Rational Algebraic Expression Math 8.ppt

Simplifying Rational
Expressions

Martin-Gay, Developmental Mathematics 3
Rational Expressions
Q
P
Rational expressions can be written in the form
where P and Q are both polynomials and Q 
0.
Examples of Rational Expressions
5
4
4
2
3 2



x
x
x
2
2
4
3
2
3
4
y
xy
x
y
x



4
3 2
x

To evaluate a rational expression for a particular
value(s), substitute the replacement value(s) into the
rational expression and simplify the result.
Evaluating Rational Expressions
Example
Evaluate the following expression for y = 2.




y
y
5
2
)
2 2
( 2
5


 





7
4
7
4

In the previous example, what would happen if we
tried to evaluate the rational expression for y = 5?




y
y
5
2 5 2
5 5


  0
3
This expression is undefined!
Evaluating Rational Expressions

We have to be able to determine when a
rational expression is undefined.
A rational expression is undefined when the
denominator is equal to zero.
The numerator being equal to zero is okay
(the rational expression simply equals zero).
Undefined Rational Expressions

Find any real numbers that make the following rational
expression undefined.
45
15
4
9 3


x
x
x
The expression is undefined when 15x + 45 = 0.
So the expression is undefined when x = 3.
Undefined Rational Expressions
Example

Simplifying a rational expression means writing it in
lowest terms or simplest form.
To do this, we need to use the
Fundamental Principle of Rational Expressions
If P, Q, and R are polynomials, and Q and R are not 0,
Q
P
QR
PR

Simplifying Rational Expressions

Simplifying a Rational Expression
1) Completely factor the numerator and
denominator.
2) Apply the Fundamental Principle of Rational
Expressions to eliminate common factors in the
numerator and denominator.
Warning!
Only common FACTORS can be eliminated from
the numerator and denominator. Make sure any
expression you eliminate is a factor.

Simplify the following expression.



x
x
x
5
35
7
2



)
5
(
)
5
(
7
x
x
x
x
7
Example






20
4
3
2
2
x
x
x
x





)
4
)(
5
(
)
1
)(
4
(
x
x
x
x
5
1


x
x
Example




7
7
y
y




7
)
7
(
1
y
y
1

Example

Multiplying and Dividing

Multiplying Rational Expressions
Multiplying rational expressions when P,
Q, R, and S are polynomials with Q  0
and S  0.
QS
PR
S
R
Q
P



Note that after multiplying such expressions, our result
may not be in simplified form, so we use the following
techniques.
Multiplying rational expressions
1) Factor the numerators and denominators.
2) Multiply the numerators and multiply the
denominators.
3) Simplify or write the product in lowest terms
by applying the fundamental principle to all
common factors.

Multiply the following rational expressions.


12
5
10
6
3
2
x
x
x
4
1













3
2
2
5
2
5
3
2
x
x
x
x
x
x
Example

Multiply the following rational expressions.





mn
m
m
n
m
n
m
2
2
)
(







)
(
)
(
)
)(
(
n
m
m
n
m
m
n
m
n
m
n
m
n
m


Example

Dividing rational expressions when P, Q, R,
and S are polynomials with Q  0, S  0
and R  0.
QR
PS
R
S
Q
P
S
R
Q
P




Dividing Rational Expressions

When dividing rational expressions, first
change the division into a multiplication
problem, where you use the reciprocal of the
divisor as the second factor.
Then treat it as a multiplication problem
(factor, multiply, simplify).

Divide the following rational expression.




25
15
5
5
)
3
( 2
x
x




15
5
25
5
)
3
( 2
x
x







)
3
(
5
5
5
5
)
3
)(
3
(
x
x
x
3

x
Example

Converting Between Units of Measure
Use unit fractions (equivalent to 1), but with
different measurements in the numerator and
denominator.
Multiply the unit fractions like rational
expressions, canceling common units in the
numerators and denominators.
Units of Measure

Convert 1008 square inches into square feet.















in
12
ft
1
in
12
ft
1
ft.
sq.
7
(1008 sq in)
(2·2·2·2·3·3·7 in · in) 


















in
ft
in
ft
3
2
2
1
3
2
2
1
Example
Units of Measure

Pop Quiz

Adding and Subtracting Rational
Expressions with the Same
Denominator and Least Common
Denominators

If P, Q and R are polynomials and Q  0,
R
Q
P
R
Q
R
P 


R
Q
P
R
Q
R
P 



Add the following rational expressions.






7
2
8
3
7
2
3
4
p
p
p
p
7
2
5
7


p
p





7
2
8
3
3
4
p
p
p
Adding Rational Expressions
Example

Subtract the following rational expressions.



