring down the next term
5. Repeat the process until the degree
of the remainder is less than the
degree of the binomial divisor.
Remainder
Synthetic Division
To find the quotient and remainder when a polynomial of
degree 1 or higher is divided by x – c, a shortened version
of long division called synthetic division may be used.
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Example 6
Use synthetic division to divide 2x3 – x2 – 13x + 1
by x – 3.
coefficients
Set divisor = 0
2
1
6
13
15
1
6
Always bring
down the first #
2
5
2
7
Multiply
Then add
Multiply
Then add …
7
2x  5x  2 
x3
3
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2
Example 7
Use synthetic division to divide x4 – 3x3 – 13x2 + 6x + 32
by x + 3.
3
1
3
3
13
18
6
15
32
27
1
6
5
9
59
x 4  3x3  13x 2  6 x  32
59
3
2
 x  6 x  5x  9 
x3
x3
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Add on to notes: Should you use long division or synthetic division?
1x4
1
1
+ 7/3 x3
7/
3
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10/
+ 3x
0
3
0
10/
3
19/
19/
19/
10/
1
1
+ 0x2
3
3
10/
3
+0
3
3
3
Remainder Theorem
If a polynomial P(x) is divided by x – c, then the remainder is P(c).
Example 8
Use the remainder theorem and synthetic division to find
P(3) if P( x)  3x6  18x5  32 x 4  15x 2 .
c
3
3
3
18 32
0
9 27 15
9 5 15
15
45
30
0
0
90 270
90 270
P(3)  270, the remainder.
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Factor Theorem
A polynomial function f(x) has a factor of x – c, if and
only if f(c) = 0.
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Example 9
3
2
Given the polynomial equation P( x)  2 x  13x  17 x  12
a. Use the Remainder Thm. to show that 4 is a solution of the equation.
b. Use the Factor Theorem to solve the polynomial equation.
proposed
solution
4
2
2
13 17 12
8 20 12
5 3
0
Because the remainder is 0, then 4
is a solution of the given equation.
remainder
2 x3  13x 2  17 x  12  0
( x  4)(2 x 2  5 x  3)  0
( x  4)(2 x  1)( x  3)  0
x4  0
or 2x 1  0 or
1
x  4 or x  
or
2
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x 3  0
x3
7.4 summary
Objectives:
Divide a polynomial by a monomial
Divide polynomials using long division and synthetic division and be able to tell
when to use which
Use the remainder theorem to evaluate polynomials
Use the factor theorem
Do the following now:
7.4 #1, 3, 5, 7, 13,16
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7.5
Solving Equations
Containing Rational
Expressions
Objectives:
Solve equations containing rational expressions.
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First note that an equation contains an equal sign and an
expression does not.
Equation
x x 3
 
2 5 4
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Expression
x x

2 5
Solving an Equation Containing Rational
Expressions
1. Multiply both sides of the equation by the LCD of all
rational expressions in the equation.
2. Simplify both sides.
3. Determine whether the equation is linear, quadratic,
or higher degree and solve accordingly.
4. Check the solution in the original equation.
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Example 1
LCD:
LCD is 6x
Restrictions:
6X ≠ 0
X≠0
5
7
1  .
Solve:
3x
6
 5
 7
6 x   1    6 x
 3x   6 
2
Check:
x
10  6x  7 x
10  x
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5
7
1 
3 10
6
5
7
1 
30
6
1
7
1 
6
6
LCD:
LCD is
6X(X+1)
Restrictions:
6X(X + 1) ≠ 0
X ≠ 0, X ≠ -1
Example 2
1
1
1

 2
.
Solve:
2 x x  1 3x  3x
 1  1 1  1   1 1 
6 x  x6x1 x 1        
 6 x x6x1 x  1
 2 x  2xx 1 x  1 3x(x3x1)
( x 1) 
3
2
3x  1  6x  2
3x  3  6x  2
3  3x  2
 3x  1
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1
x
3
Check restrictions
LCD:
LCD is 3(x+ 2)(x + 5)
Restrictions:
3(x+ 2)(x + 5) ≠ 0
x ≠ -2, x ≠ -5
Example 3
x2
1
1


Solve: 2
x  7 x  10 3x  6 x  5
3x  2  x  5  3x  2
3x  6  x  5  3x  6
3x  x  3x  5  6  6
5x  7
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7
x
5
Check restrictions
Restrictions:
(x – 1)(x + 1) ≠ 0
x ≠ 1, x ≠ -1
Example 4
Solve:
1
2

.
x 1 x 1
x  1  2x 1
x  1  2x  2
3 x
Single fraction on
each side I would
cross multiply
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Check restrictions
Example 5
Solve: 12 2  3  2 .
9a
3 a
3 a
12  33  a   23  a
12  9  3a  6  2a
21  3a  6  2a
15  5a
3a
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LCD:
LCD is (3 – a)(3 + a)
Restrictions:
(3 – a)(3 + a) ≠ 0
a ≠ 3, a ≠ -3
12
3
2


