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Unit 10
Rational Expressions and Equations
Introduction
The set of real numbers is made up of many subsets of numbers. In the last
unit we looked at how these sets of numbers can be related. The simplest set of
numbers was the Natural Numbers. If we add a zero to the set we have the Whole
Numbers. When we add the negative whole numbers to this set we get the Integers.
The next set is found by adding the fractions to the set of Integers. these fractions
are created by dividing integers. This set that has all the fractions in it and is called
the Rational Numbers.
The Rational Numbers get their name from the idea that fractions are a ratio of
two numbers. Ratios are formed by dividing two numbers. For instance when we
write the fraction below we call it four-fifths. To find its value as a decimal we divide.
4
= 5 4 = 0.8
5
There are many types of functions and we have studied several of them this year.
One type of function can be used to create a new type of function just as one set of
numbers can be built on another. The set of linear expressions or functions can create
quadratic expressions or functions as shown in the example below.
(x + 5)(x + 4)
+ 5x + 4x + 20
x2 + 9x + 20
x2
In the above example (x + 5) and (x + 4) are linear expressions and x2 + 9x + 20 is a
quadratic. If we multiply quadratics we can get a higher order polynomial as in the
example below.
(x2 + 3)(x2 – 6)
x4 – 6x2 + 3x2 – 18
x4 – 3x2 – 18
If we divide polynomials we create Rational Functions the same way that if we divide
Integers we get Rational Numbers. An example of a Rational expression is given below.
x 2  2x  3
x2 1
A quadratic or any polynomial divided by another polynomial produces a Rational
expression or function.
Unit 10
Vocabulary and Concepts
Complex Fraction
A complex fraction or rational expression has a fraction in the
numerator or denominator, or both.
Extraneous Answer An extraneous answer occurs in a problem when a potential
solution to an equation when substituted back into the original
equation fails to make the equation true. This word is based
on “extra”. It means extra but not relevant.
Proportion A proportion is two equal fractions or rational expressions.
Rational Expression A rational expression is a fraction with expressions in the
numerator and denominator.
Restriction A restriction is a value that can not be substituted into an expression
because it will make it undefined.
Undefined
An expression is undefined when it can not be evaluated. The occurs
when we try to divide by zero.
Key Concepts for Exponents and Polynomials
Finding Restrictions
This process is done by setting the denominator equal to
zero and solving to find the restrictions.
Simplifying/Reducing Rational Expressions To simplify or reduce fractional
expressions we factor the numerator and denominator
then divide out common factors.
Multiplying Rational Expressions To multiply rational expressions we factor and
both fractions completely and divide out common factors.
Dividing Rational Expressions To divide rational expressions we invert the divisor
and multiply.
Adding/Subtracting Rational Expressions To add/subtract rational expressions
we must have like denominators and then add/subtract
the numerators.
Unit 10 Section 1
Objective

The student will find restrictions of rational expressions.
In the English language restrictions are things we can not do. In mathematics restriction
has the same basic meaning. When we use rational expressions we are performing the
operation of division. There is one restriction we have to know about with division. The
restriction is that we can not divide by zero. When we try to divide by zero we say the
answer is “undefined”. The examples of division below will help us understand why
dividing by zero is undefined.
We are going to look at visual examples of division.
Example A
6
can be represented as
and be interpreted as
2
“How many groups of two can we make?”
We make our first group of two.
We make our second group of two.
We make our third group of two and we are done since there’s nothing
left in the numerator. So our answer is three.
Example B
12
can be represented as
3
“How many groups of three can we make?”
and be interpreted as
We make our first group of three.
We make our second group of three.
We make our third group of three.
We make our fourth group of three and we are done
since there’s nothing left in the numerator.
So our answer is four.
Example C
5
can be represented as
and be interpreted as
0
“How many groups of zero can we make?”
We make our first group of zero.
We make our second group of zero.
We make our third group of zero.
We make our fourth group of zero.
The question is “How long can we continue to make groups of zero?”
The answer is we can make groups of zero as long as we want and
we will never use up the five items in the numerator. We
could make 100 groups of zero and there will still be 5 items in the
numerator. We could make 200, 1000, 10000, or more and we will
never use up any of the items in the numerator. So this means there
is no answer when we divide by zero. That is why we say the answer
is “undefined” and we should never divide by zero.
Rational expressions may have variables in the denominator. When there are variables
in the denominator the expression has the possibility of being undefined. If there is a
value or values that will cause a zero in the denominator then these values are the
restrictions. These numbers would be values we do not want to use. The examples
below illustrate the concept of restrictions.
Example A
Example B
Example C
Example D
7
a
2
y
x 1
x
z3 1
z2
a0
y0
x0
z0
From these examples we should conclude that the numerator does not matter. Only
the numbers that make the denominator in the bottom of the fraction zero can be
restrictions. A second and perhaps more important conclusion is that if we have a
factor that is a single variable with or without an exponent the restriction is zero.
Example A
Example B
Example C
Example D
3
ab
1
2y
x 1
3 z 2
a  0 or b  0
y0
x0
z0
In Example A we
have two single
variable factors. It
could be written
as ‘ a  b ’. So both
or either of the
two variables
must not equal
zero.
In Example B we
have a coefficient
and a single variable
as a factor. So the
restriction is that the
single variable can
not be equal to 0.
In Example C the
expression could be
written as
1
x
The denominator
has a single variable
as a factor.
In Example D the
expression could be
written as
3
z2
The denominator
has a single variable
as a factor.
The denominators of rational expressions can be more complex as shown below.
Example A
Example B
Example C
3
x 1
y
y 5
z
x ( x  2)
( z  3) 1
x  1
y5
x  0,2
z 3
In Example A
if we substitute
the ‘-1’ in for x
we get -1 + 1.
This is equal to 0
and we can’t
divide by zero.
so the restriction
is -1.
In Example B
if we substitute
the ‘5’ in for x
we get 5 - 5.
This is equal to 0
and we can’t
divide by zero.
so the restriction
is 5.
In Example C
if we substitute the
‘-2’ in for x we get
-2 + 2 = 0.
We also have a
single variable
factor of x. 0 in it
should cause 0.
When we substitute
we get
0(0 + 2) = 0.
Example D
In Example D the
expression could be
written as
1
z 3
If we substitute the
‘3’ we get 3 – 3= 0.
Example E
Example F
Example G
Example H
3
( x  1)( x  4)
y
( y  5)( y  3.5)
z
x( x  1)( x  9)
( z  4)
( z  2)( z  2)
x  1, 4
y  5, 3.5
x  0,1, 9
z  2
In Example E we
have two binomial
factors each of
which will produce
a restriction.
In Example F we
have a decimal
value as a
restriction because
that is the value
that will cause a
zero 3.5 – 3.5 = 0.
In Example G
we have three
restrictions because
of the two binomials
and the single
variable factor.
In Example H
we have one
restriction because
the two binomials
are the same. The
‘-2’ could be listed
twice but does not
need to be.
There are times when we will need to set the denominator equal to zero and solve
the equation in order to find the restrictions.
Example A
Find restrictions for
3x + 12 = 0
-12 -12
3x = -12
3
3
x = -4
4
3 x  12
We set the denominator equal to 0.
Then we use the Subtraction Property of Equality.
To finish solving the two step equation we use
the Division Property of Equality.
This is the solution and it makes the denominator
equal to zero. So it is our restriction.
x ≠ -4
Example B
Find restrictions for
2x  4 7
4 7
= 0
4 7
2x  4 7
2
2
x= 2 7
x≠ 2 7
x 2  2x  5
2x  4 7
We set the denominator equal to 0.
Then we use the Addition Property of Equality.
To finish solving the two step equation we use
the Division Property of Equality.
This is the solution and it makes the denominator
equal to zero. So it is our restriction.
In Examples A and B we see two step equations. Solving a two-step equation is
something we did using a short cut in Unit 7 where we solved equations by factoring.
The opposite of the constant divided by the coefficient will always be our answer so
in Example C we will apply this method.
Example C Find restrictions for
-2x – 7 = 0
7
x=
2
7
x≠2
x4  x2
 2x  7
We set the denominator equal to 0.
We can always check our answer by substituting the restriction into the
denominator to see if we get a zero.
 7
 2    7  7  7  0
 2
We can use any technique to solve equations we have learned in the past units. The
examples that follow show us some of these methods being used.
Example D Find restrictions for
x2 – 7x + 12 = 0
(x – 3)(x – 4) = 0
x = 3, 4
x ≠ 3, 4
x 2  5x
x 2  7 x  12
We set the denominator equal to 0.
Then we factor the trinomial.
Our solutions to the equation are 3 and 4.
So our restrictions for the expression are 3 and 4.
Example E Find restrictions for
x2  6
x 2  11x
x2 + 11x = 0
We set the denominator equal to 0.
x(x + 11) = 0
Then we use common factoring.
x = 0, -11 Our solutions to the equation are 0 and -11.
x ≠ 0, -11 So our restrictions for the expression are 0 and -11.
x3  x
x 2  15
Example F Find restrictions for
x2 – 15 = 0
+15 +15
x2 = 15
x 2 =  15
x =  15
x≠
We set the denominator equal to 0.
We will use the square root principle to solve the
equation. Starting with the Addition Property of
Equality, then taking square roots on both sides.
15 ,  15
So the restrictions are
Example G Find restrictions for
x3 – 16x =
x(x2 – 16) =
x(x + 4)(x – 4) =
x=
15 and - 15 .
x2
x  16 x
3
0
We set the denominator equal to 0.
0
First we use common factoring.
0
We use the difference of perfect squares factoring.
0, 4, -4 These are the solutions.
x ≠ 0, 4, -4 So these are the restrictions.
Example H Find restrictions for
6x2 + 13x - 5 = 0
(2x + 5)(3x – 1) = 0
x8
6 x  13 x  5
2
We set the denominator equal to 0.
We trial and error factoring.
5 1
x =  , These are the solutions.
2 3
5 1
x≠  ,
2 3
VIDEO LINK:
So these are the restrictions.
Youtube: Finding restrictions of rational expressions
Exercises Unit 10 Section 1
1. Explain how we can find the restrictions for a rational expression.
2. What is the difference in the results from evaluating the expressions listed below
x
with x = 1
x 1
x 1
with x = 1
x
Find restrictions for each of the expressions.
3.
8
x
7. x 3
11.
5
 2ab
4.
x4
x
5.
3
a
6. z 1
8.
6
3y
9.
z5
 3z
10.
a 1
xy
a2
a  13
12.
b 2  15
15.
b  5.2
16.
19. ( x  21) 2
20.
x
x7
x7
x 2
3
y  14
5
23.
x( x  10)
y
24.
y ( y  4)
27. 5x 1 ( x  8) 1
28.
13.
z 1
z6
14.
17.
z
z 1
18. ( w  9) 1
21.
z 5
z  .5
22.
1
3
y  12
26.
y ( y  19)
x
a2  9
25.
a(a  20)
x  18
x( x  6.8)
29.
7
z2
4 z ( z  2)
30.
a  21
 2a(3a  15)
31.
(a  4) 2
7a(4a  8)
32.
(b  5) 2
11b(10b  5)
33.
a6
9a(4a  6)
34.
( z  5) 3
a(5a  1)
35.
(b  3)(b  2)
 3b(7b  28)
36.
4
 a(2a  16)
37.
a2
(a  5)( a  8)
38.
a  12
(a  14)( a  1)
39.
6
( x  3)( x  9)
40.
y
( y  1)( y  18)
41.
2
( x  16)( x  4)
42.
( x  12)( x  1)
x( x  4)( x  9)
43.
( y  5)( y  2)
3 y ( y  10)( y  9)
44.
2
x  3x
45.
x  12
x  3x  2
46.
y7
2
y  121
47.
( x  10) 2
x 2  8x
48.
x 1
x  8 x  12
49.
z 9
2 z  24 x
50.
x
x  2 x  35
51.
x4
x2  9
52.
a2
3a 2  7a
53.
x  11
x  11x  30
54.
x9
x  4x
55.
a2  9
a 2  10a  11
56.
x 1
x  x  20
57.
z 1
z  5 z  14
2
2
2
2
2
2
2
3
2
58.
a 2  14
2a 3  20a 2  50a
59.
61.
a  17
a  30a 2  30a
62.
64.
x 1
5 x  5 x 2  30 x
67.
70.
60.
y7
y  10 y 2  9
( x  2)( x  1)
3x 3  12 x 2  63 x
63.
y 1
y  13 y 2  36
65.
( x  5)( x  2)
x  12 x 2  36 x
66.
y
y  4y  4
x5
2
2 x  3x  1
68.
x4
2
5 x  11x  1
69.
y2
9 y 2  12 y  4
x
2
3x  20 x  7
71.
x 8
3
2 x  13 x 2  11x
72.
y 2  9 y  20
2 y 2  y  10
3
3
x2
5x 3  5x
3
4
4
2
Unit 10 Section 2
Objective