 2
16
2
8
y
y
y



2
16
8
y
y



2
)
2
(
8
y
y
8
Subtracting Rational Expressions
Example






 10
3
6
10
3
3
2
2
y
y
y
y
y




10
3
6
3
2
y
y
y




)
2
)(
5
(
)
2
(
3
y
y
y
5
3

y
Subtracting Rational Expressions
Example

To add or subtract rational expressions with
unlike denominators, you have to change
them to equivalent forms that have the same
denominator (a common denominator).
This involves finding the least common
denominator of the two original rational
expressions.
Least Common Denominators

To find a Least Common Denominator:
1) Factor the given denominators.
2) Take the product of all the unique factors.
Each factor should be raised to a power equal
to the greatest number of times that factor
appears in any one of the factored
denominators.

Find the LCD of the following rational expressions.
12
4
3
,
6
1

y
x
y
y
y 3
2
6 

)
3
(
2
)
3
(
4
12
4 2




 y
y
y
)
3
(
12
)
3
(
3
2
is
LCD
the
So 2



 y
y
y
y
Example

21
10
2
4
,
3
4
4
2
2




 x
x
x
x
x
)
1
)(
3
(
3
4
2




 x
x
x
x
)
7
)(
3
(
21
10
2




 x
x
x
x
7)
1)(x
3)(x
(x
is
LCD
the
So 


Example

1
2
4
,
5
5
3
2
2
2


 x
x
x
x
x
)
1
)(
1
(
5
)
1
(
5
5
5 2
2





 x
x
x
x
2
2
)
1
(
1
2 


 x
x
x
2
1)
-
1)(x
5(x
is
LCD
the
So 
Example

x
x 
 3
2
,
3
1
Both of the denominators are already factored.
Since each is the opposite of the other, you can
use either x – 3 or 3 – x as the LCD.
Example

To change rational expressions into
equivalent forms, we use the principal that
multiplying by 1 (or any form of 1), will give
you an equivalent expression.
R
Q
R
P
R
R
Q
P
Q
P
Q
P






 1
Multiplying by 1

Rewrite the rational expression as an equivalent
rational expression with the given denominator.
9
5
72
9
3
y
y


5
9
3
y

 4
4
5
8
8
9
3
y
y
y 9
4
72
24
y
y
Equivalent Expressions
Example

Adding and Subtracting
Rational Expressions with
Different Denominators

As stated in the previous section, to add or
subtract rational expressions with different
denominators, we have to change them to
equivalent forms first.
Unlike Denominators

Adding or Subtracting Rational Expressions with
Unlike Denominators
1) Find the LCD of all the rational expressions.
2) Rewrite each rational expression as an
equivalent one with the LCD as the
denominator.
3) Add or subtract numerators and write result
over the LCD.
4) Simplify rational expression, if possible.
Unlike Denominators



a
a 6
8
7
15
a
a 6
8
,
7
15






a
a 6
7
8
7
7
6
15
6


a
a 42
56
42
90

a
42
146
a
21
73
Adding with Unlike Denominators
Example

x
x 2
6
3
,
6
2
5





 x
x 2
6
3
6
2
5



 6
2
3
6
2
5
x
x

 6
2
8
x




)
3
(
2
2
2
2
x 3
4

x
Subtracting with Unlike Denominators
Example

3
and
3
2
7

x



3
3
2
7
x




 3
2
)
3
2
(
3
3
2
7
x
x
x




 3
2
9
6
3
2
7
x
x
x




3
2
9
6
7
x
x
3
2
6
16


x
x
Subtracting with Unlike Denominators
Example

6
5
,
6
4
2
2



 x
x
x
x
x





 6
5
6
4
2
2
x
x
x
x
x





 )
2
)(
3
(
)
2
)(
3
(
4
x
x
x
x
x










)
3
)(
2
)(
3
(
)
3
(
)
3
)(
2
)(
3
(
)
3
(
4
x
x
x
x
x
x
x
x
x







)
3
)(
3
)(
2
(
3
12
4 2
x
x
x
x
x
x
)
3
)(
3
)(
2
(
12
2





x
x
x
x
x
Adding with Unlike Denominators
Example

Solving Equations
Containing Rational
Expressions

Solving Equations
First note that an equation contains an equal sign
and an expression does not.
To solve EQUATIONS containing rational
expressions, clear the fractions by multiplying
both sides of the equation by the LCD of all the
fractions.
Then solve as in previous sections.
Note: this works for equations only, not
simplifying expressions.

