LESSON
9.3
Name
Solving Rational
Equations
9.3
Solve simple rational and radical equations in one variable, and give
examples showing how extraneous solutions may arise. Also A-CED.1,
A-CED.3, A-REI.1
Solving Rational Equations Graphically
x = 2 by graphing.
Solve the rational equation _
x- 3
A
Mathematical Practices
MP.3 Logic
Language Objective
Describe to a partner how to solve rational equations.
B
First, identify any excluded values.
A number is an excluded value of a
rational expression if substituting the
number into the expression results in
a division by 0, which is undefined.
Solve x - 3 = 0 for x.
So, 3 is an excluded value of the
rational equation. Rewrite the
equation with 0 on one side.
x =2
_
x- 3
x
____
-2
x=
C
© Houghton Mifflin Harcourt Publishing Company
Essential Question: What methods are
there for solving rational equations?
E
3
D
Graph the left side of the equation
as a function. Substitute y for 0 and
complete the table below.
x
y
0
-2
1
-2.5
2
-4
4
2
5
0.5
9
-0.5
(x, y)
y
4
(0, -2)
(1, -2.5)
(2, -4)
(4, 2)
(5, 0.5)
(9, -0.5)
x
-8
-4
0
-4
4
8
-8
F
Identify any x-intercepts of the graph.
There is an x-intercept at
Use the table to graph the
function.
8
Is the value of x an excluded
value? What is the solution
x = 2?
of _
x- 3
No, x = 6 is not an excluded value.
(6, 0) .
x
The solution of ____
= 2 is x = 6.
x- 3
Module 9
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 3
453
gh “File info”
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Essential
HARDCOVER PAGES 325332
Turn to these pages to
find this lesson in the
hardcover student
edition.

with 0 on
equation
x =2
_
x- 3
ed values
fy any exclud
a
First, identi
ed value of
is an exclud
tuting the
A number
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in
rational expres expression results
the
number into which is undefined.
by 0,
a division
x.
3 = 0 for
Solve x -

x -2
____
3
x-
=0
0
x- 3=
x=

3

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the equati
left side of
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-2.5
1
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to graph
8
the
y
4
(0, -2)
(1, -2.5)
(2, -4)
(4, 2)
(5, 0.5)
-0.5
9
Identify any

Use the table
function.
(x, y)
y
x
y
g Compan
A2_MNLESE385894_U4M09L3.indd 453
-8 -4
x
0
4
8
-4
-8
(9, -0.5)
.
the graph

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of x an exclud
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Is the value
is the solutio
value? What
x = 2?
of _
value.
x- 3
excluded
is not an
No, x = 6
x = 2 is x = 6.
____
ion of x - 3
The solut
Lesson 3
453
Module 9
L3.indd
4_U4M09
SE38589
A2_MNLE
Lesson 9.3
=0
x-3
x- 3=0
ENGAGE
453
Resource
Locker
A rational equation is an equation that contains one or more rational expressions. The
d
time t in hours it takes to travel d miles can be determined by using the equation t = _
r ,
where r is the average rate of speed. This equation is an example of a rational equation.
One method to solving rational equations is by graphing.
A-REI.2
View the Engage section online. Discuss the photo
and ask students to list the information they would
need to calculate the profit per item. Then preview
the Lesson Performance Task.
Solving Rational Equations
Explore
The student is expected to:
PREVIEW: LESSON
PERFORMANCE TASK
Date
Essential Question: What methods are there for solving rational equations?
Common Core Math Standards
Rational equations can be solved algebraically by
finding the LCM of the denominators of the rational
expressions and multiplying each side of the
equation by that LCM. When the equation is
simplified, the result is a polynomial equation that
can be solved by factoring, graphing, and other
methods. Rational equations also can be solved by
graphing. For example, each side of a rational
equation may be entered as a function. The solution
is the x-coordinate(s) of the point(s) at which the
graphs intersect.
Class
453
27/03/14
8:04 AM
27/03/14 8:04 AM
Reflect
1.
EXPLORE
Discussion Why does rewriting a rational equation with 0 on one side help with
solving the equation?
Rewriting the equation with 0 on one side helps because the expression on the other side
Solving Rational Equations
Graphically
can be graphed and the solution is the x-intercept.
INTEGRATE TECHNOLOGY
Explain 1
Students have the option of completing the Explore
activity either in the book or online.
Solving Rational Equations Algebraically
Rational equations can be solved algebraically by multiplying through by the LCD and solving
the resulting polynomial equation. However, this eliminates the information about the excluded
values of the original equation. Sometimes an excluded value of the original equation is a solution
of the polynomial equation, and in this case the excluded value will be an extraneous solution
of the polynomial equation. Extraneous solutions are not solutions of an equation.
Example 1

INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Students should recognize that the
Solve each rational equation algebraically.