The student will simplify rational expressions containing monomials.
Simplifying numeric rational expressions can be done in several ways. We are going to
use a method which employs factoring. This method will help us when we have to
perform other operations such as multiplication and division. Using this method is easier
than many other ways as well, since we make large numbers into smaller numbers as
we go through the process. We have factored numbers to their primes before in Unit 6.
The examples below should remind us of this process.
Example A
Example B
18
3
3
.
.
2
Example C
63
6
.
.
3
3
3
.
3
-99
-1 . 11 . 9
21
.
-1 . 11 . 3
7
.
3
The process of factoring to primes starts with any pair or factors and continues to
break down the value until all the factors are prime. Simplify numeric rational
expressions can be described by three steps.
1. Factor the numerator and denominator to primes.
2. Divide out the common factors.
3. Multiply any factors that remain.
The examples below show us how to use this process.
Example A
Simplify
30
42
1
1
30
2 . 3. 5
2. 3. 5
=
=
=
42
2 . 3. 7
2. 3. 7
Step 1
The prime
factoring
=
5
7
Step 2
Step 3
Dividing out
the common
factors
Multiply
the remaining
factors
When factors divide out they gives us the ones we see in the example.
There will be ones in both the numerator and denominator.
Example B
Simplify
36
150
1
1
36
2 . 2 3. 3
2 .2. 3. 3
6
=
=
=
=
150
2 . 3 . 5. 5
2 .3. 5. 5
25
Step 1
The prime
factoring
Example C
Simplify
Step 2
Step 3
Dividing out
the common
factors
Multiply
the remaining
factors
 50
35
1
-50
-1 . 2 . 5 . 5
-1 . 2 . 5 . 5
-10
=
=
=
=
35
5.7
5. 7
7
Step 1
The prime
factoring
Step 2
Step 3
Dividing out
the common
factors
Multiply
the remaining
factors
For these problems we can have improper fractions. If you want to convert the fractions
to mixed numbers that is acceptable but not mandatory.
Example D
Simplify
4
20
1
4
=
=
20
1
2.2
2.2
1
=
=
2 . 2. 5
2 . 2. 5
5
Step 1
The prime
factoring
Step 2
Step 3
Dividing out
the common
factors
Multiply
the remaining
factors
When factors divide out they give us ones. So when all the factors divided out of the
numerator in this example we still had the ones created by 2 ÷ 2.
We can also simplify rational expressions containing monomials.
Example E
Simplify
9x 2 y
6 xz
1
1
33 x  x  y
33 x  x  y
3 xy
9x y
=
=
=
23 x  z
23 x  z
2z
6 xz
2
We will also be asked to evaluate rational expressions. The example below reminds us
of the skills we learned early in our Algebra course.
Example F
Evaluate the expression
5 xy 2
with x = -2, y = 3, and z = 6
4z
1
5 xy
4z
2
=
1
 15
5(1)( 2)(3)(3)
5(2)(3)
=
=
4
(2)( 2)( 2)(3)
4(6)
2
VIDEO LINK: Youtube: Simplifying rational expressions with monomials
Exercises Unit 10 Section 2
Simplify the following rational values as much as possible. Leave all your answers in
fractional form. Show your factoring.
1.
28
42
2.
18
30
3.
9
12
4.
21
14
5.
3
15
6.
54
27
7.
22
99
8.
20
25
9.
 26
30
10.
13
39
11.
36
48
12.
 24
 56
13.
10
22
14.
32
80
15.
40
52
16.
30
6
17.
8x
10 x 2
18.
15ab 3
12ac
19.
 6w
18w 4
20.
20 x 4
4x
Evaluate the following with a = -2, b = 3 and c = 4. Show the substitution. All
answers must be simplified as much as possible.
21.
5a
3b
22.
 6a
4b 2
23.
9c
8b 2
24.
3a 2
7bc
25.
16b
15c
26.
a 2b
3c
Add or subtract as indicated.
27.
9z + 7z - 6z
29. -8x + 11 + x - 6 + 2x
2
2
2
28.
a - 5a + 2a
30.
x + 3z - 7x + 13z - (-3z) + z + 8
2
2
Multiply
31. x
33.
5 .
3 .
a
3
.
x
32. x x
7.
.
.
6 .
b b b a
35. (-3a2b2c)(-6ab4d)
b
2.
x
5
34. (-4x)(-2x)(-3x)(x)(x)(x)
Unit 10 Section 3
Objective

