Transcript Step 2

Warm Up
Solve each proportion.
1.
10
2.
3.
4.
2.625
4.2
2.5
5. The value of y varies directly with x, and y = –6
when x = 3. Find y when x = –4. 8
6. The value of y varies directly with x, and y = 6 when
x = 30. Find y when x = 45. 9
Inverse
Variation
A relationship that can be written in the form y =
,
where k is a nonzero constant and x ≠ 0, is an
inverse variation. The constant k is the constant of
variation. Inverse variation implies that one quantity
will increase while the other quantity will decrease
(the inverse, or opposite, of increase).
Multiplying both sides of y = by x gives xy = k.
So, for any inverse variation, the product of x and y
is a nonzero constant.
Remember!
A direct variation is an equation that can be
written in the form y = kx, where k is a nonzero
constant.
There are two methods to determine whether a
relationship between data is an inverse variation.
You can write a function rule in y = form, or you
can check whether xy is a constant for each
ordered pair.
Example 1A: Identifying an Inverse Variation
Tell whether each relationship is an inverse
variation. Explain.
Method 1 Write a function rule.
Can write in y =
form.
The relationship is an inverse
variation.
Method 2 Find xy for each ordered pair.
1(30) = 30, 2(15) = 30, 3(10) = 30
The product xy is constant, so the relationship is
an inverse variation.
Example 1B: Identifying an Inverse Variation
Tell whether each relationship is an inverse
variation. Explain.
Method 1 Write a function rule.
y = 5x
Cannot write in y =
The relationship is not an
inverse variation.
Method 2 Find xy for each ordered pair.
1(5) = 5, 2(10) = 20, 4(20) = 80
The product xy is not constant, so the
relationship is not an inverse variation.
form.
Example 1C: Identifying an Inverse Variation
Tell whether each relationship is an inverse
variation. Explain.
2xy = 28
xy = 14
Find xy. Since xy is multiplied by 2,
divide both sides by 2 to undo the
multiplication.
Simplify.
xy equals the constant 14, so the relationship is
an inverse variation.
Check It Out! Example 1a
Tell whether each relationship is an inverse
variation. Explain.
Method 1 Write a function rule.
y = –2x
Cannot write in y =
The relationship is not an
inverse variation.
Method 2 Find xy for each ordered pair.
–12 (24) = –228 , 1(–2) = –2, 8(–16) = –128
The product xy is not constant, so the
relationship is not an inverse variation.
form.
Check It Out! Example 1b
Tell whether each relationship is an inverse
variation. Explain.
Method 1 Write a function rule.
Can write in y =
form.
The relationship is an inverse
variation.
Method 2 Find xy for each ordered pair.
3(3) = 9, 9(1) = 9, 18(0.5) = 9
The product xy is constant, so the relationship is
an inverse variation.
Check It Out! Example 1c
Tell whether each relationship is an inverse
variation. Explain.
2x + y = 10
Cannot write in y =
form.
The relationship is not an inverse variation.
Helpful Hint
Since k is a nonzero constant, xy ≠ 0. Therefore,
neither x nor y can equal 0, and no solution
points will be on the x- or y-axes.
An inverse variation can
also be identified by its
graph. Some inverse
variation graphs are
shown. Notice that each
graph has two parts that
are not connected.
Also notice that none of
the graphs contain (0, 0).
This is because (0, 0) can
never be a solution of an
inverse variation equation.
Example 2: Graphing an Inverse Variation
Write and graph the inverse variation in which
y = 0.5 when x = –12.
Step 1 Find k.
k = xy
= –12(0.5)
Write the rule for constant of variation.
Substitute –12 for x and 0.5 for y.
= –6
Step 2 Use the value of k to write an inverse
variation equation.
Write the rule for inverse variation.
Substitute –6 for k.
Example 2 Continued
Write and graph the inverse variation in which
y = 0.5 when x = –12.
Step 3 Use the equation to make a table of values.
x
–4
y
1.5
–2 –1
3
0
1
6 undef. –6
2
4
–3 –1.5
Example 2 Continued
Write and graph the inverse variation in which
y = 0.5 when x = –12.
Step 4 Plot the points and connect them with
smooth curves.
●
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●
●
●
●
Check It Out! Example 2
Write and graph the inverse variation in which
y = when x = 10.
Step 1 Find k.
k = xy
Write the rule for constant of variation.
= 10
Substitute 10 for x and
for y.
= 5
Step 2 Use the value of k to write an inverse
variation equation.
Write the rule for inverse variation.
Substitute 5 for k.
Check It Out! Example 2 Continued
Write and graph the inverse variation in which
y = when x = 10.
Step 3 Use the equation to make a table of values.
x
y
–4
–2
–1
0
1
2
4
–1.25
–2.5
–5
undef.
5
2.5
1.25
Check It Out! Example 2 Continued
Write and graph the inverse variation in which
y = when x = 10.
Step 4 Plot the points and connect them with
smooth curves.
●
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●
●
●
●
Example 3: Transportation Application
The inverse variation xy = 350 relates the
constant speed x in mi/h to the time y in
hours that it takes to travel 350 miles.
Determine a reasonable domain and range and
then graph this inverse variation. Use the
graph to estimate how long it will take to
travel 350 miles driving 55 mi/h.
Step 1 Solve the function for y so you can graph it.
xy = 350
Divide both sides by x.
Example 3 Continued
Step 2 Decide on a reasonable domain and range.
x>0
Length is never negative and x ≠ 0
y>0
Because x and xy are both positive, y is
also positive.
Step 3 Use values of the domain to generate
reasonable ordered pairs.
x
y
20
40
60
80
17.5 8.75 5.83 4.38
Example 3 Continued
Step 4 Plot the points. Connect them with a
smooth curve.
●
●
●
●
Step 5 Find the y-value where x = 55. When the
speed is 55 mi/h, the travel time is about 6 hours.
Remember!
Recall that sometimes domain and range are
restricted in real-world situations.
Check It Out! Example 3
The inverse variation xy = 100 represents the
relationship between the pressure x in
atmospheres (atm) and the volume y in mm³
of a certain gas. Determine a reasonable
domain and range and then graph this inverse
variation. Use the graph to estimate the
volume of the gas when the pressure is 40
atmospheric units.
Step 1 Solve the function for y so you can graph it.
xy = 100
Divide both sides by x.
Check It Out! Example 3 Continued
Step 2 Decide on a reasonable domain and range.
x>0
Pressure is never negative and x ≠ 0
y>0
Because x and xy are both positive, y is
also positive.
Step 3 Use values of the domain to generate
reasonable pairs.
x
y
10
10
20
5
30
3.34
40
2.5
Check It Out! Example 3 Continued
Step 4 Plot the points. Connect them with a
smooth curve.
●
●
●
●
Step 5 Find the y-value where x = 40. When the
pressure is 40 atm, the volume of gas is about 2.5
mm3.
The fact that xy = k is the same for every ordered
pair in any inverse variation can help you find
missing values in the relationship.
Example 4: Using the Product Rule
Let
as x. Find
and
Let y vary inversely
Write the Product Rule for Inverse
Variation.
Substitute 5 for
3 for
and 10 for
Simplify.
Solve for
Simplify.
by dividing both sides by 5.
.
Check It Out! Example 4
Let
and
inversely as x. Find
Let y vary
Write the Product Rule for Inverse
Variation.
Substitute 2 for
–4 for
and –6 for
Simplify.
Solve for
Simplify.
by dividing both sides by –4.
Example 5: Physics Application
Boyle’s law states that the pressure of a quantity
of gas x varies inversely as the volume of the gas
y. The volume of gas inside a container is 400 in3
and the pressure is 25 psi. What is the pressure
when the volume is compressed to 125 in3?
Use the Product Rule for Inverse
Variation.
(400)(25) = (125)y2 Substitute 400 for
for
Simplify.
Solve for
125 for
and 25
by dividing both sides by 125.
Example 5 Continued
Boyle’s law states that the pressure of a quantity
of gas x varies inversely as the volume of the gas
y. The volume of gas inside a container is 400 in3
and the pressure is 25 psi. What is the pressure
when the volume is compressed to 125 in3?
When the gas is compressed to 125 in3, the
pressure increases to 80 psi.
Check It Out! Example 5
On a balanced lever, weight varies inversely as
the distance from the fulcrum to the weight.
The diagram shows a balanced lever. How much
does the child weigh?
Check It Out! Example 5 Continued
Use the Product Rule for Inverse
Variation.
Substitute 3.2 for
for
, 60 for
and 4.3
Simplify.
Solve for
Simplify.
The child weighs 80.625 lb.
by dividing both sides by 3.2.
Lesson Quiz: Part I
1. Write and graph the inverse variation in which
y = 0.25 when x = 12.
Lesson Quiz: Part II
2. The inverse variation xy = 210 relates the length
y in cm to the width x in cm of a rectangle with
an area of 210 cm2. Determine a reasonable
domain and range and then graph this inverse
variation. Use the graph to estimate the length
when the width is 14 cm.
Lesson Quiz: Part III
3. Let x1= 12, y1 = –4, and y2 = 6, and let y vary
inversely as x. Find x2.
–8
New Objectives
Identify excluded values of rational
functions.
Graph rational functions.
Vocabulary
•
•
•
•
Rational function
Excluded value
Discontinuous function
Asymptote
A rational function is a function whose rule is
a quotient of polynomials in which the
denominator has a degree of at least 1. In
other words, there must be a variable in the
denominator. The inverse variations you
studied in the previous lesson are a special
type of rational function.
Rational functions:
Not rational functions:
For any function involving x and y, an
excluded value is any x-value that makes the
function value y undefined. For a rational
function, an excluded value is any value that
makes the denominator equal to 0.
Example 1: Identifying Excluded Values
Identify the excluded value for each rational
function.
A.
Set the denominator equal to 0.
x=0
The excluded value is 0.
B.
x–2=0
Set the denominator equal to 0.
Solve for x.
x=2
The excluded value is 2.
Check It Out! Example 1
Identify the excluded value for each rational
function.
a.
Set the denominator equal to 0.
x=0
The excluded value is 0.
b.
x–1=0
Set the denominator equal to 0.
Solve for x.
x=1
The excluded value is 1.
Check It Out! Example 1
Identify the excluded value for each rational
function.
c.
x+4=0
x = –4
Set the denominator equal to 0.
Solve for x.
The excluded value is –4.
Most rational functions are discontinuous
functions, meaning their graphs contain one or
more jumps, breaks, or holes. This occurs at an
excluded value.
One place that a graph of a
rational function is discontinuous
is at an asymptote. An asymptote
is a line that a graph gets closer
to as the absolute value of a
variable increases. In the graph
shown, both the x- and y-axes