3x + 7
5x + 17
_
=_
x- 5
2x - 10
Identify any excluded values.
x- 5=0
x=5
x-intercepts represent the solutions of the equation
because they are the zeros of the related function.
Help them to make this connection, emphasizing that
a point with coordinates (x, 0) represents a value of x
for which the function is equal to 0.
2x - 10 = 0
x=5
The excluded value is 5.
Identify the LCD by finding all factors of the denominators.
2x - 10 = 2(x - 5)
The different factors are 2 and x - 5.
The LCD is 2(x - 5).
Multiply each term by the LCD.
3x + 7
5x + 17
_
∙ 2 (x - 5) = _ ∙ 2 (x - 5)
x- 5
2 (x - 5)
(3x + 7)2 = 5x + 17
Simplify.
6x + 14 = 5x + 17
Use the Distributive Property.
x + 14 = 17
Solve for x.
QUESTIONING STRATEGIES
© Houghton Mifflin Harcourt Publishing Company
Divide out common factors.
3x + 7
5x + 17
_
· 2(x - 5) = _ ∙ 2(x - 5)
x- 5
2(x - 5)
x=3
If the two sides of the equation are graphed as
separate functions, will the graphs intersect,
and if so, where? Yes; the graphs will intersect at the
point or points whose x-values are the solutions of
the equation.
The solution x = 3 is not an excluded value. So, x = 3 is the solution of the equation.
EXPLAIN 1
Solving Rational Equations
Algebraically
Module 9
454
Lesson 3
QUESTIONING STRATEGIES
PROFESSIONAL DEVELOPMENT
A2_MNLESE385894_U4M09L3.indd 454
Integrate Mathematical Practices
This lesson provides an opportunity to address Mathematical Practice MP.3,
which calls for students to “construct viable arguments.” Students learn that they
can solve rational equations graphically, by writing a related function and finding
the zeros of the function. They also learn to solve rational equations algebraically,
multiplying the equation by the LCD and converting it into an equivalent
polynomial equation. Students draw connections between the excluded values of
the expressions and the extraneous solutions of the equation.
27/03/14 7:54 AM
How does finding the LCD of the rational
expressions and multiplying each side of the
equation by the LCD turn the rational equation into a
polynomial equation? After the multiplication of
both sides by the LCD has been carried out,
common factors in the numerators and
denominators can be divided out. The resulting
denominators on each side are 1, and since the
numerators are polynomials, the equation becomes
a polynomial equation.
Solving Rational Equations
454
B
CONNECT VOCABULARY
2x - 9 + _
5
x =_
_
x-7
x-7
2
Identify any excluded values.
Have students look up the definition of extraneous
and relate its use in this context to its use in
non-mathematical contexts.
x- 7=0
x= 7
7
The excluded value is
.
Identify the LCD.
QUESTIONING STRATEGIES
The different factors are
How can you tell whether a solution of a
rational equation is extraneous? It is
extraneous if it is an excluded value for one of the
rational expressions in the equation.
The LCD is
2(x - 7)
2 and x - 7
.
.
Multiply each term by the LCD.
2x - 9 ∙
_
x-7
2(x - 7)
x∙
+_
2
2(x - 7)
5 ∙
=_
x-7
2(x - 7)
Divide out common factors.
2x - 9 ∙
_
x-7
2(x - 7)
x∙
+_
2
2(x - 7)
5 ∙
=_
x-7
2(x - 7)
(
2 (2x - 9)+ x
Simplify.
INTEGRATE TECHNOLOGY
Use the Distributive Property.
Students can use the graphing method
presented in the Explore to check their
solutions. Have them graph the corresponding
function and compare its zeros to the solutions found
algebraically.
Write in standard form.
4x - 18
x-7
+ x2 - 7x = 10
x - 3x - 28
2
(
Factor.
Use the Zero Product Property.
x-7
)(
) = 5( 2 )
x+4
=0
)=0
x - 7 = 0 or
x+4
=0
x = 7 or x = -4
Solve for x.
The solution x = 7 is extraneous because it is an excluded value. The only solution is x = -4 .
Your Turn
© Houghton Mifflin Harcourt Publishing Company
Solve each rational equation algebraically.
2.
x+ 1
8 =_
_
x+3
x+6
Excluded values: -3 and -6
LCD: (x + 3) (x + 6)
x+1
8
_
⋅ (x + 3)(x + 6) = _ ⋅ (x + 3)(x + 6)
x+6
x+3
x+1
8
_
_
⋅ (x + 3)(x + 6)
⋅ (x + 3)(x + 6) =
x+6
x+3
8 (x + 6) = (x + 1)(x + 3)
8x + 48 = x 2 + 4x + 3
0 = x 2 - 4x - 45
0 = (x - 9)(x + 5)
x = 9 or x = -5
The solutions are x = 9 or x = -5.