The student will simplify rational expressions containing polynomials.
The process of simplifying rational expressions containing polynomials uses the same
approach as our last section. We must factor both the numerator and denominator
then divide out common factors. We will not always need to multiply out the remaining
factors. We only need to do this if the directions tell us to multiply the remaining factors.
The factored forms of the polynomials have advantages when we are evaluating and
looking for restrictions so we usually do not multiply out the factors of the polynomials.
Factoring the polynomials may require using any of the methods we studied earlier. The
examples below show us how to use these techniques.
Example A
Simplify
x 2  5x
x 2  6x  5
x 2  5x
x( x  5)
x( x  5)
x
=
=
=
2
( x  1)( x  5)
( x  1)( x  5) ( x  1)
x  6x  5
We used common factoring in the numerator and trinomial factoring
in the denominator. Then we divided out the common factors of (x – 5).
Example B
Simplify
x 2  25
x 2  7 x  10
x 2  25
( x  5)( x  5)
( x  5)( x  5)
( x  5)
=
=
=
2
( x  5)( x  2)
( x  5)( x  2)
( x  2)
x  7 x  10
We used the difference of perfect squares in the numerator and trinomial
factoring in the denominator. Then we divided out the common factors
of (x + 5).
Example C
Simplify
4 x 2  8x
2 x 3  18 x  28
4 x 2  8x
2  2  x  ( x  2)
4 x( x  2)
2
=
=
=
3
2  x  ( x  7)( x  2)
2 x( x  2)( x  7)
( x  7)
2 x  18 x  28
Example D
Simplify
x2  x  2
x 4  5x 2  4
x2  x  2
=
x 4  5x 2  4
( x  2)( x  1)
( x 2  1)( x 2  4)
( x  2)( x  1)
( x  2)( x  1)
=
2
2
( x  1)( x  1)( x  2)( x  2)
( x  1)( x  4)
( x  2)( x  1)
( x  2)( x  1)
1
=
=
( x  1)( x  1)( x  2)( x  2)
( x  1)( x  1)( x  2)( x  2) ( x  1)( x  2)
In this example all the factors in the numerator divided out. When factors
divide out we are left with 1’s. As a result we have a ‘1’ in the numerator.
VIDEO LINK: Youtube: Virtualnerd Simplifying rational expressions
Exercises Unit 10 Section 3
Simplify the rational expressions. Show your factoring and dividing out.
1.
x ( x  11)
3 x 2 ( x  11)
2.
6a(a  1)
12a(a  1)( a  1)
3.
( y  8)
( y  8)( y  5)
4.
4( x  2)( x  1)
2 x( x  1)( x  4)
5.
15a 2 (a  7)
3a(a  7)(a  2)
6.
y (4 y  3)
(4 y  3)( y  1)
7.
x2
x 2  5x
8.
x 2  3x
x2  9
9.
6x 2
3x 2  12 x
x 2  100
10. 2
x  12 x  20
11.
x
x  2x
12.
a7
a  14a  49
z2  1
z 2  10 x  11
14.
y2  6y
y 3  2 y 2  24 y
15.
a 2  9a  8
a 2  3a  2
x 3  16 x
16. 3
x  10 x 2  24 x
17.
2y  3
2
2y  5y  3
b 2  81
18. 2
b  7b  18
y 2  25
y 4  26 y 2  25
20.
a 2  13a  30
2a 2  8
21.
13.
19.
2
2
x 2  8 x  16
x 2  7b  12
22.
y2  y  6
y 2  15 y  36
z 2  2 z  15
25.
z 3  9z
28.
y3  6y2  7 y
y 4  50 y 2  49
b2  7
b 3  7b
23.
5x 2  7 x
5 x 2  12 x  7
24.
26.
x 2  4x
3x 2  11x  4
y2  y 6
27.
y 4  13 y 2  36
29.
x 2  11x  30
x 2  6x
30.
b 2  6b  16
b 8
Add or subtract as indicated.
31. (3x + 5y) + (7x – 8y)
32. (5a – 2b + 9) + (-5a – 4)
33. (4a2 – a + 1) + (a – 6 + a2)
34. (9a + 11) – (a – 12)
35. (4 – 2a + b) – (8b – 5 – 2a)
36. (8x – 3x2 + y) – (2z + 8x – y + x2)
Multiply
37. (x – 4)(x – 7)
38. (y + 2)2
39. (z – 11)(z + 11)
40. (2x + 5)(3x – 8)
41. (2x – 9)2
42. (2a + b)(2a – b)
43. (y +7)(y + 10)
44. (a – 3)(a – 11)
45. (x +2y)(x – 10y)
Unit 10 Section 4
Objective

The student will multiply rational expressions.
Rational expressions can be numeric, or have monomials, or contain polynomials.
The method we will use to multiply rational expressions applies to all these different
possibilities. When we learned to multiply fractions in early grades we found out that
we simply multiply the numerators to get the numerator in the product and likewise
we multiply the denominators to get the denominator in the product. The example
below uses this process to multiply fractions.
11 . 3
11  3
33
=
=
5
7
57
35
For this problem we multiplied (11)(3) = 33 for the numerator and (5)(7) = 35 for
the denominator. This problem contained only prime numbers. Our next example
has some composite values but the process remains the same.
10 . 7
10  7
257
=
=
=
15
21  15
37 35
21
2
9
In this example we still multiplied the numerators and denominators but we left them
as a product and then factored each individual number. This is because there is no
reason to make the values bigger before we factor them. After we factored the values
we found common factors and divided them out. In our next example we will use
monomials in our rational expressions or fractions.
2 2 a  a 35b
4a 2  15b
4a 2 . 15b
=
=
=
6a
3b  23 a
3b  6a
3b
10a
3
This method for multiplying the rational expressions will also work when we have
polynomials.
x 2  1 . x 2  7 x  10
( x  1)( x  1)( x  2)( x  5)
=
2
x5
( x  5)( x  1)( x  3)
x  4x  3
First we factor the polynomials.
( x  1)( x  1)( x  2)( x  5)
( x  1)( x  1)( x  2)( x  5) Then we divide out the
=
common factors.
( x  5)( x  1)( x  3)
( x  5)( x  1)( x  3)
( x  1)( x  1)( x  2)( x  5)
( x  1)( x  2)
=
( x  5)( x  1)( x  3)
( x  3)
The remaining factors form our answer.
The examples that follow show us how to use this method.
Example A
Multiply
x 2
x 2  6x .
3
2
x  12 x 2  36 x
x 4
x 2
x 2  6x .
x( x  6)( x  2)
=
3
2
2
x  12 x  36 x
( x  2)( x  2)( x)( x  6)( x  6)
x 4
x( x  6)( x  2)
x( x  6)( x  2)
=
( x  2)( x  2)( x)( x  6)( x  6)
( x  2)( x  2)( x)( x  6)( x  6)
x( x  6)( x  2)
1
=
( x  2)( x  2)( x)( x  6)( x  6)
( x  2)( x  6)
In Example A all the factors in the numerator divided out. When factors
divide out they leave us with 1’s. This means that the numerator has
three 1’s that we multiply together and have a ‘1’ as the result in the
numerator.
Example B
x2  9 .
Multiply
3x  1
x2  9 .
3x  1
3x 2  16 x  5
x3
3x 2  16 x  5
( x  3)( x  3)(3x  1)( x  5)
=
x3
(3x  1)( x  3)
( x  3)( x  3)(3x  1)( x  5)
( x  3)( x  3)(3x  1)( x  5)
=
(3x  1)( x  3)
(3x  1)( x  3)
( x  3)( x  3)(3x  1)( x  5)
= ( x  3)( x  5)
(3x  1)( x  3)
In Example B all the factors in the denominator divided out. When factors
divide out they leave us with 1’s. This means that the denominator has
two 1’s that we multiply together and have a ‘1’ as the result in the
denominator. However we don’t have to write the ‘1’ since dividing by
one does not change the value of a number or expression.
We have been showing all the steps individually but the process can be somewhat
shorter as shown in the next example.
Example C
x 7 .
6x 2
Multiply
3x 2  9 x
x 2  5 x  14
x 7 .
6x 2
( x  7)( 2)(3)( x)( x)
2x
=
=
2
2
3x  9 x
(3)( x)( x  3)( x  7)( x  2)
( x  3)( x  7)
x  5 x  14
Your work in the exercises that follow should be similar to Example C.
Example D
Multiply
1 x .
2x  1
2 x 2  11x  5
x2  1
1 x .
2x  1
2 x 2  11x  5
=
x2  1
By factoring out a ‘-1’ from
the first numerator we can
have a positive ‘x’ term.
 1( x  1)( 2 x  1)( x  5)
(2 x  1)( x  1)( x  1)
 1( x  1)( 2 x  1)( x  5)
 1( x  1)( 2 x  1)( x  5)
 1( x  5)
=
=
(2 x  1)( x  1)( x  1)
(2 x  1)( x  1)( x  1)
( x  1)
VIDEO LINKS: Youtube: Multiplying rational expressions 1
Youtube: Virtualnerd Multiplying rational expressions
Exercises Unit 10 Section 4
Multiply the following. Show your work. All answers should be simplified as much
as possible.
1.
14 . 10
15
21
2.
8 . 3
9
4
3.
10 . 6
3
5
4.
12 . 5
25
18
5.
4 . 5
15
8
6.
16 . 21
7
4
7.
x2 . a
2x
3a
8.
 4a 3 b . 15
2ab 2
5
9.
6x .  y 2
 5y
4x 5
10.
yz 3 . 6
z
12
11.
ab . 5c
20c
3ab 2
12.
x . y
x
y
13.
5 z 2 . 12
18
3 yz
14.
a 1 .
5a
2a
2(a  1)
15.
x ( x  2) . 10
5a
x
17.
x 2  3x .
x2  4
18.
x1
x 2  5x  6 .
2
2
x  4x  4
x 1
19.
y 2  3y  2 . y  7
y2  9
y 7
20.
a 2  a  20 . a  11
a 2  25
a 2  11a
21.
x 8 .
x 2  81
x 9
x 2  2 x  48
22.
b 2  3b  10 .
4b 2
b 2  10b  25
b 2  2b
23.
x
.
x  11
24.
x  13
x 2  5x  4 .
2
2
x  7 x  12
x  13x
25.
16.
x 5 .
x2  x
x1
x 2  8 x  15
x 2
x3
x 2  12 x  11
x2  x
2 z . z 2  12 z  36
z 6
2 z 2  18 z  36 z
26.
3 x
x 2  10 x  21 .
2
2
x  11x  28
x 9
27.
10  z
z
.
z  10
2z 2  7z
28.
x 8
x 2  11x  30 .
2
2
x  x  20
x  9x  8
29.
5y
.
2
6y  y 1
30.
a 2  13a  30 . a  6
a5
a 2  36
31.
x 1
x 3  5 x 2  50 x .
2
2
x  10 x
x  2x  1
32.
1
y 4  13 y 2  36 .
2
y  5y  6
y 3
33.
x 2  x  12 . x  7
x 2  16
x 2  6x  9
9y2 1
y2
Unit 10 Section 5
Objective