9  32 3  3 3  3
12 3 2
 
0 5 0
Restriction: a ≠ 3 so the

answer is no solution.
7.5 summary
Objectives:
Solve equations containing rational expressions
•Find the LCD
•Multiply both sides by the LCD
•Solve the equation with skills from Algebra I
•Don’t forget to check for the restrictions!!!!
Do the following now:
7.5 #3, 6, 9, 12, 15
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
7.6
Rational Equations and
Problem Solving
Objectives:
Solve an equation containing rational expressions for a specified variable.
Solve problems by writing equations containing rational expressions.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Solving Equations for a Specified Variable
1.
2.
3.
4.
5.
Clear the equation of fractions or rational expressions by
multiplying each side of the equation by the least common
denominator (LCD) of all denominators in the equation.
Use the distributive property to remove grouping symbols
such as parentheses.
Combine like terms on each side of the equation.
Use the addition property of equality to rewrite the equation as
an equivalent equation with terms containing the specified
variable on one side and all other terms on the other side.
Use the distributive property and the multiplication property
of equality to get the specified variable alone.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
LCD:
LCD is RR1R2
Example 1
1 1
1
Solve: R  R  R
1
2
for R1.
R1R2  RR2  RR1
R1R2  RR1  RR2
R1 R2  R  RR2
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RR2
R1 
R2  R
Example 2
The quotient of a number and 9 times its reciprocal is 1.
Find the number.
 1
n  9    1
 n
9
n   1
n
n
n 1
9
n2  9
n  3,3
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n = the number, then
1
= the reciprocal of the number.
n
Ratio is the quotient of two numbers or two quantities.
The ratio of the numbers a and b can also be
a
written as a:b, or .
b
The units associated with the ratio are important.
The units should match.
If the units do not match, it is called a rate, rather
than a ratio.
A proportion is a mathematical statement
that two ratios are equal to each other.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 3
Solve the proportion for x.
x 1 5

x2 3
Single fraction on each
side of the equality sign,
so cross multiple
Restriction(s)
x ≠ –2
3x  1  5x  2
3x  3  5x 10
 2x  7
7
x
2
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Check restrictions
Example 4
If a 170-pound person weighs approximately 65 pounds
on Mars, how much does a 9000-pound satellite weigh
on Mars?
person
satellite
170 x  9000  65
170 x  585, 000
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
585, 000
x
170
x  3441 pounds
Example 5
An experienced roofer can roof a house in 26 hours. A
beginner needs 39 hours to do the same job. How long will
it take if the two roofers work together?
Let t = time in hours to complete the job together
Then 1/t = part of the job they each complete.
Hours to
Complete the
Job
Beginning
roofer
39
Experienced
roofer
26
Together
t
Part of the
Job
Completed in
1 Hour
1
39
1
26
1
t
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
1
1 1


26 39 t
1  1
 1
78t      78t
 26 39   t 
3t  2t  78
5t  78
t  78 / 5 or 15.6 hours
Example 6
The speed of Lazy River’s current is 5 mph. A boat travels
20 miles downstream in the same time as traveling 10 miles
upstream. Find the speed of the boat in still water.
20
10

r 5 r 5
20r  5  10r  5
20r 100  10r  50
10r  150
r  15 mph
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Rate Time
Distance
Downstream
r+5
20
r 5
20
Upstream
r–5
10
r 5
10
7.6 summary
Objectives:
Solve an equation containing rational expressions for a specified variable.
Solve problems by writing equations containing rational expressions.
Do the following now:
7.6 #4, 9, 10, 12, 13,17
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
7.7
Variation and Problem
Solving
Objectives:
Solve problems involving direct, inverse, joint, and combined variations.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Direct Variation
y varies directly as x, or y is directly proportional
to x, if there is a nonzero constant k such that
y = kx
The number k is called the constant of variation or
the constant of proportionality.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 1
If y varies directly as x, find the constant of variation k
and the direct variation equation, given that y = 5 when
x = 30.
y = kx
The constant of variation is 1/6.
1
5 = k·30
The direct variation equation is y  x.
6
k = 1/6
If y varies directly as x, and y = 48 when x = 6, then find
y when x = 15.
y = kx
So the equation is y = 8x.
48 = k·6
y = 8·15
8=k
y = 120
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 2
At sea, the distance to the horizon is directly
proportional to the square root of the elevation of the
observer. If a person who is 36 feet above water can see
7.4 miles, find how far a person 64 feet above the water
can see. Round your answer to two decimal places.
d k e
7.4  k 36
7.4  6k
7 .4
k
6
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
7 .4
So our equation is d 
e
6
7 .4
d
64
6
7.4
59.2
d
(8) 
 9.87 miles
6
6
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 3
If y varies inversely as x, find the constant of variation k
and the inverse variation equation, given that y = 63
when x = 3.
k
y
x
k
63 
3
k = 63·3
k = 189
189
.
The inverse variation equation is y 
x
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Joint Variation
If the ratio of a variable y to the product of two or
more variables is constant, then y varies jointly as, or
is jointly proportional to the other variables. If
y = kxz
then the number k is the constant or variation or the
constant or proportionality.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 4
The maximum weight that a circular column can hold is
inversely proportional to the square of its height.
If an 8-foot column can hold 2 tons, find how much
weight a 10-foot column can hold.
k
w 2
h
k
2 2
8
k
2
64
k  128
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128
So our equation is w  2 .
h
128 128
w 2 
 1.28 tons
10
100
7.7 summary
Objectives:
Solve problems involving direct, inverse, joint, and combined variations.
Directly
y = kx
Inversely
y = k/x
Jointly
y = kxz
Do the following now:
7.7 #1, 4, 6, 10, 12,13
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