are asymptotes. The graphs of
rational functions will
get closer and closer to but never
touch the asymptotes.
For rational functions,
vertical asymptotes will
occur at excluded values.
Look at the graph of y =
The denominator is 0
when x = 0 so 0 is an
excluded value. This
means there is a vertical
asymptote at x = 0. Notice
the horizontal asymptote
at y = 0.
Writing Math
Vertical lines are written in the form x = b,
and horizontal lines are written in the form
y = c.
Look at the graph of y =
Notice that the graph of the
parent function y=
has
been translated 3 units right
and there is a vertical
asymptote at x = 3. The
graph has also been
translated 2 units up and
there is a horizontal
asymptote at y = 2.
These translations lead to the following
formulas for identifying asymptotes in
rational functions.
Example 2A: Identifying Asymptotes
Identify the asymptotes.
Step 1 Write in y =
form.
Step 2 Identify the asymptotes.
vertical: x = –7
horizontal: y = 0
Example 2B: Identifying Asymptotes
Identify the asymptotes.
Step 1 Identify the vertical asymptote.
2x – 3 = 0
+3 +3
2x = 3
Find the excluded value. Set the
denominator equal to 0.
Add 3 to both sides.
Solve for x.
Is an excluded value.
Example 2B Continued
Identify the asymptotes.
Step 2 Identify the horizontal asymptote.
c=8
y=8
y=c
Vertical asymptote: x =
asymptote: y = 8
; horizontal
Check It Out! Example 2a
Identify the asymptotes.
Step 1 Identify the vertical asymptote.
x–5=0
+5 +5
x=5
x=5
Find the excluded value. Set the
denominator equal to 0.
Add 5 to both sides.
Solve for x. 5 is an excluded value.
Check It Out! Example 2a Continued
Identify the asymptotes.
Step 2 Identify the horizontal asymptote.
c=0
y=0
y=c
Vertical asymptote: x = 5; horizontal
asymptote: y = 0
Check It Out! Example 2b
Identify the asymptotes.
Step 1 Identify the vertical asymptote.
4x + 16 = 0
–16 –16
4x = –16
x = –4
Find the excluded value. Set the
denominator equal to 0.
Subtract 16 from both sides.
Solve for x. –4 is an excluded value.
Check It Out! Example 2b Continued
Identify the asymptotes.
Step 2 Identify the horizontal asymptote.
c=5
y=5
y=c
Vertical asymptote: x = –4; horizontal
asymptote: y = 5
Check It Out! Example 2c
Identify the asymptotes.
Step 1 Identify the vertical asymptote.
x + 77 = 0
–77 –77
x = –77
x = –77
Find the excluded value. Set the
denominator equal to 0.
Subtract 77 from both sides.
Solve for x. –77 is an excluded
value.
Check It Out! Example 2c Continued
Identify the asymptotes.
Step 2 Identify the horizontal asymptote.
c = –15
y = –15
y=c
Vertical asymptote: x = –77; horizontal
asymptote: y = –15
To graph a rational function in the form y =
when a = 1, you can graph the asymptotes
and then translate the parent function y = .
However, if a ≠ 1, the graph is not a
translation of the parent function. In this
case, you can use the asymptotes and a table
of values.
Example 3A: Graphing Rational Functions Using
Asymptotes
Graph the function.
Since the numerator is not 1, use the
asymptotes and a table of values.
Step 1 Identify the vertical and horizontal
asymptotes.
vertical: x = 3
Use x = b. x – 3 = 0, so b = 3.
horizontal: y = 0 Use y = c. c = 0
Example 3A Continued
Step 2 Graph the asymptotes using dashed lines.
Step 3 Make a table of values.
Choose x-values on both sides of
the vertical asymptote.
x
0
1
2
4
5
6
y
Step 4 Plot the points and connect
them with smooth curves. The
curves will get very close to the
asymptotes, but will not touch them.
x=3
● ●
●
●●
●
y=0
Example 3B: Graphing Rational Functions Using
Asymptotes
Graph the function.
Step 1 Since the numerator is 1, use the
asymptotes and translate y = .
vertical: x = –4
Use x = b. b = –4
horizontal: y = –2
Use y = c. c = –2
Example 3B Continued
Step 2 Graph the asymptotes
using dashed lines.
Step 3 Draw smooth curves
to show the translation.
Check It Out! Example 3a
Graph each function.
Step 1 Since the numerator is 1, use the
asymptotes and translate y = .
vertical: x = –7
Use x = b. b = –7
horizontal: y = 3
Use y = c. c = 3
Check It Out! Example 3a Continued
Graph each function.
Step 2 Graph the asymptotes
using dashed lines.
Step 3 Draw smooth curves
to show the translation.
Check It Out! Example 3b
Graph each function.
Since the numerator is not 1, use the
asymptotes and a table of values.
Step 1 Identify the vertical and horizontal
asymptotes.
vertical: x = 3
Use x = b. x – 3 = 0, so b = 3.
horizontal: y = 2 Use y = c. c = 2
Check It Out! Example 3b
Step 2 Graph the asymptotes using dashed
lines.
Step 3 Make a table of values.
Choose x-values on both sides of
the vertical asymptote.
x
y
0
1
2
4
5
6
4
Step 4 Plot the points and connect them with
smooth curves. The curves will get very close
to the asymptotes, but will not touch them.
Example 4: Application
Your club has $75 with which to purchase
snacks to sell at an afterschool game. The
number of snacks y that you can buy, if the
average price of the snacks is x-dollars, is
given by y =
a. Describe the reasonable domain and range
values.
Both the number of snacks purchased and
their cost will be positive values so
nonnegative values are reasonable for both
domain and range.
Example 4 Continued
b. Graph the function.
Step 1 Identify the vertical and horizontal
asymptotes.
vertical: x = 0
Use x = b. b = 0
horizontal: y = 0
Use y = c. c = 0
Step 2 Graph the asymptotes using dashed
lines. The asymptotes will be the x- and yaxes.
Example 4 Continued
Step 3 Since the domain is restricted to
nonnegative values, only choose x-values on
the right side of the vertical asymptote.
Number of snacks
Cost of snacks($)
2
37.5
4
18.75
6
12.5
8
9.38
Example 4 Continued
Step 4 Plot the points and connect them
with a smooth curve.
Check It Out! Example 4
A librarian has a budget of $500 to buy copies
of a software program. She will receive 10 free
copies when she sets up an account with the
supplier. The number of copies y of the
program that she can buy is given by
y=
+ 10, where x is the price per copy.
a. Describe the reasonable domain and range
values.
The domain would be all values greater
than 0 up to $500 dollars and the range
would be all natural numbers greater than
10.
Check It Out! Example 4 Continued
b. Graph the function.
Step 1 Identify the vertical and horizontal
asymptotes.
vertical: x = 0
Use x = b. b = 0
horizontal: y = 10 Use y = c. c = 10
Step 2 Graph the asymptotes using dashed
lines. The asymptotes will be the x- and yaxes.
Check It Out! Example 4 Continued
Step 3 Since the domain is restricted to
nonnegative values, only choose x-values on
the right side of the vertical asymptote.
Number of copies
Price ($)
20
35
40
23
60
18
80
16
Check It Out! Example 4 Continued
Step 4 Plot the points and connect them
with a smooth curve.
The table shows some of the properties of the
four families of functions you have studied
and their graphs.
Lesson Quiz: Part I
Identify the exceeded value for each rational
function.
1.
0
5
2.
3. Identify the asymptotes of
then graph the function.
x = –4; y = 0
and
Lesson Quiz: Part II
You have $ 100 to spend on CDs. A CD club
advertises 6 free CDs for anyone who
becomes a member. The number of CDs y
that you can receive is given by y =
where x is the average price per CD.
,
a. Describe the reasonable domain and range
values.
D: x > 0
R: natural numbers >
6
Lesson Quiz: Part III
b. Graph the function.
New Objectives
•Simplify rational expressions.
•Identify excluded values of
rational expressions.
A rational expression is an algebraic
expression whose numerator and
denominator are polynomials. The value of
the polynomial expression in the
denominator cannot be zero since division
by zero is undefined. This means that
rational expressions may have excluded
values.
Example 1A: Identifying Excluded Values
Find any excluded values of each rational
expression.
g+4=0
g = –4
Set the denominator equal to 0.
Solve for g by subtracting 4
from each side.
The excluded value is –4.
Example 1B: Identifying Excluded Values
Find any excluded values of each rational
expression.
x2 – 15x = 0
Set the denominator equal to 0.
x(x – 15) = 0
Factor.
x = 0 or x – 15 = 0
Use the Zero Product Property.
Solve for x.
x = 15
The excluded values are 0 and 15.
Example 1C: Identifying Excluded Values
Find any excluded values of each rational
expression.
y2 + 5y + 4 =
(y 0
+ 4)(y + 1) = 0
Set the denominator equal to 0.
Factor
y + 4 = 0 or y + 1 = 0 Use the Zero Product Property.
y = –4
or y = –1
Solve each equation for y.
The excluded values are –4 and –1.
Check It Out! Example 1a
Find any excluded values of each rational
expression.
t+5=0
t = –5
Set the denominator equal to 0.
Solve for t by subtracting 5
from each side.
The excluded value is –5.
Check It Out! Example 1b
Find any excluded values of each rational
expression.
b2 + 5b = 0
b(b + 5) = 0
Set the denominator equal to 0.
Factor.
b = 0 or b + 5 = 0 Use the Zero Product Property.
b = –5 Solve for b.
The excluded values are 0 and –5.
Check It Out! Example 1c
Find any excluded values of each rational
expression.
k2 + 7k + 12 = 0
(k + 4)(k + 3) = 0
Set the denominator equal to 0.
Factor
k + 4 = 0 or k + 3 = 0 Use the Zero Product Property.
k = –4
or k = –3
Solve each equation for k.
The excluded values are –4 and –3.
A rational expression is in its simplest form
when the numerator and denominator have
no common factors except 1. Remember
that to simplify fractions you can divide out
common factors that appear in both the
numerator and the denominator. You can do
the same to simplify rational expressions.
Example 2A: Simplifying Rational Expressions
Simplify each rational expression, if possible.
Identify any excluded values.
4
Factor 14.
Divide out common factors.
Note that if r = 0, the
expression is undefined.
Simplify. The excluded
value is 0.
Example 2B: Simplifying Rational Expressions
Simplify each rational expression, if possible.
Identify any excluded values.
Factor 6n² + 3n.
Divide out common factors. Note
that if n =
, the expression is
undefined.
3n; n ≠
Simplify. The excluded value is
.
Example 2C: Simplifying Rational Expressions
Simplify each rational expression, if possible.
Identify any excluded values.
3p – 2 =
0
3p = 2
There are no common factors.
Add 2 to both sides.
Divide both sides by 3. The
excluded value is
Caution
Be sure to use the original denominator
when finding excluded values. The excluded
values may not be “seen” in the simplified
denominator.
Check It Out! Example 2a
Simplify each rational expression, if possible.
Identify any excluded values.
Factor 15.
Divide out common factors.
Note that if m = 0, the
expression is undefined.
Simplify. The excluded