Module 9
455
Lesson 3
COLLABORATIVE LEARNING
A2_MNLESE385894_U4M09L3 455
Small Group Activity
Have students work in groups of 3–4 students. Provide each group with a different
rational equation to solve. Instruct students to create a poster showing the
equation solved using three different methods: graphically using one function,
graphically using two functions, and algebraically. Remind students to look for
and to indicate extraneous solutions. Have students share their posters with
the class.
455
Lesson 9.3
7/7/14 9:35 AM
Explain 2
Solving a Real-world Problem with a Rational Equation
EXPLAIN 2
Rational equations are used to model real-world situations. These equations can be solved algebraically.
Example 2

Use a rational equation to solve the problem.
Solving a Real-World Problem with a
Rational Equation
Kelsey is kayaking on a river. She travels 5 miles upstream
and 5 miles downstream in a total of 6 hours. In still water,
Kelsey can travel at an average speed of 3 miles per hour.
What is the average speed of the river’s current?
AVOID COMMON ERRORS
Analyze Information
Students may need to be reminded to check to see
whether the solutions of their equations are
extraneous solutions. Remind them to also consider
restrictions that are imposed on the variable due to
the context of the application.
Identify the important information:
• The answer will be the average speed
of
the current
• Kelsey spends
• She travels
.
6 hours
5 miles
• Her average speed in still water is
kayaking.
upstream and
5 miles
downstream.
3 miles per hour .
QUESTIONING STRATEGIES
• Formulate a Plan
How could you solve the problem
graphically? Graph each side of the equation
as a function and find the points of intersection of
the two graphs. The x-values of the points of
intersection will be the solutions of the equation.
Let c represent the speed of the current in miles per hour. When Kelsey is going
upstream, her speed is equal to her speed in still water minus c. When Kelsey is
© Houghton Mifflin Harcourt Publishing Company • Image Credit: ©Robert
Michael/Corbis
going downstream, her speed is equal to her speed in still water plus c.
positive real numbers
The variable c is restricted to
.
Complete the table.
Distance (mi)
Average speed (mi/h)
Upstream
5
3-c
Downstream
5
3+c
Time (h)
5
3-c
5
3+c
_
_
Use the results from the table to write an equation.
total time = time upstream + time downstream
5
5
_
_
6 =
+
3-c
3+c
Module 9
456
Lesson 3
DIFFERENTIATE INSTRUCTION
A2_MNLESE385894_U4M09L3.indd 456
Multiple Representations
27/03/14 7:55 AM
Point out to students that they can use the table feature of a graphing calculator to
check their solutions to rational equations. They can enter the left side of the
equation as Y1 and the right side as Y2. Then, by using the table feature, they can
verify that Y1 and Y2 have the same values of x that they determined by
using algebra.
Solving Rational Equations
456
Solve
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Discuss with students how the rational
3-c=0
3+c=0
3 =c
c = -3
3 and -3
Excluded values:
LCD: (3 - c)(3 + c)
expressions used in the example model the situation.
Have students describe what each numerator and
denominator represents, and why the equation
models the real-world relationship described in
the problem.
Divide out common
factors.
5 ∙ (
5 ∙ (
6 ∙ (3 - c)(3 + c) = _
3 - c)(3 + c) + _
3 - c)(3 + c)
3-c
3+c
5 ∙ (
5 ∙ (3 - c)(3 + c)
6 ∙ (3 - c)(3 + c) = _
3 - c)(3 + c) + _
3-c
3+c
Simplify.
6 ∙ (3 - c)(3 + c) = 5 ∙
Multiply by the LCD.
54 - 6c
Use the Distributive
Property.
2
3+c
6 c 2 - 24
0=
Factor.
0 = 6 (c + 2)
c + 2 = 0 or
Use the Zero Product
Property.
Solve for c.
15 - 5c
= 15 + 5c +
Write in standard form.
3-c
+5∙
(
c-2
c-2
=0
)
c = -2 or c = 2
are
no
There
extraneous solutions. The solutions are c = -2 or c = 2 .
Justify and Evaluate
The solution c = -2 is unreasonable because the speed of the current cannot
negative, but the solution c = 2 is reasonable because the speed of the
current can be positive . If the speed of the current is 2 miles per hour , it
© Houghton Mifflin Harcourt Publishing Company
be
would take Kelsey
hour(s) to go upstream and
downstream, which is a total of 6
hours.
3.
hour(s) to go
Why does the domain of the variable have to be restricted in real-world problems that can be modeled with
a rational equation?
The variable must make sense in a real-world context. The speed of the current cannot be
negative or 0, so the domain of c had to be restricted to positive real numbers.