The student will divide rational expressions.
Dividing rational expressions uses the same process as dividing fractions. We need to
compare the processes of multiplication and division to understand the process.
Division
Multiplication
6 . 1
6
=
=3
1
2
2
We can see here that the answers are the same, but more importantly the multiplication
problem turned into the division problem shown by the circled items.
6÷2 =3
6÷2 =3
6 . 1
6
=
=3
1
2
2
We can see this happening over and over again in the problems below. The multiplication
problem turned into the division problem shown by the circled items.
14 ÷ 7 = 2
14 . 1
14
=
=2
1
7
7
18 ÷ 3 = 6
18 . 1
18
=
=6
1
3
3
36 ÷ 4 = 9
36 . 1
36
=
=9
4
1
4
The summary of what we see in the examples is given below.
1
Dividing by 2 is the same as multiplying by
2
1
Dividing by 7 is the same as multiplying by
7
1
Dividing by 3 is the same as multiplying by
3
1
Dividing by 4 is the same as multiplying by
4
Our conclusion should be dividing by a number is the same thing as multiplying by the
reciprocal.
To divide rational expressions we multiply by the reciprocal of the divisor.
The examples below show us how to divide rational expressions.
Example A
Divide
6
2
÷
25
5
6
2
6 . 5
235
3
÷
=
=
=
25
5
25
2
552
5
We change the
division to
multiplying by
the reciprocal.
We turned the
divisor upside
down.
Example B
Divide
We divided out
the common
factors.
14a 2
7a 2
÷
3b
9b 3
2  7  a  a 3b
2a
14a 2
14a 2 . 3b
7a 2
÷
=
=
=
2
3
3
33b b b  7  a
3b 2
3b
7a
9b
9b
We change the
division to
multiplying by
the reciprocal.
We turned the
divisor upside
down.
Example B
Divide
We divided out
the common
factors.
x 2  5x
x2
÷ 2
x2  4
x  6x  8
2
x 2  5x
x( x  4)
x2
x2
x 2  ( x  2)( x  4)
. x  6x  8 =
÷
=
=
2
2
2
2
x 4
x 4
( x  2)( x  2)  x  ( x  5) ( x  2)( x  5)
x  6x  8
x  5x
We change the
division to
multiplying by
the reciprocal.
We turned the
divisor upside
down.
We divided out
the common
factors.
The process for dividing rational expression adds one step to multiplying. We start
the division by changing the problem into multiplication and using what we already
know how to do. The only change then from multiplication to division is to find the
reciprocal of the divisor and change the problem to multiplication.
VIDEO LINK: Youtube: Viirtualnerd Dividing rational expressions
Exercises Unit 10 Section 5
Divide the following. Show your factoring and dividing out of common factors. All
answers should be simplified as much as possible.
1.
9
3
÷
14
7
2.
4
12
÷
15
5
3.
10
5
÷
14
21
4.
11
22
÷
6
9
5.
2
4
÷
25
5
6.
18
3
÷
5
10
10.
 ab
a2
9.
÷
5
15
8a
4a 3
8.
÷
7b
21b 2
2z
4
7.
÷
2
3y
9y
16ac
2c
÷
15b
9
11.
 3x
33 x 3
÷
10 y 2
 5y2
12. 
ab
a2
÷ 
7
14
13.
x ( x  7)
x2
÷
( x  5)
( x  1)( x  5)
15.
( x  2)( x  3)
( x  2)
÷
(2 x  9)
(2 x  9)
16.
x 2  36
( x  6)
÷
( x  1)
x2  1
17.
( x  2)
x 2  2x
÷
2 x  22
( x  11)
18.
x3
x 2  6x  9
÷
4 x 2  25
(2 x  5)
19.
x7
x 2  2 x  35
÷ 2
x  2x  8
( x  4)
20.
a 2  10a  21
a 2  7a  12
÷
a2
a2  4
22.
(a 2  8a  7) ÷
24.
a 2  6a  7
a6
y 2  3y
21.
÷ ( y 2  9)
y5
23.
y 2  3 y  10
y2
÷
2
y5
y  25
z 2  4a  32
z 8
÷
z 1
z2  1
25.
z 2  4z  3
z 2  5z  4
÷
z 2  z  20
z 2  2 z  15
26.
9x 2  1
3x 2  10 x  7
÷
4 x  14
2x  7
27.
x 2  64
8 x
÷
x 2  13x
x2
28.
2x 2  x  3
2 z 2  7 z  15
÷
z 7
z 2  49
29.
x3
9  x2
÷
3
x2
x
30.
x 2  3x  2
÷ ( x  2)
z 4
31.
( x 2  6 x  5) ÷
x2 1
x
Unit 10 Section 6
Objective

The student will simplify complex rational expressions.
Complex rational expressions are formed by dividing rational expressions vertically
instead of horizontally. There are several methods for simplifying these types of
expressions but we will relay on the method that turns our problems into the division
problems we studied in the previous section. The examples that follow show us how
to change the vertical form into the horizontal form.
Example A
Simplify
5
9
10
3
=
5
9
10
3
5
10
÷
9
3
First we turn
the vertical
problem into a
horizontal
problem.
Example B
Simplify
5a 2
10 a
7b
=
5 . 3
9 10
=
=
53
1
=
33 25
6
Next we find
the reciprocal
of the divisor
and change
the problem to
multiplication.
Our last step
is to divide out
the common
factors.
5a 2
10 a
7b
5a 2 ÷
10 a
7b
First we turn
the vertical
problem into a
horizontal
problem.
=
5a 2 . 7 b
10 a
1
We turn the
first monomial
into a fraction.
Then we find
the reciprocal
of the divisor
and change
the problem to
multiplication.
=
5a a 7b
7ab
=
25a
2
Our last step
is to divide out
the common
factors.
Example C
a 2  49
a3
a 2  8a  7
Simplify
a 2  49
a3
a 2  8a  7
=
a 2  49
÷ (a 2  8a  7)
a3
First we write the problem horizontally.
a 2  49
a 2  49
a 2  8a  7
÷ (a 2  8a  7) =
÷
1
a3
a3
Next we write the divisor
as a fraction.
a 2  49
a 2  8a  7
÷
1
a3
=
a 2  49 .
1
2
a  8a  7
a3
Here we multiply by the
reciprocal of the divisor.
a 2  49 .
1
2
a  8a  7
a3
=
(a  7)( a  7)
(a  3)( a  1)( a  7)
Then we factor the numerator
and denominator.
=
(a  7)( a  7)
(a  3)( a  1)( a  7)
We divide out the common factors.
(a  7)( a  7)
(a  3)( a  1)( a  7)
( a  7)
(a  3)( a  1)
So this is our simplified form.
In Example C all the steps are shown in detail with explanations. When we are doing our
problems we do not have to show the work in two columns. The example below shows
how our work can look.
Example D
Simplify
a 2  9a  20
a4
2
a  25
We change the problem
to horizontal form.
Next we multiply by the
reciprocal of the divisor.
Then we factor the
numerator and
denominator and
divide out common
factors.
a 2  9a  20
a 2  9a  20
a 2  25
a 2  9a  20 .
1
(a  4)( a  5)
a4
=
÷
=
=
2
2
a  25
a4
1
a4
(a  4)( a  5)( a  5)
a  25
So our answer is
1
a5
Exercises Unit 10 Section 6
Simplify the complex rational expressions below. Show your work. All fractions must be
simplified as much as possible.
1.
9
10
3
4
5.
6
5
3
10
9.
a2
b
a
b2
2.
6.
10.
4a 2
13. 9
1
3
a 2  16
a 1
a4
a 1
17.
20.
14.
z  z  12
z4
z 1
2
3
8
15
4
4.
2
9
2
3
8.
4
9
8
3
11.
a2
6x
3a
2
12.
5a 2
16b
15a 3
8b
15.
12 xy
5
 4x
16.
3.
9
10
6
xy
z2
x3
z2
4z 2
2z
3
18.
21.
7.
x5
x 2  8x
x 2  6x  5
x
x7
x
2
2 x  14 x
21
10
7
8
10
5
6
19.
22.
5a
10a 2
2
y 3
y  2y 1
y2  9
y 1
2
w 2  25
w 2  4w  4
w 2  7 x  10
w 2  3x  2
Unit 10 Section 7
Objective