value is 0.
Check It Out! Example 2b
Simplify each rational expression, if possible.
Identify any excluded values.
Factor the numerator.
Divide out common factors. Note
that the expression is not
undefined.
Simplify. There is no excluded
value.
Check It Out! Example 2c
Simplify each rational expression, if possible.
Identify any excluded values.
The numerator and denominator
have no common factors. The
excluded value is 2.
From now on in this chapter, you may
assume that the values of the variables that
make the denominator equal to 0 are
excluded values. You do not need to include
excluded values in your answers unless they
are asked for.
Example 3: Simplifying Rational Expressions with
Trinomials
Simplify each rational expression, if possible.
A.
Factor the numerator B.
and the denominator
when possible.
Divide out common
factors.
Simplify.
Check It Out! Example 3
Simplify each rational expression, if possible.
a.
b.
Factor the numerator
and the denominator
when possible.
Divide out common
factors.
Simplify.
Opposite binomials can help you factor
polynomials. Recognizing opposite binomials can
also help you simplify rational expressions.
Consider
The numerator and denominator
are opposite binomials. Therefore,
Example 4: Simplifying Rational Expressions Using
Opposite Binomials
Simplify each rational expression, if possible.
A.
B.
Factor.
Identify opposite
binomials.
Rewrite one
opposite
binomial.
Example 4 Continued
Simplify each rational expression, if possible.
Divide out
common
factors.
Simplify.
Check It Out! Example 4
Simplify each rational expression, if possible.
a.
b.
Factor.
Identify opposite
binomials.
Rewrite one
opposite
binomial.
Check It Out! Example 4 Continued
Simplify each rational expression, if possible.
Divide out
common
factors.
Simplify.
Check It Out! Example 4 Continued
Simplify each rational expression, if possible.
c.
Factor.
Divide out common
factors.
Example 5: Application
A theater at an amusement park is shaped like
a sphere. The sphere is held up with support
rods.
a. What is the ratio of the theater’s volume to
its surface area? (Hint: For a sphere, V =
and S = 4r2.)
Write the ratio of volume to
surface area.
Divide out common factors.
Example 5 Continued
Use the Property of Exponents.
Multiply by the reciprocal of 4.
Divide out common factors.
Simplify.
Example 5 Continued
b. Use this ratio to find the ratio of the
theater’s volume to its surface area when
the radius is 45 feet.
Write the ratio of volume to surface
area. Substitute 45 for r.
Check It Out! Example 5
Which barrel cactus has less of a chance to
survive in the desert, one with a radius of 6
inches or one with a radius of 3 inches?
Explain.
The barrel cactus with a radius of 3 inches
has less of a chance to survive. Its surfacearea-to-volume ratio is greater than for a
cactus with a radius of 6 inches.
Remember!
For two fractions with the same numerator,
the value of the fraction with a greater
denominator is less than the value of the
other fraction.
9>3
Lesson Quiz: Part I
Find any excluded values of each rational
expression.
1.
0
2.
0, 2
Simplify each rational expression, if possible.
3.
5.
4.
6. Calvino is building a rectangular tree house.
The length is 10 feet longer than the width.
His friend Fabio is also building a tree house,
but his is square. The sides of Fabio’s tree
house are equal to the width of Calvino’s
tree house.
a. What is the ratio of the area of Calvino’s
tree house to the area of Fabio’s tree
house?
b. Use this ratio to find the ratio of the
areas if the width of Calvino’s tree
house is 14 feet.
Warm Up
Multiply.
1. 2x2(x + 3)
2x3 + 6x2
2. (x – 5)(3x + 7) 3x2 – 8x – 35
3. 3x(x2 +2x + 2) 3x3 + 6x2 + 6x
4. Simplify
.
Warm Up
Divide. Simplify your answer.
5.
6.
7.
8.
New Objective
Multiply and divide rational
expressions.
The rules for multiplying rational
expressions are the same as the rules for
multiplying fractions. You multiply the
numerators, and you multiply the
denominators.
Example 1A: Multiplying Rational Expressions
Multiply. Simplify your answer.
Multiply the numerators and
denominators.
Factor.
Divide out the common factors.
Simplify.
Example 1B: Multiplying Rational Expressions
Multiply. Simplify your answer.
Multiply the numerators and
the denominators. Arrange
the expression so like
variables are together.
Simplify.
Divide out common factors. Use
properties of exponents.
Simplify. Remember that z0 = 1.
Example 1C: Multiplying Rational Expressions
Multiply. Simplify your answer.
Multiply. There are no common
factors, so the product cannot be
simplified.
Remember!
The Quotient of Powers Property
Check It Out! Example 1a
Multiply. Simplify your answer.
Multiply the numerators and
the denominators. Arrange
the expression so like
variables are together.
Simplify.
Divide out common factors.
Use properties of
exponents.
Check It Out! Example 1b
Multiply. Simplify your answer.
Multiply the numerators and
the denominators. Arrange
the expression so like
variables are together.
Simplify.
Divide out common factors.
Use properties of
exponents.
Example 2: Multiplying a Rational Expression by a
Polynomial.
Multiply
answer.
. Simplify your
Write the polynomial over 1.
Factor the numerator and
denominator.
Divide out common factors.
Multiply remaining factors.
Check It Out! Example 2
Multiply
answer.
Simplify your
Write the polynomial over 1.
Factor the numerator and
denominator.
Divide out common factors.
Multiply remaining factors.
Remember!
Just as you can write an integer as a
fraction, you can write any expression as a
rational expression by writing it with a
denominator of 1.
There are two methods for simplifying
rational expressions. You can simplify first
by dividing out and then multiply the
remaining factors. You can also multiply first
and then simplify. Using either method will
result in the same answer.
Example 3: Multiplying a Rational Expression
Containing Polynomial
Multiply
answer.
. Simplify your
Method 1 Simplify first.
Factor.
Divide out common factors
Then multiply.
Simplify.
Example 3 Continued
Method 2 Multiply first.
Multiply.
Distribute.
Example 3 Continued
Then simplify.
Factor.
Divide out common
factors.
Simplify.
Check It Out! Example 3a
Multiply
. Simplify your answer.
Simplify first.
Factor.
Divide out common factors
Then multiply.
Simplify.
Check It Out! Example 3b
Multiply
. Simplify your answer.
Simplify first.
Factor.
Divide out common factors.
Then multiply.
p
Simplify.
The rules for dividing rational expressions
are the same as the rules for dividing
fractions. To divide by a rational expression,
multiply by its reciprocal.
Example 4A: Dividing by Rational Expressions and
Polynomials
Divide. Simplify your answer.
Write as multiplication by the
reciprocal.
Multiply the numerators and the
denominators.
Divide out common factors.
Simplify.
Example 4B: Dividing by Rational Expressions and
Polynomials
Divide. Simplify your answer.
Write as multiplication by the
reciprocal.
Factor. Rewrite one opposite
binomial.
Example 4B Continued
Divide. Simplify your answer.
Divide out common factors
Simplify.
Example 4C: Dividing by Rational Expressions and
Polynomials
Divide. Simplify your answer.
Write the binomial over 1.
Write as multiplication by the
reciprocal.
Multiply the numerators and
the denominators.
Example 4C Continued
Divide. Simplify your answer.
Divide out common factors.
Simplify.
Check It Out! Example 4a
Divide. Simplify your answer.
Write as multiplication by the
reciprocal.
Multiply the numerators and
the denominators.
Simplify. There are no
common factors.
Check It Out! Example 4b
Divide. Simplify your answer.
Write as multiplication by the
reciprocal.
Multiply the numerators and
the denominators and
cancel common factors.
Simplify.
Check It Out! Example 4c
Divide. Simplify your answer.
Write the trinomial over
1.
Write as multiplication by
the reciprocal.
Multiply.
Check It Out! Example 4c Continued
Divide. Simplify your answer.
Factor. Divide out common
factors.
Simplify.
Tanya is playing a carnival game. She needs to
pick 2 cards out of a deck without looking. The
deck has cards with numbers and cards with
letters. There are 6 more letter cards than
number cards.
a. Write and simplify an expression that represents
the probability that Tanya will pick 2 number
cards.
Let x = the number cards.
number + letter = total
x
+ x + 6 = 2x + 6
Write expressions for
the number of
each kind of card
and for the total
number of items.
Example 5 Continued
The probability of picking a letter card and
then another letter card is the product of the
probabilities of the individual events.
1st pick letter
1st pick: total items
2nd pick letter
2nd pick: total items
b. What is the probability that Tanya picks 2
number cards if there are 25 number cards
in the deck before her first pick? Round
your answer to the nearest hundredth.
Use the probability of picking two number
cards. Since x represents the number of
number cards, substitute 25 for x.
P(number, letter)
Substitute.
Use the order
of
operations
to simplify.
Example 5 Continued
The probability of picking 2 number cards if
there are 25 number cards in the deck of 51
before the first pick is approximately 0.19.
Check It Out! Example 5
What if…? There are 50 blue items in the bag
before Marty’s first pick. What is the
probability that Marty picks two blue items?
Use the probability of picking two blue items.
Since x represents the number of blue items,
substitute 50 for x.
P(blue, blue)
Substitute.
Use the order of
operations to
simplify.
The probability of picking two blue items is
about 0.23.
Lesson Quiz: Part I
Multiply. Simplify your answer.
1.
2
2.
3.
Divide. Simplify your answer.
4.
5.
Lesson Quiz: Part II
6. A bag contains purple and green toy cars.
There are 9 more purple cars than green
cars.
a. Write and simplify an expression to
represent the probability that someone will
pick a purple car and a green car.
b. What is the probability of someone picking
a purple car and a green car if there are
12 green cars before the first pick? Round
to the nearest hundredth.
0.24
Warm Up
Add. Simplify your answer.
1.
2.
3.
4.
Subtract. Simplify your answer.
5.
6.
7.
8.
Objectives
•Add and subtract rational
expressions with like
denominators.
•Add and subtract rational
expressions with unlike
denominators.
The rules for adding rational expressions are
the same as the rules for adding fractions. If
the denominators are the same, you add the
numerators and keep the common
denominator.
Example 1A: Adding Rational Expressions with Like
Denominators
Add. Simplify your answer.
Combine like terms in the
numerator. Divide out common
factors.
