Transcript Intermediate Algebra - Seminole State College

Intermediate Algebra
Exam 2 Material
Rational Expressions
Rational Expression
• A ratio of two polynomials where the denominator is
not zero (an “ugly fraction” with a variable in a
denominator)
• Example:
x2  x  2
x3
• Will the value of the denominator ever be zero?
If x = - 3, then the denominator becomes 0, so we say
that – 3 is a restricted value of x
• What is the “domain” of the rational expression (all
acceptable values of the variable)?
Domain is the set of all real numbers except - 3.
Domain: {x | x ≠ -3}
Finding Restricted Values and
Domains of Rational Expressions
• Completely factor the denominator
• Make equations by setting each factor of
the denominator equal to zero
• Solve the equations to find restricted
values
• The domain will be all real numbers that
are not restricted
Example
3x  5
x2  4
• Factor the denominator:
(x – 2)(x + 2)
• Set each factor equal to zero and solve the equations:
x – 2 = 0 and x + 2 = 0
x=2
and x = -2
(Restricted Values)
• Domain:
{x | x ≠ -2, x ≠ 2}
Find the domain:
Evaluating Rational Expressions
• To “evaluate” a rational expression means
to find its “value” when variables are
replaced by specific “unrestricted”
numbers inside parentheses
2x  3
• Example:
Evaluate
for x  -2
2
2 2  3

2
 2  2 2  3
x  2x  3
43

443
7

3
7
3
Fundamental Principle of
Fractions
• If the numerator and denominator of a
fraction contain a common factor, that
factor may be divided out to reduce the
fraction to lowest terms:
1
ab

1 ac
b
c
1
1
12
2 23 2


18 1 2 13  3 3
Whencommonfactorsare divided out,"1"is left in each place.
1
1
5

15 x
x
Reducing Rational Expressions
to Lowest Terms
• Completely factor both numerator and
denominator
• Apply the fundamental principle of
fractions: divide out common factors
that are found in both the numerator and
the denominator
Example of Reducing Rational
Expressions to Lowest Terms
• Reduce to lowest terms:
3 x 3  24
3x  6
• Factor top and bottom:
3 x3  8
3x  2

1


3x  2 x 2  2 x  4
3x  2
1
1
• Divide out common factors to get:
1
x2  2x  4

Example of Reducing Rational
Expressions to Lowest Terms
• Reduce to lowest terms:
• Factor top and bottom:
x3
3 x
x3
1x  3

 x3
 1x  3
1x  3

 1x  3
• Divide out common factors to get:
1
1
Equivalent Forms
of Rational Expressions
• All of the following are equivalent:
p
p
p


q
q
q
• In words this would say that a negative
factor in the numerator or denominator can
be moved, respectively, to a negative
factor in the denominator or numerator, or
can be moved to the front of the fraction,
or vice versa
Example of Using Equivalent
Forms of Rational Expressions
• Write equivalent forms of:
x 5
  x  5



x2
x2
 x5
x2
x5
x 5



 x  2
x2
x5
x2
Homework Problems
• Section: 6.1
• Page: 401
• Problems: Odd: 3 – 9, 13 – 23, 27 – 63,
67 – 73
• MyMathLab Homework Assignment 6.1 for
practice
• MyMathLab Quiz 6.1 for grade
Multiplying Rational Expressions
(Same as Multiplying Fractions)
• Factor each numerator and denominator
• Divide out common factors
• Write answer (leave polynomials in factored
form)
• Example:
1 1
1
5
2  2 35
4 15




21
33 2  2  7
9 28
1
1 1
Example of Multiplying
Rational Expressions
3x 2  2 x  8 3x  2

2
3x  14x  8 3x  4
Completely factor each top and bottom:
1
3x  4x  2  3x  2
3x1 2x  4 3x  4
1
1
Divide out common factors:
x  2
x  4
Dividing Rational Expressions
(Same as Dividing Fractions)
• Invert the divisor and change problem to
multiplication
a c
 
b d
a d
ad
 
b c
bc
• Example:
2 3
 
3 4
2 4
 
3 3
8
9
Example of Dividing
Rational Expressions
2y 8y  4y
2y
27


 5

3
9
27
9 8y  4y
2
5
3
2
1 1
2y
27
 3 2

9 4 y 2 y 1
1
3
2 y 2 y2 1


3
2
2 y


Homework Problems
• Section: 6.2
• Page: 408
• Problems: Odd: 3 – 25, 29 – 61
• MyMathLab Homework Assignment 6.2 for
practice
• MyMathLab Quiz 6.2 for grade
Finding the Least Common
Denominator, LCD, of Rational
Expressions
• Completely factor each denominator
• Construct the LCD by writing down each
factor the maximum number of times it is
found in any denominator
Example of Finding the LCD
• Given three denominators, find the LCD:
2
2
6
x