The student will add and subtract rational expressions with like
denominators.
For this section we need to be reminded how fractions are added. The problem and
its diagrams below should help us follow the process.
Example
Add the following
2
3
+
7
7
This problem can be represented by the diagram below.
+
Each of the rectangles represents a whole or ‘1’ unit. The rectangles have
been divided into seven equal parts. The first rectangle has two parts
shaded so it represents two-sevenths, the first fraction. The second
rectangle also has seven parts and three of them are shaded representing
three-sevenths, the second fraction. When we add the fractions we add
the shaded sections. So the result is
2
7
+
+
3
7
=
5
7
=
The method that we have used in the past, will continue to be used with
rational expressions is adding the numerators of fractions with like
denominators.
To add or subtract rational expressions with like denominators we add or
or subtract the numerators and keep the denominator the same.
The examples below show us how to add rational expressions of various types and
simplify or reduce the answer to lowest terms.
Example A
Add
5
1
+
9
9
5
1
6
23
2
+
=
=
=
9
9
9
3
33
Example B
Add
7
3
−
10
10
7
3
4
22
2
−
=
=
=
10
10
10
5
25
Example C
Add
a
a
+
4
4
a
a
2a
2a
a
+
=
=
=
4
4
4
2
22
Example D
Add
2a
2b
−
ab
ab
2a
2b
−
=
ab
ab
Example E
Add
2a  2b
2( a  b )
=
= 2
ab
( a  b)
x
3
+ 2
x  5x  6
x  5x  6
2
x
3
x3
( x  3)
1
+ 2
= 2
=
=
x  5x  6
x  5x  6
x  5x  6
( x  3)( x  2)
( x  2)
2
Example F
Add
x  20
x2
− 2
2
x  2 x  15
x  2 x  15
x  20
x2
x 2  x  20
( x  4)( x  5)
( x  4)
−
=
=
=
2
2
2
x  2 x  15
( x  3)( x  5)
( x  3)
x  2 x  15
x  2 x  15
In this example when we subtract the numerators we must be careful to
subtract both of the terms in the numerator of the second fraction.
 2x
x2
Add 2
− 2
x  12 x  20
x  12 x  20
Example G
 2x
x2
x 2  2x
x( x  2)
x
−
=
=
=
2
2
2
x  12 x  20
( x  2)( x  10)
( x  10)
x  12 x  20
x  12 x  20
In this example when we subtract the ‘-2x’ in the second fraction’s numerator it
becomes a positive term.
VIDEO LINK: Khan Academy Adding and Subtracting Rational Expressions 1
Exercises Unit 10 Section 7
Add or subtract the rational expressions and simplify or reduce to lowest terms.
1.
3
5
+
11
11
2.
4
1
−
5
5
3.
2
5
−
7
7
4.
3
5
+
8
8
5.
5
1
−
16
16
6.
2
7
+
15
15
7.
5
5
−
6
6
8.
7
1
+
4
4
9.
2
2
+
3
3
10.
5x
x
−
6
6
11.
b
b
+
2a
2a
12.
 2z
5z
+
3
3
13.
8
4 11
−
+
5a 5a 5a
14.
x
2
+
x2
x2
15.
3
a
−
a3
a3
16.
12 x
3x 2
−
x4 x4
17.
18.
8w
16
−
w2
w2
z
6
+ 2
z  6z
z  6z
2
36
a 2  12a
+
a6
a6
19.
25
x2
−
x5 x5
22.
x
6
+ 2
x  5x  6 x  5x  6
23.
24.
9
y 2  10 y
+ 2
2
y  81
y  81
25.
26.
10 x  11
x2
− 2
2
x  13x  22 x  13 x  22
27.
20.
2
21.
40
y 2  3y
−
y 8
y 8
a
2
+ 2
a  4a  12 a  4a  12
2
14a  49
a2
+ 2
2
a  9a  14 a  9a  14
4
a
− 2
a  a  12 a  a  12
2
Find restrictions for each of the expressions.
a2
a
28.
4
x
29.
32.
x
x  12
33. (a  1) 1
36.
a2
(a  4)(2a  7)
y
2
y  7y  6
39.
30. z 2
y
y ( y  9)
34.
37.
a
(a  15)( a  8)
40.
x 2  2x
x 3  10 x 2 25 x
38.
31.
3 z
xy
35.
a2
 3a(4a  20)
6
x  100
2
Simplify the following rational values as much as possible. Leave all your answer in
fractional form. Show your factoring.
41.
24
42
42.
20
35
43.
10
12
44.
6a
4
45. Given the diagram to the right, the expression for the
perimeter in lowest terms is
a.
b.
c.
d.
9a
4
18a
8
9a
2
9a
54
66
3a
4
Unit 10 Section 8
Objective

The student will find the least common denominator for a set of
fractions.
All the fractions we saw in the last section had common denominators. But there are
many times when our rational expressions will have different denominators. Before
we can add them we have to find common denominators. Finding the lowest common
denominator can be accomplished by factoring. The examples below show us the process.
Example A
Find the lowest common denominator (LCD) for
30
42
2 . 15
2 . 21
2. 3 . 5
2 .3.7
7
5
,
30 42
First we factor the two denominators.
30 = 2
42 = 2
.3.5
.3.7
Once we have the denominators’ factors we start
to find the LCD by identifying the common factors.
30 = 2
42 = 2
. 3.5
.3.7
We put the common factors into the LCD, one from
each pair.
LCD = 2 . 3
Next we put the factors that were not in common into our LCD and multiply.
LCD = 2 . 3 . 5 . 7 = 210
Example B
Find the lowest common denominator (LCD) for
14
9
2 . 7
3. 3
14 = 2 . 7
9= 3.3
3
5
,
14 9
First we factor the two denominators
Sometimes there are no common factors so we put
all the factors into the LCD and multiply.
LCD = 2 . 3 . 3 . 7 = 126
When we know there are no common factors as in Example B we can just
multiply denominators.
Example C
2z
3z
,
2
25a b 10 ab
Find the lowest common denominator (LCD) for
25a2b
10ab
5.5.a.a.b
2.5.a.b
First we factored the denominators.
25a2b = 5 . 5 . a . a . b
10ab = 2 . 5 . a . b
LCD = 5 . a . b . 2 . 5 . a
LCD = 50a2b
Example D
Find the lowest common denominator (LCD) for
x2 + 3x
2x2 − 10x
x . (x + 3)
x2 + 3x
x 1
x2
,
x  3 x 2 x 2  10 x
2
2 . x . (x − 5)
=
First we factored the denominators.
x . (x + 3)
2x2 − 10x = 2 . x . (x − 5)
LCD = x . (x + 3) . 2 . (x − 5)
LCD = 2x(x + 3)(x − 5)
Example E
Find the LCD for
x
3x
, 2
x  3x  2 x  4 x  4
2
x2 + 3x + 2
x2 + 4x + 4
(x + 2)(x + 1)
(x + 2)(x + 2)
First we factored the denominators.
x2 + 3x + 2 = (x + 2) (x + 1)
x2 + 4x + 4 = (x + 2) (x + 2)
LCD = (x + 2) . (x + 1) . (x + 2)
LCD = (x + 1)(x + 2)2
VIDEO LINK: Khan Academy Adding and Subtracting Rational Expressions 2
Exercises Unit 10 Section 8
Find lowest common denominators (LCD) for the following.
1.
7
5
,
12 18
2.
3
1
,
7 11
3
1
,
16 4
4.
1
7
,
8 10
5.
1
2
,
12 9
6.
2 5
,
3 6
7.
5 1
,
4 6
8.
11
2
,
30 45
9.
5
7
,
4a 6a 2
10.
2
1
,
5 xy xz
11.
2
1
,
2
9ab
6a 2 b
12.
z
a
,
2z
2a
13.
3a
5
,
8bc 4b
14.
4 7
1
, ,
w y wy
15.
3
9
,
3
5ab
2ab 2
16.
x
b
,
3a 5 z
17.
x5
x 1
,
( x  1)( x  6) ( x  2)( x  6)
3.
18.
2
5
,
a ( a  4) 3a(a  8)
19.
x7
x2
,
x( x  2) 3x( x  2)
2
20.
x
3
,
x6
x  6x
21.
5
4
,
a 3 3a
22.
x 1
x  11
, 2
2
2x  8x x  4x
23.
y5
2y
,
2
y 4
y2
24.
z2
z
, 2
z  5z  4 z  7 z  6
25.
x4
x 1
, 2
2
x  9 x  6 x  36
26.
a3
2a  5
, 2
2
a  9a a  10a  9
27.
y6
y 1
, 2
y  8 y  16 y  16
28.
x4
x8
, 3
3
x  x x  6 x 2  5x
2
2
2
Simplify the rational expressions. Show your factoring and dividing out.
29.
x 2 ( x  3)
4 x( x  3)
30.
9a(a  1)
12(a  1)( a  1)
31.
( y  9)
( y  7)( y  9)
32.
4( x  2)( x  5)
2 x( x  5)( x  2)
33.
( a  7)
(7  a )
34.
y2  y
6( y  1)
35.
x 2  25
x 2  5x
36.
x 2  3x  2
x2  4
37.
x 2  2 x  15
3x 2  15 x
Unit 10 Section 9
Objective