Simplify.
Example 1B: Adding Rational Expressions with Like
Denominators
Add. Simplify your answer.
Combine like terms
in the numerator.
Factor. Divide out
common factors.
Simplify.
Example 1C: Adding Rational Expressions with Like
Denominators
Add. Simplify your answer.
Combine like terms
in the numerator.
Factor. Divide out
common factors.
Simplify.
Check It Out! Example 1a
Add. Simplify your answer.
Combine like terms in the
numerator. Divide out common
factors.
=2
Simplify.
Check It Out! Example 1b
Add. Simplify your answer.
Combine like terms
in the numerator.
Factor. Divide out
common factors.
Simplify.
To subtract rational expressions with like
denominators, remember to add the
opposite of each term in the second
numerator.
Example 2: Subtracting Rational Expressions with
Like Denominators
Subtract. Simplify your answer.
Subtract numerators.
Combine like terms.
Factor. Divide out
common factors.
Simplify.
Caution
Make sure you add the opposite of all the
terms in the numerator of the second
expression when subtracting rational
expressions.
Check It Out! Example 2a
Subtract. Simplify your answer.
Subtract
numerators.
Combine like
terms.
Factor. Divide out
common factors.
Simplify.
Check It Out! Example 2b
Subtract. Simplify your answer.
Subtract
numerators.
Combine like
terms.
Factor. There are no
common factors.
As with fractions, rational expressions must have
a common denominator before they can be added
or subtracted. If they do not have a common
denominator, you can use the least common
multiple, or LCM, of the denominators to find one.
To find the LCM, write the prime factorization of
both expressions. Use each factor the greatest
number of times it appears in either expression.
Example 3A: Identifying the Least Common Multiple
Find the LCM of the given expressions.
12x2y, 9xy3
Write the prime
9xy3 =
33x
y  y  y factorization of
each expression.
12x2y = 2  2  3 
xxy
Use every factor
LCM = 2  2  3  3  x  x  y  y  y of both
expressions the
greatest number
= 36x2y3
of times it
appears in either
expression.
Example 3B: Identifying the Least Common Multiple
Find the LCM of the given expressions.
c2 + 8c + 15, 3c2 + 18c + 27
3c2
+ 18c + 27 =
3(c2
+6c +9)
= 3(c + 3)(c + 3)
c2 + 8c + 15 =
(c + 3)(c + 5)
LCM = 3(c + 3)2(c + 5)
Factor each
expression.
Use every factor
of both
expressions the
greatest
number of times
it appears in
either
expression.
Check It Out! Example 3a
Find the LCM of the given expressions.
5f2h, 15fh2
5f2h =
5ffh
15fh2 = 3  5  f 
hh
LCM = 3  5  f  f  h  h
=
15f2h2
Write the prime
factorization of each
expression. Use every
factor of both
expressions the
greatest number of
times it appears in
either expression.
Check It Out! Example 3b
Find the LCM of the given expressions.
x2 – 4x – 12, (x – 6)(x +5)
x2 – 4x – 12 = (x – 6)(x + 2)
Factor each
expression.
Use every factor
(x – 6)(x +5) = (x – 6)(x + 5)
of both
expressions
LCM = (x – 6)(x + 5)(x + 2)
the greatest
number of
times it
appears in
either
expression.
The LCM of the denominators of fractions or
rational expressions is also called the least
common denominator, or LCD. You use the
same method to add or subtract rational
expressions.
Example 4A: Adding and Subtracting with Unlike
Denominators
Add or subtract. Simplify your answer.
Identify the LCD.
5n3 =
5nnn
Step 1 2n2 = 2
nn
LCD = 2  5  n  n  n = 10n3
Multiply each
Step 2
expression by an
appropriate form of 1.
Write each expression
Step 3
using the LCD.
Example 4A Continued
Add or subtract. Simplify your answer.
Step 4
Add the numerators.
Step 5
Factor and divide out
common factors.
Step 6
Simplify.
Example 4B: Adding and Subtracting with Unlike
Denominators.
Add or subtract. Simplify your answer.
Step 1 The denominators are opposite
binomials. The LCD can be either w – 5 or 5 –
w.
Identify the LCD.
Step 2
Step 3
Multiply the first expression by
to get an LCD of w – 5.
Write each expression
using the LCD.
Example 4B Continued
Add or Subtract. Simplify your answer.
Step 4
Subtract the numerators.
Step 5, 6
No factoring needed, so just
simplify.
Check It Out! Example 4a
Add or subtract. Simplify your answer.
3d
3d
Identify the LCD.
Step 1 2d3 = 2 
ddd
LCD = 2  3 d  d  d = 6d3
Step 2
Step 3
Multiply each
expression by an
appropriate form of
1.
Write each expression
using the LCD.
Check It Out! Example 4a Continued
Add or subtract. Simplify your answer.
Step 4
Step 5
Step 6
Subtract the numerators.
Factor and divide out
common factors.
Simplify.
Check It Out! Example 4b
Add or subtract. Simplify your answer.
Step 1
Factor the first term. The
denominator of second
term is a factor of the first.
Step 2
Add the two fractions.
Step 3
Divide out common factors.
Step 4
Simplify.
Example 5: Recreation Application
Roland needs to take supplies by canoe to
some friends camping 2 miles upriver and
then return to his own campsite. Roland’s
average paddling rate is about twice the speed
of the river’s current.
a. Write and simplify an expression for how
long it will take Roland to canoe round trip.
Step 1 Write expressions for the distances
and rates in the problem. The distance in both
directions is 2 miles.
Example 5 Continued
Let x represent the rate of the current, and
let 2x represent Roland’s paddling rate.
Roland’s rate against the current is 2x – x, or x.
Roland’s rate with the current is 2x + x, or 3x.
Step 2 Use a table to write expressions for time.
Direction
Distance
(mi)
Rate
(mi/h)
Upstream
(against current)
2
x
Downstream
(with current)
2
3x
Time (h) =
Distance
rate
Example 5 Continued
Step 3 Write and simplify an expression for
the total time.
total time = time upstream + time downstream
total time =
Step 4
Step 5
Step 6
Substitute known values.
Multiply the first fraction
by a appropriate form
of 1.
Write each expression
using the LCD 3x.
Add the numerators.
Example 5 Continued
b. If the speed of the river’s current is 2.5
miles per hour, about how long will it
take Roland to make the round trip?
Substitute 2.5 for x. Simplify.
It will take Roland
of an hour or 64
minutes to make the round trip.
Check It Out! Example 5
What if?...Katy’s average paddling rate
increases to 5 times the speed of the current.
Now how long will it take Katy to kayak the
round trip?
Step 1 Let x represent the rate of the
current, and let 5x represent Katy’s paddling
rate.
Katy’s rate against the current is 5x – x, or 4x.
Katy’s rate with the current is 5x + x, or 6x.
Check It Out! Example 5 Continued
Step 2 Use a table to write expressions for
time.
Direction
Distance
(mi)
Rate
(mi/h)
Upstream
(against current)
1
4x
Downstream
(with current)
1
6x
Time (h) =
distance
rate
Check It Out! Example 5 Continued
Step 3 Write and simplify an expression for
the total time.
total time = time upstream + time downstream
total time =
Step 4
Step 5
Step 6
Substitute known values.
Multiply each fraction by
a appropriate form of
1.
Write each expression
using the LCD 24x.
Add the numerators.
Check It Out! Example 5 Continued
b. If the speed of the river’s current is 2
miles per hour, about how long will it
take Katy to make the round trip?
Substitute 2 for x. Simplify.
It will take Katy
of an hour or 12.5
minutes to make the round trip.
Lesson Quiz: Part I
Add or subtract. Simplify your answer.
1.
2.
3.
4.
5.
Lesson Quiz: Part II
6. Vong drove 98 miles on interstate highways
and 80 miles on state roads. He drove 25%
faster on the interstate highways than on
the state roads. Let r represent his rate on
the state roads in miles per hour.
a. Write and simplify an expression that
represents the number of hours Vong
drove in terms of r.
b. Find Vong’s driving time if he averaged 55
miles per hour on the state roads.
about 2 h 53 min
Warm Up
Divide.
1. m2n ÷ mn4
2. 2x3y2 ÷ 6xy
3. (3a + 6a2) ÷ 3a2b
Factor each expression.
4. 5x2 + 16x + 12
5. 16p2 – 72p + 81
New Objective
Divide a polynomial by a monomial
or binomial.
To divide a polynomial by a monomial, you
can first write the division as a rational
expression. Then divide each term in the
polynomial by the monomial.
Example 1: Dividing a Polynomial by a Monomial
Divide (5x3 – 20x2 + 30x) ÷ 5x
Write as a rational
expression.
Divide each term in the
polynomial by the
monomial 5x.
Divide out common
factors.
x2 – 4x + 6
Simplify.
Check It Out! Example 1a
Divide.
(8p3 – 4p2 + 12p) ÷ (–4p2)
Write as a rational
expression.
Divide each term in the
polynomial by the
monomial –4p2.
Divide out common
factors.
Simplify.
Check It Out! Example 1b
Divide.
(6x3 + 2x – 15) ÷ 6x
Write as a rational
expression.
Divide each term in the
polynomial by the
monomial 6x.
Divide out common
factors.
Simplify.
Division of a polynomial by a binomial is
similar to division of whole numbers.
Example 2A: Dividing a Polynomial by a Binomial
Divide.
Factor the numerator.
Divide out common factors.
x+5
Simplify.
Example 2B: Dividing a Polynomial by a Binomial
Divide.
Factor both the numerator
and denominator.
Divide out common factors.
Simplify.
Helpful Hint
Put each term of the numerator over the
denominator only when the denominator is
a monomial. If the denominator is a
polynomial, try to factor first.
Check It Out! Example 2a
Divide.
Factor the numerator.
Divide out common factors.
k+5
Simplify.
Check It Out! Example 2b
Divide.
Factor the numerator.
Divide out common factors.
b–7
Simplify.
Check It Out! Example 2c
Divide.
Factor the numerator.
Divide out common factors.
s+6
Simplify.
Recall how you used long division to
divide whole numbers as shown at
right. You can also use long division to
divide polynomials. An example is
shown below.
(x2 + 3x + 2) ÷ (x + 2)
Divisor
x+1
x + 2) x2 + 3x + 2
x2 + 2x
x+2
x+2
0
Quotient
Dividend
Example 3A: Polynomial Long Division
Divide using long division.
(x2 +10x + 21) ÷ (x + 3)
Write in long division
Step 1 x + 3)
+ 10x + 21
form with
expressions in
standard form.
x
Step 2 x + 3) x2 + 10x + 21 Divide the first term of
the dividend by the
first term of the
divisor to get the first
term of the quotient.
x2
Example 3A Continued
Divide using long division.
(x2 +10x + 21) ÷ (x + 3)
x
Multiply the first term of
the quotient by the
Step 3 x + 3) x2 + 10x + 21
binomial divisor.
x2 + 3x
Place the product
under the dividend,
x
aligning like terms.
Step 4 x + 3) x2 + 10x + 21 Subtract the product
from the dividend.
–(x2 + 3x)
0 + 7x
Example 3A Continued
Divide using long division.
x
Step 5 x + 3) x2 + 10x + 21
–(x2 + 3x)
7x + 21
Bring down the next
term in the dividend.
x+7
Step 6 x + 3) x2 + 10x + 21 Repeat Steps 2-5 as
necessary.
2
–(x + 3x)
7x + 21
–(7x + 21)
The remainder is 0.
0
Example 3A Continued
Check: Multiply the answer and the divisor.
(x + 3)(x + 7)
x2 + 3x + 7x + 21