A2_MNLESE385900_U4M09L3.indd 457
Lesson 9.3
1
Reflect
Module 9
457
5
457
Lesson 3
19/03/14 2:17 PM
Your Turn
4.
ELABORATE
Kevin can clean a large aquarium tank in about
7 hours. When Kevin and Lara work together,
they can clean the tank in 4 hours. Write and
solve a rational equation to determine how long,
to the nearest tenth of an hour, it would take
Lara to clean the tank if she works by herself.
Explain whether the answer is reasonable.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 Discuss with students how the algebraic
1
Kevin’s rate: _
of the tank per hour
7
1
Lara’s rate: of the tank per hour,
t
where t is the time in hours needed to
_
process of converting a rational equation into a
polynomial equation can introduce extraneous
solutions. Lead them to recognize that polynomial
expressions themselves do not have excluded values
and, therefore, the solutions to a polynomial equation
will all satisfy the polynomial equation. However, if
the polynomial equation is derived from a rational
equation, any excluded values of the rational equation
that are solutions of the polynomial equation will be
extraneous.
clean the tank alone.
The domain of t must be positive real numbers.
Kevin’s rate ⋅ hours worked + Lara’s rate ⋅ hours worked = 1 complete job
_1 (4) + _1 (4) = 1
t
7
Excluded value: 0; LCD: 7t
_1 (4) + _1 (4) = 1
t
7
_1 (4) ∙ 7 ∙ t + _1 (4) ∙ 7 ∙ t = 1 ∙ 7 ∙ t
t
4t + 1 ∙ 4 ∙ 7 = 1 ∙ 7 ∙ t
4t + 28 = 7t
28 = 3t
1
9 =t
3
It would take Lara about 9 hours and 20 minutes to clean the tank by herself. The answer is
1
1
1 ( )
reasonable because 9 is positive and (4) +
4 = 1.
1
7
3
9
3
7
_
_
_
_
Elaborate
5.
Why is it important to check solutions to rational equations?
To make sure that there are no extraneous solutions.
6.
Why can extraneous solutions to rational equations exist?
Because the process of multiplying through by the LCD may eliminate the information
about excluded values.
7.
SUMMARIZE THE LESSON
© Houghton Mifflin Harcourt Publishing Company • Image Credit: ©Atlantide
Phototravel/Corbis
_
How can you use the LCD of the rational
expressions in an equation to solve the
equation? You can multiply each term in the
equation by the LCD. This will change the rational
equation into a polynomial equation. You can then
solve the polynomial equation, being careful to
check whether any of the solutions are extraneous.
Essential Question Check In How can you solve a rational equation without graphing?
Rational equations can be solved algebraically by finding the LCM of the denominators of
the rational expressions and multiplying each side of the equation by that LCM. When the
equation is simplified, the result is a polynomial equation that can be solved.
Module 9
458
Lesson 3
LANGUAGE SUPPORT
A2_MNLESE385894_U4M09L3 458
7/7/14 9:36 AM
Communicate Math
Have students work in pairs to complete a chart shown on solving rational
equations.
Type of Equation
Rational: Write an
equation
Rational: Write an
equation
Type of
Solution
Graphical
Algebraic
Notes Explaining Steps
For example, rewrite with 0 on one side
and graph the function on the other side.
For example, find the LCD by
factoring, etc.
Solving Rational Equations
458
EVALUATE
Evaluate: Homework and Practice
• Online Homework
• Hints and Help
• Extra Practice
Solve each rational equation by graphing using a table of values.
1.
x
_____
= -3
x+4
x
Excluded value: x = -4
x
____
= -3
x+4
x
____
+3= 0
ASSIGNMENT GUIDE
Concept and Skills
x+4
Explore
Solving Rational Equations
Graphically
Exercises 1–2
Example 1
Solving Rational Equations
Algebraically
Exercises 3–8
Example 2
Solving a Real-World Problem with a
Rational Equation
-8
5
-6
6
-5
8
-3.5
-2
Practice
0
-4
2
3
(x, y)
8
(-8, 5)
(-6, 6)
(-5, 8)
2.
x
_
=3
2x - 10
x
0
x
_____
=3
2x - 10
x
_____
-3= 0
Exercises 9–16
3
4
2x - 10
5.5
7
10
AVOID COMMON ERRORS
4
x
-8
(-3.5,-4)
(-2, 2)
(0, 3 )
0
-4
4
8
4
8
-8
8
(0,-3)
-3
y
4
-3.75 (3, -3.75)
x
(4, -5)
(5.5, 2.5)
2.5
-1.25 (7, -1.25)
-2
(10, -2)
-5
-8
-4
0
-4
-8
© Houghton Mifflin Harcourt Publishing Company
The x-intercept is at (6, 0). The solution is x = 6.
Solve each rational equation algebraically.
9 -_
13
5 = -_
3. _
4x
12x
6
4.