The student will add or subtract rational expressions with unlike denominators.
Adding or subtracting rational expression requires that we find the least common
denominator for the expressions or fractions. Once we have found the common
denominators we must multiply each fraction by a '1' created from the LCD in order
to change the way the expression looks but not the value of the expression. Finally we
will be able to add or subtract the numerators. The examples below model this process.
Example A
Add
1
3
+
3
7
The LCD = 21 this means we want to have a '21' in the denominator of
both fractions. So the first fraction's denominator must be multiplied
by '7', and the second must be multiplied by '3'. But we must multiply
by 1's in order not to change the value of the rational expressions.
7 1
3 3
∙
+
∙
=
7 3
7 3
7
9
16
+
=
21
21
21
In the above example we multiplied the first fraction by 7 over 7 which is
equal to 1 and the second fraction by 3 over 3 also a 1. Once both
denominators were the same we added the numerators.
Example B
Subtract
5
1
−
6
4
The LCD = 12 this means we want to have a '12' in the denominator of
both fractions. So the first fraction's denominator must be multiplied
by '2', and the second must be multiplied by '3'. But we must multiply
by 1's in order not to change the value of the rational expressions.
2 5
1 3
∙
−
∙
=
2 6
4 3
10
3
7
−
=
12
12
12
In the above example we multiplied the first fraction by 2 over 2 which is
equal to 1 and the second fraction by 3 over 3 also a 1. Once both
denominators were the same we subtracted the numerators.
Example C
Add
3
5
+
4a
2b
The LCD = 4ab this means we want to have a '4ab' in the denominator of
both fractions. So the first fraction's denominator must be multiplied
by 'b', and the second must be multiplied by '2a'. But we must multiply
by 1's in order not to change the value of the rational expressions.
b 3
5 2a
∙
+
∙
=
b 4a
2b 2 a
3b
10a
10a  3b
+
=
4ab
4ab
4ab
In the above example we multiplied the first fraction by b over b which is
equal to 1 and the second fraction by 2a over 2a also a 1. Once both
denominators were the same we added the numerators.
Example D
4
3
−
x2
x 1
Subtract
The LCD = (x+2)(x-1) this means we want to have (x+2)(x-1) in the
denominator of both fractions. So the first fraction's denominator must
be multiplied by (x - 1), and the second must be multiplied by (x+2).
But we must multiply by 1's in order not to change the value of the
rational expressions.
x 1
4
3
x2
∙
−
∙
=
x 1 x  2
x 1 x  2
4x  4
3x  6
−
( x  1)( x  2)
( x  1)( x  2)
4x  4
3x  6
x  10
−
=
( x  1)( x  2)
( x  1)( x  2)
( x  1)( x  2)
Example E
x
x
− 2
x  3x
x 9
Subtract
2
When we factor the denominators we get
x
x
x
x
=
and 2
=
x  3x
x 9
x( x  3)
( x  3)( x  3)
2
The LCD = x(x+3)(x-3) this means we want to have x(x+3)(x-3) in the
denominator of both fractions. So the first fraction's denominator must
be multiplied by (x + 3), and the second must be multiplied by x.
But we must multiply by 1's in order not to change the value of the
rational expressions.
x3
x
x
x
∙
−
∙
=
x  3 x( x  3)
( x  3)( x  3) x
x 2  3x
x2
−
x( x  3)( x  3)
x( x  3)( x  3)
Example E (continued)
3x
x 2  3x
x2
−
=
x( x  3)( x  3)
x( x  3)( x  3)
x( x  3)( x  3)
3x
3x
3
=
=
x( x  3)( x  3)
x( x  3)( x  3)
( x  3)( x  3)
In Example E the 'x2' terms in the numerators subtracted and gave us a 0.
There was also a factor of 'x' left in the numerator and denominator
so we divided this factor out.
Exercises Unit 10 Section 9
Add or subtract as indicated. Show your work. Simplify your answers as much as
possible.
1.
2
1
+
3
6
2.
3
1
+
5
10
3.
5
1
+
8
6
4.
3
1
−
4
8
5.
1
3
−
25
10
6.
1
5
−
8
10
7.
a
a
−
4
6
8.
1
1
+
2a
9a
9.
2a
2b
+
b
a
10.
a2
a2
+
4
6
11.
a 1
1
−
2
2a
a
12.
2
2
+
x 1
x3
13.
a 1
a3
−
2
4
14.
2a  1
1
−
2
a
a 1
15.
2
2
+
z4
z4
16.
19.
22.
24.
1
1
−
17.
z 3
z3
1
1
+ 2
20.
2
x x
x 1
1
x
x
+
−
2
x2
x2
x 4
1
1
+
2x
x2
2
2
4
2b
+
18.
−
a2
b 1
a ( a  2)
(b  1)(b  2)
y 1
y
2x
1
−
21.
− 2
2
2
x 1
x  2x  1
y  7 y  12
y 9
2x
x
x
23.
+ 2
+ 2
2
x  11x  18
x 4
x  7 x  18
25.
1
x 1
− 3
x 4
x  4x
2
Unit 10 Section 10
Objective

The student will solve equations with rational expressions.
Equations with rational expressions will frequently have three or more fractions or
rational expressions in them. The first thing we must be able to do is find the lowest
common denominator for three fractions.
Example A
Find the LCD for
3 5
1
,
,
8 6 15
First we factor the denominators.
8=2.2.2
6=2.3
15 = 5 . 3
To make the LCD we find pairs or triples of factors the
denominators we have in common and put them into the LCD.
LCD = 2 . 3
Then we put all the other factors into the LCD
LCD = 2 . 3 . 2 . 2 . 5 = 120
Example B
Find the LCD for
b
1
a
,
,
2 a 6b 12
First we factor the denominators.
2a = 2 . a
6b = 2 . 3 . b
12 = 2 . 3 . 2
To make the LCD we find pairs or triples of factors the
denominators we have in common and put them into the LCD.
LCD = 2 . 3
Then we put all the other factors into the LCD
LCD = 2 . 3 . 2 . a . b = 12ab
When we are finding pairs or triples the values must come from different factors. As a
result we can only use the three circled 2’s and can’t include the second 2 in 12.
The process of solving an equation containing rational expressions or fractions can be
listed as a set of steps or algorithm. These steps are given below.
Step 1. Find the LCD for all the fractions or expressions.
Step 2. Use the Multiplicative Property of Equality and multiply by the
LCD on both sides of the equation. We will also have to use
the Distributive Property frequently.
Step 3. Simplify the equation.
Step 4. Use an appropriate method and solve the equation.
The examples that follow use this algorithm.
Example A
Solve
x
x
+ 14 =
10
3
The LCD = 30 or
30
1
Step 1. Find the LCD.
30
x
x 30
∙(
+ 14) =
∙
10
3
1
1
30x
30x
+ 420 =
10
3
3x + 420 = 10x
3x + 420 = 10x
-3x
-3x
420 = 7x
7
7
Step 2. Use the Multiplication Property
of Equality.
Use the Distributive Property
Step 3. Simplify the equation.
Step 4. Solve the equation.
60 = x
When solving equations with rational expressions we should check
our answers because there can be “extraneous” solutions. The
term extraneous means that solving equation might produce answers
which give us undefined or non-real values if substituted back into
the equation. So we check our answer.
x
x
+ 14 =
10
3
60
60
+ 14 =
10
3
6 + 14 = 20
20 = 20
The answer checks, we have valid solution.
Example B
10
6
44
+
= 2
x
x 1
x
Solve
x 2 ( x  1)
The LCD = x ( x  1) or
1
2
Step 1. Find the LCD.
10
6
44 x 2 ( x  1) Step 2. Use the Multiplication Property
x 2 ( x  1)
∙(
+
)= 2 ∙
of Equality.
x
x 1
1
1
x
10 x 2 ( x  1) 6 x 2 ( x  1)
44 x 2 ( x  1)
+
=
x
x 1
x2
10 x( x  1)  6 x 2  44( x  1)
10 x  10 x  6 x
2
2
Use the Distributive Property
Step 3. Simplify the equation.
 44 x  44
10 x 2  10 x  6 x 2  44 x  44
Step 4. Solve the equation.
16 x  10 x  44 x  44
2
 44 x  44 x  44
 44
16 x  54 x  44  0
2(8x 2  27 x  22)  0
2(8 x  11)( x  2)  0
11
x  2,
8
2
This time we solve the equation
by factoring.
We need to check our answers.
10
44
6
+
= 2
x
x 1
x
10
6
44
+
= 2
2
2 1
2
5  6  11
11  11
10
6
44
+
= 2
x
x 1
x
44
10
6
+
=
2
11
11
 11 
1


8
8
8
80
48
2816
+
=
3
121
11
7
Both answers check and are valid.
3
3
+ 16 = 23
11
11
3
3
23 = 23
11
11
There is a special kind of equation that has rational expressions. This type of
equation is called a proportion. Proportional reasoning is a key concept in
mathematics. People reason proportionally without realizing it. Proportions are used
in many ways. Models of buildings, ships, and other things have scale factors that
tell us how large they are in relation to the original.
Example A
The Statue of Liberty has a scale factor of 1/25. The statue from the bottom of Lady
Liberty's feet to the top of her head measures close to 125 ft. If we know the height
of the statue and the scale factor we should be able to find the height of the woman
the creator used as his model.
First we let x = the height of the woman. This is what we want to find.
When we set up a proportion we want set up two fractions that are equal. The way we
will do this is shown below.
The small value in the scale
The large value in the scale
=
The size of the woman (small number)
The size of the statue (large number)
In the fractions above we can see that the small measurements are located in the top
of the fractions while both the large measurements are located in the denominators.
Also the scale is treated as one fraction while the actual measurements are in the
second fraction. This is generally the way we want to set up a proportion. We will
finish the problem by substituting the values we know and our variable 'x'.
1
x
=
25
125
The LCD = 125 so we will use the Multiplication Property of Equality and multiply by the
LCD on both sides.
125 1
x 125


=
1 25
125 1
125
125x
=
25
125
5 ft. = x
There is a short cut we can use to solve proportions. This short cut is called cross
multiplying. The derivation that follows on the next page shows us why and how this
works.
Given the proportion
a
c
=
b
d
( two equal fractions )
If we want to clear the denominators ( get rid of the fractions ) we use the Multiplication
Property of Equality on both sides with the LCD. The LCD here is bd .
bd a
c bd