x2 + 10x + 21
Example 3B: Polynomial Long Division
Divide using long division.
x – 4) x2 – 2x – 8
Write in long division form.
x+ 2
x – 4) x2 – 2x – 8
–(x2 – 4x)
2x – 8
–(2x – 8)
x2 ÷ x = x
Multiply x  (x – 4 ). Subtract.
0
Bring down the 8. 2x ÷ x =2.
Multiply 2(x – 4). Subtract.
The remainder is 0.
Example 3B Continued
Check: Multiply the answer and the divisor.
(x + 2)(x – 4)
x2 + 2x – 4x – 8
x2 – 2x + 8
Check It Out! Example 3a
Divide using long division.
(2y2 – 5y – 3) ÷ (y – 3)
Step 1
y – 3)
Step 2
2y
y – 3) 2y2 – 5y – 3
2y2
– 5y – 3
Write in long division
form with
expressions in
standard form.
Divide the first term of
the dividend by the
first term of the
divisor to get the first
term of the quotient.
Check It Out! Example 3a Continued
Divide using long division.
(2y2 – 5y – 3) ÷ (y – 3)
Step 3
2y
y – 3) 2y2 – 5y – 3
2y2 – 6y
2y
Step 4
y – 3) 2y2 – 5y – 3
–(2y2 – 6y)
0+ y
Multiply the first term of
the quotient by the
binomial divisor.
Place the product
under the dividend,
aligning like terms.
Subtract the product
from the dividend.
Check It Out! Example 3a Continued
Divide using long division.
2y
Bring down the next
Step 5 y – 3) 2y2 – 5y – 3
term in the dividend.
–(2y2 – 6y)
y –3
Step 6
2y + 1
y – 3) 2y2 – 5y – 3
–(2y2 – 6y)
y–3
–(y – 3)
0
Repeat Steps 2–5 as
necessary.
The remainder is 0.
Check It Out! Example 3a Continued
Check: Multiply the answer and the divisor.
(y – 3)(2y + 1)
2y2 + y – 6y – 3
2y2 – 5y – 3
Check It Out! Example 3b
Divide using long division.
(a2 – 8a + 12) ÷ (a – 6)
a – 6) a2 – 8a + 12
a– 2
a – 6) a2 – 8a + 12
–(a2 – 6a)
–2a + 12
–(–2a + 12)
0
Write in long division form.
a2 ÷ a = a
Multiply a  (a – 6 ). Subtract.
Bring down the 12. –2a ÷ a = –2.
Multiply –2(a – 6). Subtract.
The remainder is 0.
Check It Out! Example 3b Continued
Check: Multiply the answer and the divisor.
(a – 6)(a – 2)
a2 – 2a – 6a + 12
a2 – 8a + 12