12
,
,
4
x
 16x  16
3x  12
• Factor each denominator:


3x 2 12  3 x 2  4  3x  2x  2
6x 12  6x  2  2  3x  2
2
4 x 2  16x  16  4 x  4 x  4  2  2x  2x  2


• Construct LCD by writing each factor the maximum
number of times it’s found in any denominator:
LCD  2  2  3x  2x  2x  2
LCD  12x  22 x  2
Equivalent Fractions
• The fundamental principle of fractions,
mentioned earlier, says:
ab
b

ac
c
• In words, this says that when numerator and
denominator of a fraction are multiplied by the
same factor, the result is equivalent to the
original fraction
2
62


3
63
12
18
.
Writing Equivalent Fractions
With Specified Denominator
• Given a fraction and a desired denominator for
an equivalent fraction that is a multiple of the
original denominator, write an equivalent fraction
by multiplying both the numerator and
denominator of the original fraction by all factors
of the desired denominator not found in the
original denominator
• To accomplish this goal, it is usually best to
completely factor both the original denominator
and the desired denominator
Example
Write an equivalent fraction to the given
fraction that has a denominator of 24:
Factor each denominato r :
5
?

6 24
5 225
20


6 226
24
6  23
24  2  2  2  3
Example
Write an equivalent rational expression to
the given one that has a denominator of
Factor each denominato r :
2 y3  4 y 2  2 y :
y2
?
 3
2
2
y  y 2y  4y  2y
y 2  y  y y 1
2 y 3  4 y 2  2 y  2 y y 1 y 1


y2
y2

y  22 y  1
2 y2  y  2




2
y  y y  y  1 y  y  12 y  1
y y  12 y  1
2 y2  2 y  4
2 y3  4 y 2  2 y
Homework Problems
• Section: 6.3
• Page: 414
• Problems: Odd: 5 – 43, 51 – 69
• MyMathLab Homework Assignment 6.3 for
practice
• MyMathLab Quiz 6.3 for grade
Adding and Subtracting Rational
Expressions (Same as Fractions)
• Find a least common denominator, LCD,
for the rational expressions
• Write each fraction as an equivalent
fraction having the LCD
• Write the answer by adding or
subtracting numerators as indicated,
and keeping the LCD
• If possible, reduce the answer to lowest
terms
Example
y2
3y
1
 2

2
y  y 2y  4y  2 y
•

y2
3y
1


y  y  1 2 y  1 y  1 y
Find a least common denominator, LCD, for the rational expressions:
y y 1
2 y  1 y 1
y
LCD 
2 y y 1 y 1
•
Write each fraction as an equivalent fraction having the LCD:
•
Write the answer by adding or subtracting numerators as indicated, and keeping the
LCD:
2 y  2 y  1
3y  y
2 1 y  1 y  1


2 y y  1 y  1 2 y y  1 y  1 2 y y  1 y  1




2 y2  2 y  4  3y2  2 y2  4 y  2
2 y2  y  2  3y2  2 y2  2 y 1






2 y y 1 y 1
2 y y  1 y  1
•
If possible, reduce the answer to lowest terms
y2  2 y  2
2 y y  1 y  1
Since top won't factor,fraction won't reduce!
Homework Problems
• Section: 6.4
• Page: 422
• Problems: Odd: 9 – 21, 25 – 47,
51 – 71
• MyMathLab Homework Assignment 6.4 for
practice
• MyMathLab Quiz 6.4 for grade
Complex Fraction
• A “fraction” that contains a rational expression
in its numerator, or in its denominator, or both
• Example:
1
2
3x
5
6y
• Think of it as “fractions inside of a fraction”
• Every complex fraction can be simplified to a
rational expression (ratio of two polynomials)
Two Methods for Simplifying
Complex Fractions
• Method One
– Do math on top to get a single fraction
– Do math on bottom to get a single fraction
– Divide top fraction by bottom fraction
• Method Two (Usually preferred)
– Find the LCD of all of the “little fractions”
– Multiply the complex fraction by “1” where “1”
is the LCD of the little fractions over itself
Method One Example of
Simplifying a Complex Fraction
1
2
3x
5
6y
• Do math on top to get single fraction:
1
1 2
1 6x
1 6 x
2
 


3x
3x 1
3x 3x
3x
• Do math on bottom to get single fraction:
In thiscase, bottomis alreadysingle fraction:
• Top fraction divided by bottom:
2
1 6x 5
1 6x 6 y




3x
6y
3x
5
2 y  12 xy
5x
5
6y
Method Two Example of
Simplifying a Complex Fraction
1
2
3x
5
6y
• Find the LCD of all of the “little fractions”:
6 xy
• Multiply the complex fraction by “1” where “1” is the LCD
of the little fractions over itself
 1
 6 xy