=
1 b
d 1
abd
bcd
=
b
d
ad = bc
Once the denominators have been cleared the equation can be solved with methods we
have previously studied. The proportion we started with can however be simplified
quickly by cross multiplying as shown below.
ad
=
bc
a
c
=
b
d
The cross multiplying follows the arrows. We multiply along both arrows producing the
same result from using the full process of solving an equation with rational expressions.
There are two key facts to remember for proportions.
Definition
A Proportion is two equal ratios or fractions.
The short cut to clear denominators in a proportion is to cross multiply.
We must keep in mind that cross multiplying can ONLY be used with proportions.
The examples that follow on the next page show us how to solve proportions with
cross multiplication.
Example A
Solve
x
6
=
12
9
x
6
=
12
9
( Cross multiplying can be going up or down. )
72 = 9x
9
9
8=x
Example B
Solve
x 1
4
=
20
5
x 1
4
=
20
5
When we cross multiply we get (4)(2) = 80
and (5)(x+1) which means we will use the
Distributive Property and get 5x + 5.
5x + 5 = 80
-5
-5
5x = 75
5
5
x = 15
Example C
Solve
3
2x  1
=
4
x  12
3
2x  1
=
4
x  12
8x − 4 = 3x + 36
-3x
-3x
5x − 4 = 36
+4 +4
5x = 40
5
5
x =8
When we cross multiply we will use the
Distributive Property twice. 4(2x-1) = 8x-4
and 3(x+12) = 3x +36.
Example D
Solve
3
x
=
x
x6
When we cross multiply we will use the
Distributive Property with 3(x+6) = 3x+18
and (x)(x) = x2.
3
x
=
x
x6
x2 = 3x + 18
-3x
-3x -18
-18
2
x – 3x – 18 = 0
(x – 6)(x+3) = 0
x = 6, -3
This equation can be solved by factoring.
These are our answers.
We will check the answers by substituting our answers into the
original equation.
3
x
=
x
x6
3
6
=
6
66
1
6
=
2
12
3
x
=
x
x6
3
3
=
3
36
3
1 =
3
1
1
=
2
2
1 = 1
Both answers check and so are valid.
Example E
Solve
2
x2
=
x
x
When we cross multiply we will use the
Distributive Property with x(x+2) = x2+2x
and then multiply (2)(x) = 2x.
2
x2
=
x
x
x 2  2x  2x
 2x  2x
x2  0
x2 
0
This equation can be solved by taking
square roots on both sides.
x  0
This is our answer.
Next we check our answer
2
x2
=
x
x
2
02
=
0
0
This answer is not valid!
2
is undefined. There is no solution.
0
Example F
Solve
x
12
=
4
x
x
12
=
4
x
When we cross multiply we will have
(4)(12) = 48 and (x)(x) = x2.
x 2  48
We take square roots on both sides next.
x2 
48
x
48
We need to simplify the radical.
This is our answer.
x 4 3
Next we check our answer.
4 3
12
=
4
4 3
3 =
3
3 =
3
3 =
3 3
3
3 =
We first divide out common factors.
We have to rationalize the denominator.
3
3
∙
3
3
3
This checks and our answer is valid.
VIDEO LINK: Khan Academy Adding and Subtracting Rational Expressions 2
Exercises Unit 10 Section 10
Solve the following equations. Show your work. Check for extraneous solutions.
Simplify all radicals completely.
1.
x
x
+
= 15
4
6
2.
a
a
−
=4
2
3
3.
3=
2x
x
−
5
3
4.
3y
y
=
+ 22
4
5
5.
a 1
a
−
=5
2
5
6.
4=
3x
x
−
2
10
7.
y6
y
=
+7
2
4
8.
5z
z3
−
= 23
3
2
9.
3x
11x
x
=
−
10
2
5
10.
b7
b
+
=0
3
4
11.
z
z  12
5z
+
=
3
9
6
12.
x 3
1
x
+
=
2
4
3
13.
30
12
+
=8
x 1
x
14.
12
18
3
=
+
a
2a
a
15.
1
1
5
+
=
3
x
6
16.
6
8
+
=5
z2
z2
17.
2
1
x 1
−
=
3
x
12
18.
20
2
15
−
=
z
z
z3
Solve the following proportions. Show your work. Check for extraneous solutions.
Simplify all radicals completely.
19.
x
5
=
12
6
20.
30
3
=
5
a
21.
8
2x
=
3
15
22.
 3x
9
=
20
5
23.
1
32
=
x
2
24.
21
3y
=
8
4
25.
1
y
=
10
5
26.
x
32
=
x
2
27.
2
z
=
3
z5
28.
2x
x2
=
11
5
29.
2
1
=
a
5a
30.
5
4
=
3a
a  14
31.
2
3
=
z
z
32.
4
y5
=
3
y 1
33.
a5
a8
=
3
5
34.
x
1
=
15
x2
35.
y
14
=
7
y
36.
4
x
=
3
3x
Unit 10 Section 11
Objective

The student will solve contextual problems with rational equations
and proportions.
Rational equations and proportions can be applied to many types of contextual
problems. Some of our problems will involve direct translation, scale factors, distance,
costs and work. We will start with examples of direct translation.
Example A
The sum of the reciprocals of two consecutive even integers
is 17 , find the numbers.
144
Let
x = the first number
then x+2 = the second number
1
1
and
x
x2
The key word in the problem is "sum" so we can create our equation
The reciprocals of the two numbers are:
1
17
1
+
=
x
x2
144
The LCD is 144x(x+1)
144 x( x  2)
∙
1
(1
x
+
1
) = 17 ∙ 144 x( x  2)
x2
144
1
144 x( x  2)
(17)(144) x( x  2)
144 x( x  2)
+
=
x
144
( x  2)
144( x  2) + 144 x = 17 x( x  2)
We multiply on both
sides by the LCD.
We used the Distributive
Property with the first
fraction.
We simplified the fractions.
144 x  288 + 144x = 17 x 2  34 x We have a quadratic equation to solve.
288x + 288 = 17 x 2  34 x
 288 x  288
- 288x - 288 =
We get the quadratic in standard form.
2
0 = 17 x  254 x  288
0 = (17 x  18)( x  16)
We factor the quadratic.
 18
These are the answer to the quadratic. BUT we are
, 16 = x
17
looking for INTEGERS. The only answer for x is 16.
Our consecutive integers are 16 and 18.
Example B
If the difference between one-third of a number and one-fifth of
number is
8
, find the number.
5
First we have to pick a variable for the value we must find.
Let y = the number
The key word her is difference, so we must subtract.
x
x
8
−
=
3
5
5
The LCD = 15
15 x
x
8 15
∙( − )=
∙
1
3
5
5
1
15x
15x
120
−
=
3
5
5
5x − 3x
2x
2
x
=
=
=
=
We multiply on both sides with
the LCD. We must use the
Distributive Property as well.
Then we simplify the equation.
24
24
2
12
Example C After Sam got his weekly paycheck, he spent one-fourth of it on gas and
one-eighth on at the movies that night. He was left with $75. What was
his weekly paycheck?
When we approach this problem we should try to think about the
arithmetic operations we need to do to find the $75. Sam starts
with x number of dollars, then pays for gas so we should subtract.
After going to the movies his amount goes down again. So we must
once more subtract.
We can pick a variable to start with.
If we let x = the amount he is paid then
x
= the amount spent on gas and
4
x
= the amount spent on the movie.
8
So now we can write the equation
x−
The LCD = 8
x
x
−
= 75
4
8
Example C
(continued)
x−
x
x
−
= 75
4
8
8
x
x
8
∙ (x−
− ) = 75 ∙
1
4
8
1
We multiply on both sides with
the LCD. We must use the
Distributive Property as well.
8x
8x
−
=
4
8
Next we simplify the equation.
8x −
600
8x − 2x − x = 600
5x = 600
5
5
x = $120
Example D
A rectangle’s width is one-tenth of the perimeter. The length is
twelve centimeters more than the width. Find the perimeter of
the rectangle.
First we list the variable for the quantity we want to find.
Let p = the perimeter
Because this is a geometric problem we will draw and label a diagram.
p
 12
10
This is the length.
p
10
This is the width.
To find the perimeter we must add up al the sides.
p =
p
p
p
p
 12 +
 12 +
+
10
10
10 10
If we multiply all the terms by 10 we can clear the denominators.
10 ∙ p =
10 p
10 p
 4p
6p
6
(
p
p
p
p
 12 +
 12 +
+
) ∙ 10
10
10
10 10
1
= p  120 + p  120 + p + p
= 4 p  240
 4p
= 240
6
p = 40 cm
Our next example shows us how to use a scale factor.
Example E
The traveling Vietnam War memorial is built with a 3/5 scale. If the original
memorial is 500 ft. long and 10 ft. high find the dimensions of the replica.
When we have scale factors we will set up proportions. From the previous
section in Unit 10 we should remember how to set up a proportion. The
general format is given below.
When we set up a proportion we want set up two fractions that are equal.
The way we will do this is shown below.
The small value in the scale
=
The large value in the scale
The size of the small item
The size of the large item
When we apply it in this problem we will use it twice, once for
each dimension.
The small value in the scale
=
The large number in the scale
The small value in the scale
The large number in the scale
The length of the replica
The length of the original
=
The height of the replica
The height of the original
We need to pick variables for the values we are supposed to find.
Let l = length
Let w = width
Now we can put all our values and variables where they belong in the
proportion based on the descriptions above.
Finding the length
Finding the width
3
l