Sometimes the divisor is not a factor of
the dividend, so the remainder is not 0.
Then the remainder can be written as a
rational expression.
Example 4: Long Division with a Remainder
Divide (3x2 + 19x + 26) ÷ (x + 5)
x + 5) 3x2 + 19x + 26 Write in long division form.
3x + 4
3x2 ÷ x = 3x.
x + 5) 3x2 + 19x + 26 Multiply 3x(x + 5). Subtract.
–(3x2 + 15x)
Bring down the 26. 4x ÷ x = 4.
+
26
4x
–(4x + 20) Multiply 4(x + 5). Subtract.
6
The remainder is 6.
Write the remainder as a
rational expression using
the divisor as the
denominator.
Example 4 Continued
Divide (3x2 + 19x + 26) ÷ (x + 5)
Write the quotient with the
remainder.
Check It Out! Example 4a
Divide.
m + 3) 3m2 + 4m – 2
Write in long division form.
3m – 5
m + 3) 3m2 + 4m – 2
–(3m2 + 9m)
3m2 ÷ m = 3m.
Multiply 3m(m + 3). Subtract.
Bring down the –2.
–5m ÷ m = –5 .
–5m – 2
–(–5m – 15) Multiply –5(m + 3). Subtract.
13
The remainder is 13.
Check It Out! Example 4a Continued
Divide.
Write the remainder as a
rational expression using
the divisor as the
denominator.
Check It Out! Example 4b
Divide.
y – 3) y2 + 3y + 2
y+ 6
y – 3) y2 + 3y + 2
–(y2 – 3y)
6y +
–(6y2–18)
20
y+6+
Write in long division form.
y2 ÷ y = y.
Multiply y(y – 3). Subtract.
Bring down the 2. 6y ÷ y = 6.
Multiply 6(y – 3). Subtract.
The remainder is 20.
Write the quotient with the
remainder.
Sometimes you need to write a
placeholder for a term using a zero
coefficient. This is best seen if you write
the polynomials in standard form.
Example 5: Dividing Polynomials That Have a Zero
Coefficient
Divide (x3 – 7 – 4x) ÷ (x – 3).
(x3 – 4x – 7) ÷ (x – 3)
x – 3) x3 + 0x2 – 4x – 7
Write in standard format.
Write in long division form.
Use 0x2 as a placeholder
for the x2 term.
Example 5: Dividing Polynomials That Have a Zero
Coefficient
Divide (x3 – 7 – 4x) ÷ (x – 3).
(x3 – 4x – 7) ÷ (x – 3)
x – 3) x3 + 0x2 – 4x – 7
x2
x – 3) x3 + 0x2 – 4x – 7
–(x3 – 3x2)
3x2 – 4x
Write the polynomials in
standard form.
Write in long division form.
Use 0x2 as a placeholder
for the x2 term.
x3 ÷ x = x2
Multiply x2(x – 3). Subtract.
Bring down –4x.
Example 5 Continued
x2 + 3x + 5
x – 3) x3 + 0x2 – 4x – 7
–(x3 – 3x2)
3x2 – 4x
–(3x2 – 9x)
5x – 7
–(5x – 15)
8
(x3 – 4x – 7) ÷ (x – 3) =
3x3 ÷ x = 3x
Multiply x2(x – 3). Subtract.
Bring down – 4x.
Multiply 3x(x – 3). Subtract.
Bring down – 7.
Multiply 5(x – 3). Subtract.
The remainder is 8.
Remember!
Recall from Chapter 7 that a polynomial in
one variable is written in standard form
when the degrees of the terms go from
greatest to least.
Check It Out! Example 5a
Divide (1 – 4x2 + x3) ÷ (x – 2).
(x3 – 4x2 + 1) ÷ (x – 2) Write in standard format.
x – 2) x3 – 4x2 + 0x + 1 Write in long division form.
Use 0x as a placeholder
for the x term.
x2 – 2x – 4
3 ÷ x = x2
x
3
2
x – 2) x – 4x + 0x + 1
–(x3 – 2x2)
Multiply x2(x – 2). Subtract.
– 2x2 + 0x
2 ÷ x = –2x.
Bring
down
0x.
–
2x
–(–2x2 + 4x)
Multiply –2x(x – 2). Subtract.
– 4x + 1
–(–4x + 8) Bring down 1.
–7 Multiply –4(x – 2). Subtract.
Check It Out! Example 5a Continued
Divide (1 – 4x2 + x3) ÷ (x – 2).
(1 – 4x2 + x3) ÷ (x – 2) =
Check It Out! Example 5b
Divide (4p – 1 + 2p3) ÷ (p + 1).
(2p3 + 4p – 1) ÷ (p + 1)
p + 1) 2p3 + 0p2 + 4p – 1
Write in standard format.
Write in long division form.
Use 0p2 as a placeholder
for the p2 term.
p3 ÷ p = p2
2p2 – 2p + 6
p + 1) 2p3 – 0p2 + 4p – 1
–(2p3 + 2p2)
Multiply 2p2(p + 1). Subtract.
– 2p2+ 4p
2 ÷ p = –2p.
Bring
down
4p.
–
2p
–(–2p2 – 2p)
Multiply –2p(p + 1). Subtract.
6p –1
–(6p + 6) Bring down –1.
Multiply 6(p + 1). Subtract.
–7
Check It Out! Example 5b Continued
(2p3 + 4p – 1) ÷ (p + 1) =
Lesson Quiz: Part I
Add or Subtract. Simplify your answer.
1. (12x2 – 4x2 + 20x) ÷ 4x)
2.
2x + 3
3.
x–2
4.
x+3
3x2 – x + 5
Lesson Quiz: Part II
Divide using long division.
5. (x2 + 4x + 7)  (x + 1)
6. (8x2 + 2x3 + 7)  (x + 3)
Warm Up
1. Find the LCM of x, 2x2, and 6.
2. Find the LCM of p2 – 4p and p2 – 16.
Multiply. Simplify your answer.
3.
5.
4.
New Objectives
•Solve rational equations.
•Identify extraneous solutions.
A rational equation is an equation that
contains one or more rational expressions. If a
rational equation is a proportion, it can be
solved using the Cross Product Property.
Example 1: Solving Rational Equations by Using
Cross Products
Solve
. Check your answer.
Use cross
products.
Check
5x = (x – 2)(3) Distribute 3 on the
right side.
5x = 3x – 6
2x = –6
x = –3
Subtract 3x from
both sides.
–1 –1
Check It Out! Example 1a
Solve
. Check your answer.
Check
Use cross
products.
3n = (n + 4)(1) Distribute 1 on the
right side.
3n = n + 4
Subtract n from both sides.
2n = 4
n=2
Divide both sides by 2.