Excluded value: x = –1
Excluded value: x = 0
LCD: 7(x + 1)
3
2
∙ 7(x + 1) + ∙ 7(x + 1)
x+1
7
= 2 ∙ 7(x + 1)
LCD: 2 ⋅ 2 ⋅ 3 ⋅ x = 12x
9
5
13
⋅ 12x - ⋅ 12x = ⋅ 12x
4x
12x
6
9 ⋅ 3 - 5 ⋅ 2x = -13
_
_
3 +_
2 =2
_
7
x+1
_
_
_
3 ∙ 7 + 2(x + 1) = 2 ∙ 7(x + 1)
27 - 10x = -13
- 10x = -40
21 + 2x + 2 = 14x + 14
-12 x + 9 = 0
The solution is x = 4.
Module 9
Exercise
A2_MNLESE385894_U4M09L3.indd 459
Lesson 9.3
-4
(x, y)
y
x=4
459
y
The x-intercept is at (-3, 0). The solution is x = -3.
Excluded value: x = 5
If students do not obtain an appropriate graph, they
may have made errors either in entering the function
(or functions) or in choosing an appropriate viewing
window. The most common error in entering a
rational function is to omit the parentheses around
the complete numerator and the complete
denominator. Check students’ calculators to catch
errors and guide them in correcting any errors.
y
-12 x = -9
3
x=
4
_
Lesson 3
459
Depth of Knowledge (D.O.K.)
Mathematical Practices
1–2
2 Skills/Concepts
MP.6 Precision
3–8
1 Recall of Information
MP.2 Reasoning
9–16
2 Skills/Concepts
MP.4 Modeling
17
1 Recall of Information
MP.2 Reasoning
18
3 Strategic Thinking
MP.2 Reasoning
19
3 Strategic Thinking
MP.6 Precision
27/03/14 7:55 AM
5.
7
6 =_
56
__
−_
x−5
x+3
x 2 − 2x − 15
CRITICAL THINKING
x - 2 x - 15 = 0
2
Ask students to consider whether it is possible to
solve a rational equation by multiplying each of its
terms by a common denominator that is not the least
common denominator of the rational expressions.
Have them predict how the result of doing this would
compare to using the LCD, and test their predictions
using one or more of the exercises.
(x - 5)(x + 3) = 0
x - 5 = 0 or (x + 3) = 0
x = 5 or x = -3
Excluded values: 5 and -3
LCD: (x - 5)(x + 3)
6
56
7
__
∙ (x - 5)(x + 3) - _ ∙ (x - 5)(x + 3) = _ ∙ (x - 5)(x + 3)
x+3
(x - 5)(x + 3)
x-5
56 - 6(x -5) = 7(x + 3)
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Students can check their solutions by
56 - 6x + 30 = 7x + 21
65 = 13x
5=x
The solution x = 5 is extraneous. There is no solution.
6.
substituting the values into the original equations and
verifying that they make the equation true. Students
should observe that any extraneous solutions would
make one or more of the rational expressions
undefined.
x − 29
6 +_
5
__
=_
x−7
x−3
x 2 − 10x + 21
2
x 2 - 10x + 21 = 0
(x - 7)(x - 3) = 0
x - 7 = 0 or x - 3 = 0
x = 7 or x = 3
Excluded values: 7 and 3
__
_
_
__
_
_
© Houghton Mifflin Harcourt Publishing Company
LCD: (x - 7)(x - 3)
6
5
x 2 - 29
∙ (x - 7)(x - 3) =
∙ (x - 7)(x - 3) +
∙ (x - 7)(x - 3)
x-7
x-3
(x - 7)(x - 3)
6
5
x 2 - 29
∙ (x - 7)(x - 3) +
∙ (x - 7)(x - 3)
∙ (x - 7)(x - 3) =
x-7
x-3
(x - 7)(x - 3)
x 2 - 29 = 6(x - 3) + 5(x - 7)
x 2 - 29 = 6x - 18 + 5x - 35
x 2 - 29 = 11x - 53
x 2 - 11x + 24 = 0
(x - 3) (x - 8) = 0
x - 3 = 0 or x - 8 = 0
x = 3 or x = 8
The solution x = 8. The solution x = 3 is extraneous.
Module 9
Lesson 3
460
Exercise
A2_MNLESE385894_U4M09L3 460
Depth of Knowledge (D.O.K.)
Mathematical Practices
20
3 Strategic Thinking
MP.3 Logic
21
3 Strategic Thinking
MP.5 Using Tools
16/10/14 2:20 PM
Solving Rational Equations
460
QUESTIONING STRATEGIES
7.
5
1 =_
2
_
−_
x+4
2x + 6
6
Excluded values: -3 and -4
How is the LCD of two rational expressions
related to each denominator? It is a multiple
of each of the denominators.