2


3x
 1



6 xy
 5 
 
1
 6y 
6 xy 12xy

3x
1 
30xy
6y
2 y  12 xy
5x
Homework Problems
• Section: 6.5
• Page: 431
• Problems: Odd: 7 – 35
• MyMathLab Homework Assignment 6.5 for
practice
• MyMathLab Quiz 6.5 for grade
Other Types of Equations
• Thus far techniques have been discussed
for solving all linear and some quadratic
equations
• Now address techniques for identifying
and solving “rational equations”
Rational Equations
• Technical Definition: An equation that
contains a rational expression
• Practical Definition: An equation that has
a variable in a denominator
• Example:
1
5
2


2
x  2x  3 x 1 x  3
Solving Rational Equations
1. Find “restricted values” for the equation by
setting every denominator that contains a
variable equal to zero and solving
2. Find the LCD of all the fractions and multiply
both sides of equation by the LCD to eliminate
fractions
3. Solve the resulting equation to find apparent
solutions
4. Solutions are all apparent solutions that are
not restricted
Example
RV
1
5
2


2
x  2x  3 x 1 x  3
1
5
2


x  1x  3 x  1 x  3
x2  2x  3  0
x 1x  3  0
x  1  0 OR x  3  0
x  3
x 1
x  1  0 AlreadySolved
x  3  0 AlreadySolved
LCD

1
5
2  LCD
x 1x  3
 x  1x  3  x  1  x  3  1


 16  3x
1  5x  3  2x 1
16
1  5 x  15  2 x  2
x
Not
RV!
1  3 x  17
3
Example
RV
2
1
1
m 1  0
 
2

m 1m  1  0
m 1 2 m 1
m  1  0 OR m  1  0
m  1
2
1
1
m 1
 
m  1  0 AlreadySolved
m  1m  1 2 m  1

2
1
1  LCD
LCD
 m  1m  1  2  m  1 1
2m  1m  1


4  m 1m  1  2m  1 0  m  3m  1
4  m2 1  2m  2
m  3  0 or m  1  0
2
4  m  1  2m  2
m  3 or m  1
2
0  m  2m  3
2


Formula
• Any equation containing more than one
variable
• To solve a formula for a specific variable
we must use appropriate techniques to
isolate that variable on one side of the
equal sign
• The technique we use in solving depends
on the type of equation for the variable for
which we are solving
Example of Different Types of
Equations for the Same Formula
2
4
B
• Consider the formula: A  3 
C 1
• What type of equation for A?
Linear (variable to first power)
• What type of equation for B?
Quadratic (variable to second power)
• What type of equation for C?
Rational (variable in denominator)
Solving Formulas Involving
Rational Equations
•
Use the steps previously discussed for solving
rational equations:
1. Find “restricted values” for the equation by setting
every denominator that contains the variable
being solved for equal to zero and solving
2. Find the LCD of all the fractions and multiply both
sides of equation by the LCD to eliminate fractions
3. Solve the resulting equation to find apparent
solutions
4. Solutions are all apparent solutions that are not
restricted
Solve the Formula for C:
4B 2
A3 
C 1
Since the formula is rational for C, find RV:
C 1
C 1  0
Multiply both sides by LCD: C  1
2
4B
C 1 A  3 
C 1
C 1
AC  3C  A  3  4B 2
Example Continued
Solve resulting equation and check
apparent answer with RV:
Now linearfor C
AC  3C  A  3  4B 2
AC  3C  4B  A  3
2
 A  3C  4B  A  3
2
 A  3C  4B  A  3
 A  3
 A  3
4B 2  A  3
C
A3
2
Not RV
Homework Problems
• Section: 6.6
• Page: 439
• Problems: Odd: 17 – 69, 73 – 87
• MyMathLab Homework Assignment 6.6 for
practice
• ( No MyMathLab Quiz until we finish
Section 6.7 )
Applications of Rational
Expressions
• Word problems that translate to rational
expressions are handled the same as all
other word problems
• On the next slide we give an example of
such a problem
Example
When three more than a number is divided by
twice the number, the result is the same as the
original number. Find all numbers that satisfy
these conditions.
RV :
Unknowns :
The number x
x3
x
2x
 x 3
2 x
  2 x x 
 2x 
x  3  2x2
0  2x2  x  3
0  2 x  3x  1
2x  0
x0
2 x  3  0 or x  1  0
2 x  3 or x  1
x
3
or x  1
2
.
Homework Problems
• Section: 6.7
• Page: 449
• Problems: Odd: 3 –9
• MyMathLab Homework Assignment 6.7 for
practice
• MyMathLab Quiz 6.6 - 6.7 for grade