Check It Out! Example 1b
Solve
. Check your answer.
Check
Use cross
products.
4h = (h + 1)(2) Distribute 2 on the
right side.
4h = 2h + 2
2h = 2
Subtract 2h from both sides.
h=1
Divide both sides by 2.

Check It Out! Example 1c
Solve
. Check your answer.
Check
Use cross
products.
21x = (x – 7)(3) Distribute 3 on the
right side.
21x = 3x –21
18x = –21
x=
Subtract 3x from both sides.
Divide both sides by 18.

Some rational equations contain sums or
differences of rational expressions. To
solve these, you must find the LCD of all
the rational expressions in the equation.
Example 2A: Solving Rational Equations by Using the
LCD
Solve each equation. Check your answer.
Step 1 Find the LCD
2x(x + 1)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute
on the left
side.
Example 2A Continued
Step 3 Simplify and solve.
Divide out common
factors.
(2x)(2) +6(x +1) = 5(x +1) Simplify.
4x + 6x + 6 = 5x + 5
10x + 6 = 5x + 5
5x = –1
Distribute and multiply.
Combine like terms.
Subtract 5x and 6
from both sides.
Divide both sides by 5.
Example 2A Continued
Check Verify
that your
solution is
not
extraneous.

Example 2B: Solving Rational Equations by Using the
LCD
Solve each equation. Check your answer.
Step 1 Find the LCD
(x2)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute
on the left
side.
Example 2B Continued
Step 3 Simplify and solve.
4x – 3 = x2
–x2 + 4x – 3 = 0
x2 – 4x + 3 = 0
(x – 3)(x – 1) = 0
x = 3, 1
Divide out
common
factors.
Simplify.
Subtract x2 from both
sides.
Multiply by – 1.
Factor.
Solve.
Example 2B Continued
Check Verify that your solution is not
extraneous.