LCD: 2 ∙ 3 ∙ (x + 3)(x + 4)
5
1
2
_
∙ 2 ∙ 3 ∙ (x + 3)(x + 4) - _ ∙ 2 ∙ 3 ∙ (x + 3)(x + 4) = _ ∙ 2 ∙ 3 ∙ (x + 3)(x + 4)
x+4
2∙3
2(x + 3)
5
1
2
_
∙ 2 ∙ 3 ∙ (x + 3)(x + 4) - _ ∙ 2 ∙ 3 ∙ (x + 3)(x + 4) = _ ∙ 2 ∙ 3 ∙ (x + 3)(x + 4)
When is the LCD of two rational expressions
the product of the denominators? When the
denominators have no common factors.
x+4
2∙3
2(x + 3)
5 ∙ 3 ∙ (x + 4) -1 ∙ (x + 3)(x + 4) = 2 ∙ 2 ∙ 3 ∙ (x + 3)
15x + 60 - x 2 - 7x - 12 = 12x + 36
-x 2 + 8x + 48 = 12x + 36
0 = x 2 + 4x - 12
0 = (x - 2)(x + 6)
x = 2 or x = -6
The solution is x = 2 or x = -6.
8.
x+6
5
1 =_
_
−_
x−2
3x − 3
x 2 − 3x + 2
Excluded values: 1 and 2
LCD: 3(x - 1)(x - 2)
x+6
5
1
∙ 3(x - 1)(x - 2) =
∙ 3(x - 1)(x - 2) ∙ 3(x - 1)(x - 2)
x-2
(x - 1)(x - 2)
3(x - 1)
x+6
5
1
∙ 3(x - 1)(x - 2) =
∙ 3(x - 1)(x - 2) ∙ 3(x - 1)(x - 2)
x-2
(x - 1)(x - 2)
3(x - 1)
_
__
_
_
© Houghton Mifflin Harcourt Publishing Company
15 - 3x + 3 = x 2 + 4x - 12
0 = x 2 + 7x - 30
0 = (x + 10)(x - 3)
x = -10 or x = 3
The solution is x = -10 or x = 3.
A2_MNLESE385894_U4M09L3 461
Lesson 9.3
_
5 ∙ 3 - 1 ∙ 3(x - 1) = (x + 6)(x - 2)
Module 9
461
__
461
Lesson 3
16/10/14 2:24 PM
For 15 and 16, write a rational equation for each real-world
application. Do not solve.
9.
COMMUNICATING MATH
Call upon students to present their solutions to the
class. Ask them to describe how they determined the
LCD and how they transformed the rational equation
into a polynomial equation. Have them describe the
steps of the solution process, and tell how they
identified any extraneous solutions.
A save percentage in lacrosse is found by dividing the number of saves
by the number of shots faced. A lacrosse goalie saved 9 of 12 shots. How
many additional consecutive saves s must the goalie make to raise his
save percentage to 0.850?
Number of saves: 9 + s
Number of shots: 12 + s
9+s
_
= 0.850
12 + s
10. Jake can mulch a garden in 30 minutes. Together, Jake and Ross can
mulch the same garden in 16 minutes. How much time t, in minutes, will
it take Ross to mulch the garden when working alone?
1 (16)
Jake’s rate ⋅ minutes worked = _
30
1 (16)
Ross’ rate ⋅ minutes worked = _
t
1 (16) + _
1 (16) = 1
_
t
30
11. Geometry A new ice skating rink will be approximately rectangular in shape and
will have an area of 18,000 square feet. Using an equation for the perimeter P, of the
skating rink in terms of its width W, what are the dimensions of the skating rink if
the perimeter is 580 feet?
(
© Houghton Mifflin Harcourt Publishing Company • Image Credit: ©Don Kelly
Photo/Corbis
L ⋅ W = 18,000
18,000
L=_
W
P = 2(L + W)
18,000
580 = 2 _ + W
W
580W = 2(18,000 + W 2)
)
290W = 18,000 + W 2
0 = W 2 - 290W + 18,000
W = 200 or W = 90
The dimensions are 200 feet by 90 feet.
12. Water flowing through both a small pipe and a large pipe can fill a water tank in
9 hours. Water flowing through the large pipe alone can fill the tank in 17 hours.
Write an equation that can be used to find the amount of time t, in hours, it would
take to fill the tank using only the small pipe.
1 ( )
large pipe’s rate ⋅ hours worked =
9
17
1
Small pipe’s rate ⋅ hours worked = _(9)
t
1 (9) + 1 (9) = 1
_
t
17
_
_
Module 9
A2_MNLESE385900_U4M09L3.indd 462
462
Lesson 3
19/03/14 2:15 PM
Solving Rational Equations
462
13. A riverboat travels at an average of 14 km per hour in still
water. The riverboat travels 110 km east up the Ohio River and
110 km west down the same river in a total of 17.5 hours. To
the nearest tenth of a kilometer per hour, what was the speed of
the current of the river?