6
7
1
3
5


x
x
x
x 6
6
7
1
3
5
6 













x
x 7
6
10 

x

10
7
3 6
10
5
1
 

6
7
1
30
5


6
7
1
6
1

 true
Solve the following rational equation.
Check in the original
equation.
Solving Equations
Example

x
x
x
x 3
3
1
1
1
2
1
2




   
1
6
)
1
(
3
1
1
1
2
1
1
6 


















 x
x
x
x
x
x
x
x
  2
6
1
3 

 x
x
2
6
3
3 

 x
x
2
3
3 
 x
1
3 

 x
3
1

x
Solving Equations
Example
Continued.

       
2
1 1 1 1
3 3 3 3
1 1 1
2 1 3 3
 
 
1
3
1
1
4
3
2
3



4
3
4
3
4
6

 true
Substitute the value for x into the original
equation, to check the solution.
So the solution is 3
1

x
Solving Equations
Example Continued

Solving Equations
Example
Continued.
5
1
6
3
1
10
7
2
2







x
x
x
x
x
     
5
2
3
5
1
6
3
1
10
7
2
5
2
3 2






















 x
x
x
x
x
x
x
x
x
     
2
3
5
2
3 



 x
x
x
6
3
5
6
3 



 x
x
x
6
6
5
3
3 



 x
x
x
7
5 

x
5
7


x

Solving Equations
Example Continued
5
18
1
6
5
21
1
10
5
49
25
49
5
3






true
7


x
       
2
7 2 1 1
5
3 6 5
7 10
7 7
7 7
5 5
5 5
 
 
 
   
 
18
5
9
5
18
5 


Solving Equations
Example
Continued.
1
2
1
1


 x
x
     
1
1
1
2
1
1
1
1 

















 x
x
x
x
x
x
 
1
2
1 

 x
x
2
2
1 

 x
x
x

3

Solving Equations
Example Continued
3 3
1 2
1 1

 
4
2
2
1
 true
So the solution is x = 3.

Solving Equations
Example
Continued.
a
a
a 



 3
2
3
3
9
12
2
     
a
a
a
a
a
a
a 



















 3
3
3
2
3
3
9
12
3
3 2
   
a
a 


 3
2
3
3
12
a
a 2
6
3
9
12 



a
a 2
6
3
21 


a
5
15 
a

3

Solving Equations
Example Continued
Since substituting the suggested value of a into the
equation produced undefined expressions, the
solution is .
2
12 3 2
3
9 3
3 3
3
 
 

0
2
5
3
0
12



Solving an Equation With Multiple Variables for
One of the Variables
1) Multiply to clear fractions.
2) Use distributive property to remove
grouping symbols.
3) Combine like terms to simplify each side.
4) Get all terms containing the specified
variable on the same side of the equation,
other terms on the opposite side.
5) Isolate the specified variable.
Solving Equations with Multiple Variables

2
1
1
1
1
R
R
R


2
1
2
1
2
1
1
1
1
R
RR
R
R
R
R
RR 















1
2
2
1 RR
RR
R
R 

2
1
2
1 RR
RR
R
R 

  2
2
1 RR
R
R
R 

R
R
RR
R


2
2
1
Solve the following equation for R1
Example
Solving Equations with Multiple Variables

Problem Solving with
Rational Equations

Ratios and Rates
Ratio is the quotient of two numbers or two
quantities.
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.
The ratio of the numbers a and b can also be
written as a:b, or .
b
a

Proportion is two ratios (or rates) that are
equal to each other.
d
c
b
a

We can rewrite the proportion by multiplying
by the LCD, bd.
This simplifies the proportion to ad = bc.
This is commonly referred to as the cross product.
Proportions

Solve the proportion for x.
3
5
2
1



x
x
   
2
5
1
3 

 x
x
10
5
3
3 

 x
x
7
2 
 x
2
7


x
Solving Proportions
Example
Continued.

3
5
2
3
2
5



true
7


x
7
2
7
1 5
3
2
2





Example Continued
Solving Proportions

If a 170-pound person weighs approximately 65 pounds
on Mars, how much does a 9000-pound satellite weigh?
Mars
on
satellite
pound
-
x
Mars
on
person
pound
-
65
Earth
on
satellite
pound
-
9000
Earth
on
person
pound
-
170

000
,
585
65
9000
170 


x
pounds
3441
170
/
585000 

x
Solving Proportions
Example

Given the following prices charged for
various sizes of picante sauce, find the best
buy.
• 10 ounces for $0.99
Solving Proportions
Example
Continued.