5 500
3 w

5 10
Next we cross multiply and solve the equations.
5l  1500
5w  30
5
5
5
l  300 ft
5
w  6 ft
Example F shows us how to a ratio for a specification.
Example F
Building codes are specifications that contractors must know in order to
construct safe and efficient structures. A building code in Phoenix could
specify the insolation requirements. A building code in Alaska might
specify the slant of a roof so that snow will slide off without damaging
someone's home. If the building code in Juno Alaska specifies that
residential roofs must have a slant given by the ratio of 1.1 ft of rise (up)
for every 2 ft of run (across), how tall must the peak of a roof be if the
width of the house is 42 ft? ( See the diagram below for how the
measurements are used. )
Distance to the peak
Width of house - the run is half the width.
The way we want to set up our proportion is similar to using a scale
factor. The general outline for how we should set up our proportion is
given below.
The ratio's value for the first quantity
The ratio's value for the second quantity
=
First quantity's value
Second quantity's value
We can use this outline to help us understand how to set up our
proportion.
The value in feet for the peak in the ratio
The value in feet for the run in the ratio
=
The value of the peak
The value of the run
We need to pick a variable for the value we are to find.
Let p = the height of the peak.
Now we can fill in our proportion based on the above description..
1.1
p

2
21
Now we cross multiply and solve the proportion.
2 p  23.1
2
2
p  11.55 ft
The next examples show us how percentage problems can be turned into proportions.
The key concept with percentages is that they represent a part out of one hundred.
This means the 35% can be written as a fraction of 35 .
100
When we create a proportion based on a percent we can set up our fractions as
shown below.
percent
part

100
whole
Example G
Find the number that is 80% of 600.
In this problem our variable x is the part. so we set up our proportion
as shown below.
80
x

100
600
48000 = 100x
100
100
We get this equation by cross multiplying.
480 = x
Example H
Juan sold 18 raffle tickets for the band fund raiser. If he had 30
tickets originally, what percentage has he sold.
In this problem we are to find the percent so x is the percent.
We should set up our proportion as shown below.
x
18

100
30
30x = 1800
30
30
We get this equation by cross multiplying.
x = 60%
Example I
Maria earned a 87.5% on the last test. If she got 42 questions correct,
how many questions were on the test.
In this problem we are to find the ‘whole’. ( How many questions were
on the whole test? ) We should set up our proportion as shown below.
87.5
42

100
x
87.5x = 4200
87.5
87.5
We get this equation by cross multiplying.
x = 48 questions
Distance problems are also at times best solved by using proportions. The example
below gives us a way to set up the proportions.
Example J
Jim flies from New York to Atlanta non-stop. He then flies home to
Phoenix. The planes flew at the same rate of speed on both parts of
his trip. The flight from Atlanta to Phoenix took 1 hour and 40 minutes
longer than the flight from New York to Atlanta. The distances are
New York to Atlanta 745 miles and Atlanta to Phoenix 1600 miles.
What is the time for the flight from New York to Atlanta?
There is a formula we must remember to solve this problem. The
formula relates distance, rate of speed, and time. The formula is “d = rt”.
We find the time so we let t = time from New York to Atlanta in
minutes. The time from Atlanta to Phoenix = t + 100. We chose the
time in minutes because an hour is 60 minutes and it’s easy to represent
the hour and 40 minutes as 60 minutes + 40 minutes.
Using the d=rt formula we can create expressions for both rates.
For New York to Atlanta d = rt
745 = r(t)
r = 745
t
For Atlanta to Phoenix
d = rt
1600 = r(t+100)
r = 1600
t  100
The key concept in the problem is that the speed is that same for
both flights. So we can generalize our equation as:
rate to Atlanta = rate to Phoenix
The we can set up our proportion with this concept.
745
1600

t
t  100
745t  74500  1600t
 745t
 745t
74500  855t
855  855
87.13 min = t
This answer converts to 1 hour and 27 minutes rounded to the nearest minute.
Rational equations can also be applied to work problems. The key idea with work
problems can be the a completed task represents one (1) job. The example below
is a sample work problem.
Example K
Jack is a painting contractor. He is going to work on a large project.
He has hired his friend Paul to help him. If Jack were to do the job
by himself it would take him 12 days. If Paul were to do the job
by himself it would take 15 days. How many days will it take the
two people working together to finish the job.
We have to find how many days it takes to do the job, so our variable
is d = number of days.
One key idea here is that we need to know how much of the job each
person finished in one day. Since it takes Jack 12 days he must be
finishing one-twelfth of the job each day. Paul will finish one-fifteenth
of the job.
Another key idea or word is “together” this tells us we should add the
parts that Jack and Paul do and get 1 full job done.
If Jack works 5 days he will get 5 ∙ 1 or 5 of the job done.
12
12
If Jack works ‘d’ days he will get d ∙ 1 or d of the job done.
12
12
The expressions for Paul will represented the same way.
Using these ideas our equation is:
d d

 1
12 15
Our LCD = 60
60
d d
∙( 
1
12 15
) = 1 ∙ 60
60d 60d

= 60
12
15
5d  4d = 60
9d = 60
9
9
d = 6.6 days
Exercises Unit 10 Section 11
Solve the following problems. Show the equations or proportions you created and
the work in solving the equations. List all answers with units where appropriate.
1. One-half a number plus one-third of the number is 15. Find the number.
2. The difference of one-fourth of a number and one-fifth of a number is five-fourths.
Find the number.
16
3. The difference between a number and its reciprocal is
. Find the number.
15
4. When a number and the reciprocal of the sum of the number plus one are
added together the total is one. Find the number.
5. The sum of the reciprocals of two consecutive odd integers is
12
. Find the integers.
35
6. A business has earned a profit for the current quarter. The owner decides to buy
new equipment with one-third of the profits and then pay his employees a bonus
with one-tenth of the profits. After this is done the owner puts the remaining
$6800 in the company’s checking account. What was the amount of the profit
originally?
7. In a recent election for student body offices three people ran for president.
Candidate A won with five-eighths of the vote, candidate B was second with
one-fourth of the votes. If candidate C got 61 votes how many people voted
in the election?
8. The first side of a triangle is one-fourth of the perimeter. The second side of the
triangle is one-third of the perimeter. The third side is 100 cm. Find the
perimeter of the triangle and the lengths of the other two sides.
9. A rectangle has a width that is one-eighth the perimeter. The length of the
rectangle is 14 cm more than the width. Find the perimeter and the dimensions
of the rectangle.
10. A triangle has sides of 12 cm, and 21 cm. The third side is one-fourth the
perimeter. Find the perimeter and the missing side.
11. An architect has built a scale model of a house for a client. The scale he used
was 1/20. If the model house has a rectangular shape and is 21 in. wide
by 39 in. long, what are the measurements in feet and inches for the dimensions
of the house?
12. A company is designing a new plane. They are currently testing a model in a wind
tunnel. The model is built with a 3/25 scale factor. If the wing span of the new
plane will be 32 m, what should the wingspan of the model be?
13. A map of the United States has a scale of 1 inch to 300 miles. How wide is the
state of Texas, if the distance from East to West on the map is 3 3 in.?
8
14. An artist is painting a school logo on a wall of a stadium. He has a picture to
of the logo. The scale factor he must work with is 4 cm on the picture will be
1 meter on the wall. If the logo in the picture is 22 cm tall, how tall will the
logo be on the wall?
15. In general the ratio of the height of a person to length of a person's arm
(from shoulder to tip of index finger) is 3/7. If a person is 6 ft. tall, how long
will his arm be in inches? (Round your answer to the nearest tenth of an inch.)
16. If the ratio of boys to girls in a class is 3/4 and there are 16 boys in the class,
how many girls are in the class and what is the total class size?
17. In general a high jumper using proper technique can clear a bar that has a
ratio of 8 cm for every 7 cm of his height. If an athlete is 195 cm tall, what
would we predict for the measurement of his jump? (Round your answer to
nearest tenth of a centimeter.)
18. A typical solar panel produces about 31 volts for every 8 ft2. Maria needs a
system that can produce 1800 volts to run her business. How many square
feet must she use? (Round your answer to the nearest square foot.)
19. Find the number that is 35% of 104.
20. 22 is what percent of 65. ( Round your answer to the nearest tenth of a percent.)
21. 32 is 8% of what number?
22. Alice has a bowling average of 136. Her last game was 152. What percent is
the 152 of her average? (Round your answer to the nearest hundredth.)
23. Sam is laying a block fence. He is 55% done. If the fence is going to be
480' long, how much of the fence has he completed?
24. Mike is paying off his new truck with monthly payments. He has made 18 payments
of $215. He has made 40% of the payments. What is the total amount of money
he will pay on the truck?
25. In the morning Juan hikes up a trail to a mountain pass. After lunch he hikes back
to his starting point. Juan can go 1 km per hour faster going downhill than he
can going uphill. The distance on the trail is 10 km. The total time he spent
hiking (not including his break for lunch) was 6 hours and 51 minutes. Find
the rate at which he was able to hike up hill.
26. Jason is an avid bicyclist. On his day off he starts a long ride. In the morning
he took a scenic route to visit a friend. If the afternoon he took a more direct route
back home. The route back home was 12 miles shorter. He was able to ride at
18 mph for the full trip. If the total time traveled was 3 hours and 40 minutes,
what was the distance he rode in the morning?
27. Sam is a contractor who has been hired to build a block fence around a property.
If Sam works by himself it will take ten days to finish. He hires Alex to help
him. Alex could finish the fence by himself in 12 days.
a. How much of the fence can each person finish in one day?
b. How long will it take for the two people to finish the fence together?
(Round your answer to the nearest tenth of a day.)
28. Pump A can drain a large tank in 16 hours. Pump B can drain the tank in
10 hours. How long will it take to drain the tank if we use both pumps?
(Round your answer to the nearest minute.)
29. Paul is counting the votes in a recent election. If he were to do all the counting
by himself it would take 130 minutes. Sarah is helping him. Together they count
all the votes in 70 minutes. How long would it take Sarah to count the votes all
by herself?