AVOID COMMON ERRORS
When solving a rational equation that models a
real-world situation, students may erroneously
believe that any negative solution must be an
extraneous solution. Help them to see that this is not
necessarily the case, and remind them to reflect upon
the rational expressions contained in the original
equation and their excluded values.
The variable c is the speed of the current in km per hour.
110
Time upstream: _
14 - c
110
Time downstream: _
14 + c
110
110
_
+ _ = 17.5
14 - c 14 + c
Excluded values: -14, 14
LCD: (14 - c)(14 + c)
110
110 (
_
14 - c)(14 + c) + _ (14 - c) (14 + c) = 17.5(14 - c)(14 + c)
14 + c
14 - c
110(14 + c) + 110(14 - c) = 17.5(14 - c)(14 + c)
3080 = 3430 - 17.5c 2
-350 = -17.5c 2
20 = c 2
±4.5 ≈ c
© Houghton Mifflin Harcourt Publishing Company • Image Credit: ©Wm. B
Baker/GhostWorx Images/Alamy
The solution c = -4.5 does not make sense in context because the speed
cannot be negative. So, the speed of the current is about 4.5 km/h.
14. A baseball player’s batting average is equal to the number of hits divided by the
number of at bats. A professional player had 139 hits in 515 at bats in 2012 and
167 hits in 584 at bats in 2013. Write and solve an equation to find how many
additional consecutive hits h the batter would have needed to raise his batting average
in 2012 to be at least equal to his average in 2013.
2012: ______
515 + h
139 + h
167
2013: ___
587
139
+h
167
______
= ___
584
515 + h
Excluded value: -515
LCD: 584(515 + h)
A2_MNLESE385894_U4M09L3.indd 463
Lesson 9.3
139
+h
167
______
⋅ 584(515 + h) = ___
⋅ 584(515 + h)
584
515 + h
(139 + h)584 = 167(515 + h)
81,176 + 584h = 86,005 + 167h
417h = 4829
h ≈ 11.58
The number of hits must be a whole number. The player would need
12 consecutive hits to raise his batting average in 2012 to be at least
equal to his average in 2013.
Module 9
463
139
+h
167
______
⋅ 584(515 + h) = ___
⋅ 584(515 + h)
584
515 + h
463
Lesson 3
27/03/14 7:55 AM
15. The time required to deliver and install a computer network at a customer’s location
2d
is t = 5 + __
r , where t is time in hours, d is the distance (in miles) from the warehouse
to the customer’s location, and r is the average speed of the delivery truck. If it takes
8.2 hours for an employee to deliver and install a network for a customer located
80 miles from the warehouse, what is the average speed of the delivery truck?
PEER-TO-PEER DISCUSSION
Ask students to discuss with a partner how solving a
proportion by cross-multiplying is similar to solving a
rational equation by multiplying through by the LCD.
Students should recognize that cross-multiplying
produces the same result as if they had multiplied
both sides of the equation by the product of the
denominators. The product of the denominators may
not be the least common denominator, but it will
transform the rational equation into a polynomial
equation.
2d
t = 5 + __
r
Excluded value: r ≠ 0
8.2 = 5 + ____
r
2(80)
LCD: r
8.2 ∙ r = 5 ∙ r + ____
r ∙r
2(80)
8.2r = 5r + 160
3.2r = 160
r = 50
The average speed of the truck is 50 mph.
16. Art A glassblower can produce several sets of simple glasses in about
3 hours. When the glassblower works with an apprentice, the job takes about
2 hours. How long would it take the apprentice to make the same number of
sets of glasses when working alone?
The variable t is time in hours it would take for the apprentice.
_1(2) + _1(2) = 1
t
3
Excluded value: 0
LCD: 3t
_1(2) + _1(2) = 1
t
3
_1(2) ∙ 3t + _1(2) ∙ 3t = 1 ∙ 3t
3
t
2t + 2 ∙ 3 = 3t
6=t
© Houghton Mifflin Harcourt Publishing Company • Image Credit: ©Picavet/
Getty Images
It would take the apprentice about 6 hours.
17. Which of the following equations have at least two excluded values? Select
all that apply.
3 +_
1 =1
excluded value: 0
A. _
x
5x
3 =_
5
x- 4 +_
excluded values: 0 and 2
B. _
x
6
x-2
5
x +1=_
excluded value: 6
C. _
2x - 12
x-6
2x - 3
3 =_
1
excluded value: 5
D. __
+_
7
x-5
x 2 - 10x + 25
3x - 4 = 9
7 +_
excluded values: -2 and -3
E. _
x+2
x 2 + 5x + 6
​
Module 9
A2_MNLESE385894_U4M09L3 464
464
Lesson 3
09/07/14 12:51 AM
Solving Rational Equations
464
JOURNAL
H.O.T. Focus on Higher Order Thinking
18. Critical Thinking An equation has the form __ax + __bx = c, where a, b, and c are
constants and b ≠ 0. How many solutions could this equation have? Explain.