Check It Out! Example 2a
Solve each equation. Check your answer.
Step 1 Find the LCD
a(a +1)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute
on the left
side.
Check It Out! Example 2a Continued
Step 3 Simplify and solve.
Divide out
common
factors.
3a = 4(a +
1)
3a = 4a + 4
–4 = a
Simplify.
Distribute the 4.
Subtract the 4 and 3a
from both sides.
Check It Out! Example 2a Continued
Check Verify that your solution is not
extraneous.

Check It Out! Example 2b
Solve each equation. Check your answer.
Step 1 Find the LCD
2j(j +2)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute on the
left side.
Check It Out! Example 2b Continued
Solve each equation. Check your answer.
Divide out common
terms.
12j – 10(2j + 4) = 4j + 8
12j – 20j – 40 = 4j + 8
–12j = 48
j = –4
Simplify.
Distribute 10.
Combine like terms.
Check It Out! Example 2b Continued
Check Verify that your solution is not
extraneous.

Check It Out! Example 2c
Solve each equation. Check your answer.
Step 1 Find the LCD
t(t +3)
Include every factor of the denominator.
Step 2 Multiply both sides by the LCD
Distribute on
the right
side.
Check It Out! Example 2c Continued
Solve each equation. Check your answer.
8t = (t + 3) + t(t + 3)
8t = t + 3 + t2 + 3t
Divide out
common
terms.
Simplify.
0 = t2 – 4t + 3
Distribute t.
Combine like terms.
0 = (t – 3)(t – 1)
Factor.
t = 3, 1
Check It Out! Example 2c Continued
Check Verify that your solution is not
extraneous.


Example 3: Problem-Solving Application
Copy machine A can make 200 copies in
60 minutes. Copy machine B can make
200 copies in 10 minutes. How long will
it take both machines working together
to make 200 copies?
1
Understand the Problem
The answer will be the number of minutes m
machine A and machine B need to print the
copies.
List the important information:
• Machine A can print the copies in 60
minutes, which is
of the job in 1 minute.
• Machine B can print the copies in 10
minutes, which is
of the job in 1 minute.
2
Make a Plan
The part of the copies that machine A
can print plus the part that machine B
can print equals the complete job.
Machine A’s rate times the number of
minutes plus machine B’s rate times the
number of minutes will give the
complete time to print the copies.
(machine m + (machine m = complet
e job
A’s rate)
B’s rate)
m
+
m
=
1
3
Solve
Multiply both sides by
the LCD, 60.
1m + 6m = 60
7m = 60
Distribute 60 on the left
side.
Combine like terms.
Divide both sides by 7.
Machine A and Machine B working together
can print the copies in a little more than 8.5
minutes.
Machine A prints
of the copies per minute
and machine B prints
of the copies per
minute. So in
minutes, machine A prints
of the copies and machine B prints
of the copies. Together, they print
Check It Out! Example 3
Cindy mows a lawn in 50 minutes. It
takes Sara 40 minutes to mow the same
lawn. How long will it take them to mow
the lawn if they work together?
1
Understand the Problem
The answer will be the number of
minutes m Sara and Cindy need to mow
the lawn.
1
Understand the Problem
The answer will be the number of
minutes m Sara and Cindy need to mow
the lawn.
List the important information:
• Cindy can mow the lawn in 50 minutes,
which is
of the job in 1 minute.
• Sara can mow the lawn in 40 minutes,
which is
of the job in 1 minute.
2
Make a Plan
The part of the lawn Cindy can mow plus
the part of the lawn Sara can mow
equals the complete job. Cindy’s rate
times the number of minutes plus Sara’s
rate times the number of minutes will
give the complete time to mow the lawn.
(Cindy’s
rate)
m + (Sara’s
rate)
m = lawn mowed
m
m
+
=
1
3
Solve
Multiply both sides by
the LCD, 200.
5m + 4m = 200
9m = 200
Distribute 200 on the
left side.
Combine like terms.
Divide both sides by 9.
Cindy and Sara working together can mow
the lawn in a little more than 22.2 minutes.
4
Look Back
Cindy mows
of the lawn in 1 minute and
Sara mows
of the lawn in 1 minute. So, in
minutes Cindy mows
and Sara mows
Together they mow
of the lawn
of the lawn.
lawn.
When you multiply each side of an
equation by the LCD, you may get an
extraneous solution. Recall from Chapter
11 that an extraneous solution is a
solution to a resulting equation that is not
a solution to the original equation.
Helpful Hint
Extraneous solutions may be introduced by
squaring both sides of an equation or by
multiplying both sides of an equation by a
variable expression.
Example 4: Extraneous Solutions
Solve
solutions.
. Identify any extraneous
Step 1
Use cross products.
Solve.
Distribute 2 on the left side.
2(x2 – 1) = (x + 1)(x – 6)
Multiply the right side.
2x2 – 2 = x2 – 5x – 6 Subtract x2 from both sides.
Add 5x and 6 to both
x2 + 5x + 4 = 0
sides.
Factor
the quadratic
(x + 1)(x + 4) = 0
expression.
Use
the Zero Product
Property.
Solve.
x = –1 or x = –4
Example 4 Continued
Solve
solutions.
. Identify any extraneous
Step 2 Find extraneous
solutions.

Because
and
are undefined –1 is
not a solution.

The only solution is – 4, so – 1 is an extraneous
solution.
Check It Out! Example 4a
Solve. Identify any extraneous solutions.
Step 1
Use cross products.
Solve.
(x – 2)(x – 7) = 3(x – 7)Distribute 3 on the right side.
Multiply the left side.
2x2 – 9x + 14 = 3x – 21 Subtract 3x from both sides.
Add 21 to both sides.
X2 – 12x + 35 = 0
Factor the quadratic
(x – 7)(x – 5) = 0
expression.
Use
the Zero Product
Property.
Solve.
x = 7 or x = 5
Check It Out! Example 4a Continued
Step 2 Find extraneous
solutions.

Because
and
are undefined 7 is
not a solution.
The only solution is 5, so 7 is an extraneous
solution.

Check It Out! Example 4b
Solve. Identify any extraneous solutions.
Step 1
Use cross products.
Solve.
(x + 1)(x – 3) = 4(x – 2) Distribute 4 on the right side.
Multiply the left side.
x2 – 2x – 3 = 4x – 8
Subtract 4x from both sides.
Add 8 to both sides.
X2 – 6x + 5 = 0
Factor the quadratic
(x – 1)(x – 5) = 0
expression.
Use
the Zero Product
Property.
Solve.
x = 1 or x = 5
Check It Out! Example 4b Continued
Step 2 Find extraneous
solutions.
1 and 5 are
solutions.

The solutions are 1 and 5, there are no
extraneous solutions.

Check It Out! Example 4c
Solve. Identify any extraneous solutions.
Step 1
Solve.
6(x2 + 2x) = 9(x2)
3x2 – 12x = 0
3x(x – 4) = 0
3x = 0, or x – 4 = 0
x = 0 or x = 4
Use cross products.
Distribute 6 on the left side.
Multiply the right side.
Subtract 9x2 from both
Multiply
sides. through with – 1.
Factor the quadratic
expression.
Use the Zero Product
Property.
Solve.
Check It Out! Example 4c Continued
Step 2 Find extraneous
solutions.

Because
and
are undefined 0 is
not a solution.
The only solution is 4, so 0 is an extraneous
solutions.

Lesson Quiz: Part I
Solve each equation. Check your answer.
1.
24
2.
–4, 3
3.
4. Pipe A can fill a tank with water in 4
hours. Pipe B can fill the same tank in 5
hours.
How long will it take both pipes working
together to fill the tank?
Lesson Quiz: Part II
5. Solve
.
Identify any extraneous solutions.
–5; 3 is extraneous.