Size Price Unit Price
10 ounces $0.99 $0.99/10 = $0.099
16 ounces $1.69 $1.69/16 = $0.105625
30 ounces $3.29 $3.29/30  $0.10967
The 10 ounce size has the lower unit price, so it is the
best buy.
Example Continued
Solving Proportions

In similar triangles, the measures of
corresponding angles are equal, and
corresponding sides are in proportion.
Given information about two similar triangles,
you can often set up a proportion that will
allow you to solve for the missing lengths of
sides.
Similar Triangles

Given the following triangles, find the unknown
length y.
10 m
12 m
5 m
y
Similar Triangles
Example
Continued

1.) Understand
Read and reread the problem. We look for the corresponding
sides in the 2 triangles. Then set up a proportion that relates
the unknown side, as well.
Example
Continued
Similar Triangles
2.) Translate
By setting up a proportion relating lengths of corresponding
sides of the two triangles, we get
y
10
5
12


Example continued
3.) Solve
Continued
Similar Triangles
6
25
12
50 

y meters
50
10
5
12 


y
y
10
5
12


Example continued
4.) Interpret
Similar Triangles
Check: We substitute the value we found from
the proportion calculation back into the problem.
25
60
6
25
10
5
12

 true
State: The missing length of the triangle is 6
25 meters

Finding an Unknown Number
Example
Continued
The quotient of a number and 9 times its reciprocal
is 1. Find the number.
Read and reread the problem. If we let
n = the number, then
= the reciprocal of the number
n
1
1.) Understand

Continued
2.) Translate
Example continued
The quotient of

a number
n
and 9 times its reciprocal






n
1
9
is
=
1
1

Example continued
3.) Solve
Continued
1
1
9 








n
n
1
9








n
n
1
9


n
n
9
2

n
3
,
3 

n

Example continued
4.) Interpret
Check: We substitute the values we found from the
equation back into the problem. Note that nothing in
the problem indicates that we are restricted to positive
values.
1
3
1
9
3 








1
3
3 

1
3
1
9
3 










1
3
3 



State: The missing number is 3 or –3.
true true

Solving a Work Problem
Example
Continued
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?
Read and reread the problem. By using the times for each
roofer to complete the job alone, we can figure out their
corresponding work rates in portion of the job done per hour.
1.) Understand
Experienced roofer 26 1/26
Beginner roofer 39 /39
Together t 1/t
Time in hrs Portion job/hr

Continued
2.) Translate
Example continued
t
1
39
1
26
1


Since the rate of the two roofers working together
would be equal to the sum of the rates of the two
roofers working independently,

Example continued
3.) Solve
Continued
t
1
39
1
26
1


t
t
t 78
1
39
1
26
1
78 













78
2
3 
 t
t
78
5 
t
hours
15.6
or
5
/
78

t

Example continued
4.) Interpret
Check: We substitute the value we found from the
proportion calculation back into the problem.
State: The roofers would take 15.6 hours working
together to finish the job.
5
78
1
39
1
26
1


78
5
78
2
78
3

 true

Solving a Rate Problem
Example
Continued
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.
Read and reread the problem. By using the formula d=rt, we
can rewrite the formula to find that t = d/r.
We note that the rate of the boat downstream would be the rate
in still water + the water current and the rate of the boat
upstream would be the rate in still water – the water current.
1.) Understand
Down 20 r + 5 20/(r + 5)
Up 10 r – 5 10/(r – 5)
Distance rate time = d/r

Continued
2.) Translate
Example continued
Since the problem states that the time to travel
downstairs was the same as the time to travel
upstairs, we get the equation
5
10
5
20


 r
r

Example continued
3.) Solve
Continued
5
10
5
20


 r
r
     
5
5
5
10
5
20
5
5 

















 r
r
r
r
r
r
   
5
10
5
20 

 r
r
50
10
100
20 

 r
r
150
10 
r
mph
15

r

Example continued
4.) Interpret
Check: We substitute the value we found from the
proportion calculation back into the problem.
5
15
10
5
15
20



10
10
20
20
 true
State: The speed of the boat in still water is 15 mph.

Complex Rational Fractions
Complex rational expressions (complex
fraction) are rational expressions whose
numerator, denominator, or both contain one or
more rational expressions.
There are two methods that can be used when
simplifying complex fractions.

Simplifying a Complex Fraction (Method 1)
1) Simplify the numerator and denominator of
the complex fraction so that each is a single
fraction.
2) Multiply the numerator of the complex
fraction by the reciprocal of the denominator
of the complex fraction.
3) Simplify, if possible.
Simplifying Complex Fractions




2
2
2
2
x
x



2
4
2
2
4
2
x
x



2
4
2
4
x
x
4 2
2 4
x
x

 
 4
4


x
x
Example

Method 2 for simplifying a complex fraction
1) Find the LCD of all the fractions in both the
numerator and the denominator.
2) Multiply both the numerator and the
denominator by the LCD.
3) Simplify, if possible.

6
5
1
3
2
1
2


y
y 2
2
6
6
y
y
  2
2
5
6
4
6
y
y
y


Example

Rational Algebraic Expression Math 8.ppt

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