Have students make a list of the steps involved in
solving a rational equation algebraically.
0, 1, or 2; the equation becomes x 2 - bcx + ab = 0 after multiplying
by the LCD and putting the equation in standard form. A quadratic
equation can have 0, 1, or 2 solutions. The number of solutions of
this equation depends on the values of a, b, and c.
19. Multiple Representations Write an equation whose graph is a straight line, but
with an open circle at x = 4.
(x - 4)(x - 3)
x 2 - 7x + 12
= ________
has the
Possible answer: The graph of y = _________
x-4
x-4
same graph as y = x - 3 but has an open circle at x = 4.
20. Justify Reasoning Explain why the excluded values do not change when
multiplying by the LCD to add or subtract rational expressions.
The function was defined not to exist at those points, so if an alternate form of the
equation is to be equivalent, the new form must not exist at those points either.
21. Critical Thinking Describe how you would find the inverse of the rational function
x - 1 , x ≠ 2. Then find the inverse.
ƒ(x) = _
x-2
Substitute y for x in the expression, and substitute x for f(x). Then solve for y.
y-1
x - 1 ⎯→ x = _
y=_
y-2
x-2
x(y - 2) = (y - 1)
Multiply.
y x - 2x = y - 1
© Houghton Mifflin Harcourt Publishing Company
y x - y = 2x - 1
_
A2_MNLESE385894_U4M09L3 465
Lesson 9.3
Collect y terms.
y(x - 1) = 2x - 1
Factor.
2x - 1
y=
Divide.
x-1
2x
- 1 , x ≠ 1.
The inverse function f -1(x) is f -1(x) = _
x-1
Module 9
465
Distribute.
465
Lesson 3
7/7/14 9:45 AM
Lesson Performance Task
CONNECT VOCABULARY
Students may not understand the concept of profit.
Explain that the money you receive when selling an
item is called the sales income, but this is not the same
thing as profit. To calculate the profit, first find how
much it costs to produce the item. The profit is the
money left over after subtracting the costs from
the income.
Kasey creates comedy sketch videos and posts them on a popular video website and is selling
an exclusive series of sketches on DVD. The total cost to make the series of sketches is $989.
The materials cost $1.40 per DVD and the shipping costs $2.00 per DVD. Kasey plans to sell
the DVDs for $12 each.
a. Let d be the number of DVDs Kasey sells. Create a profit-peritem model from the given information by writing a rule for
C(d), the total costs in dollars, S(d), the total sales income in
dollars, P(d), the profit in dollars, and P PI(d), the profit per
item sold in dollars.
b. What is the profit per DVD if Kasey sells 80 DVDs? Does this
value make sense in the context of the problem?
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Reasoning
MP.2 Have students graph the profit per item P PI
c. Then use the function P PI(d) from part a to find how many
DVDs Kasey would have to sell to break even. Identify all
excluded values.
a. C(d) = 989 + 1.4d + 2d = 989 + 3.4d
S(d) = 12d
(d)for the number of DVDs sold, d. Have them
P(d) = 12d -( 989 +3.4d) = 8.6d - 989
discuss the maximum profit Kasey can make and
what limits the profit. Have students explain how the
profit per item changes as the number of items sold
increases. The profit will approach $8.60 per item,
but it will never reach it because the cost of
materials per DVD must be subtracted.
- 989
________
P PI(d) = 8.6d
d
8.6(80) - 989
(80)
b. P PI(80) = __________
-301
= ____
80
≈ -3.76
Kasey would have a profit of -$3.76 per DVD sold. This value makes
sense because the costs are greater than the amount of sales.
© Houghton Mifflin Harcourt Publishing Company
c. The excluded value is d = 0 .
- 989
_______
0 = 8.6d
d
- 989
_______
0 ∙ d = 8.6d
∙d
d
0 = 8.6d - 989
989 = 8.6d
115 = d
Kasey would have to sell 115 DVDs to break even.
Module 9
466
Lesson 3
EXTENSION ACTIVITY
A2_MNLESE385900_U4M09L3.indd 466
Have students consider starting a bakery business to sell bread at a farmer’s
market. Have students choose a price per loaf and estimate the number of loaves
they would have to sell in order to make a profit, assuming they start by buying a
50-pound bag of flour. Then have students create a model for the profit per loaf of
bread, taking into account all relevant costs. After inputting all known values,
have students discuss whether they would adjust the price or the sales goal in
order to make a profit.
19/03/14 2:13 PM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Solving Rational Equations
466