SS6.1 Simplifying Rational Expressions
Recall that a rational number is a quotient of integers. A rational expression
is a quotient of polynomials, such as:
P
Q
where Q0 and P and Q are polynomials
A rational expression can be evaluated just as any polynomial, except that a
rational expression can be undefined when the denominator is equal to
zero.
Example: Evaluate
x2  3
x + 9
Example: For what number(s) is
when x = 5
x2  3
x2  3x  10
undefined
1) Find zeros of the polynomial in the denominator, by solving as a
quadratic
Example: Are there any zeros for
x2 + 3x  10
x2 + 5
?
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Example: Evaluate
x2  3
x + 9
when x = -9
Recall that
-a = a = _ a when b0
b
-b
b
This is also true of polynomials. My suggestion is to simply pull any
negative signs out and consider the entire expression as either a positive or
negative expression.
Example:
-(x + 3) = _
x + 3
2
2
x  3x + 2
x  3x + 2
Just as when we were dealing with fractions, if you multiply the
numerator and denominator by the same thing the resulting expression
is equivalent. This is called the Fundamental Principle of Rational
Expressions, when we are discussing a fraction of polynomials.
PR = P
QR
Q
Concept Example:
if P, Q and R are polynomials and
Q&R0
15 = 3  5 = 3
35
7  5
7
In order to simplify rational expressions we will use the Fundamental
Principle of Rational Expressions just as we used the Fundamental Principle
of Fractions to simplify fractions.
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Steps to Simplifying a Rational Expression
1) Factor the numerator and the denominator completely
2) Cancel common factors
15x2y
35xy2
Recall that another way of using the principle is division!
Example:
Simplify
Example:
Simplify
x2
x2 + 2x
Example:
Simplify
x + 5
x + 2x  15
2
Example:
Simplify
x2  x  2
x2 + 5x  14
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Example:
Simplify
Your Turn
Example: Simplify
Example:
Simplify
-x + y
x  y
x3 + 8
x2  5x  14
- x2  3x  2
x2  x  2
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SS6.2 Multiplying and Dividing Rational Expressions
Steps for Multiplying Rational Expressions
1) Factor numerators and denominators completely
2) Cancel if possible
3) Multiply numerators and denominators
Concept Example:
Example:
2  3 = 1
3
8
4
x + 2

x + 7x + 6
2
Example:
Your Turn
Example:
x + 1
x3 + 8
7x2  -3y
15y3
14x3
x2  5x  14
x3  1
 x2  8x + 7
x2 + x  2
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Example:
-7x  -6 y4
24y3
15x5
Steps for Dividing Rational Expressions
1) Take the reciprocal of the divisor.
2) Multiply the dividend and the reciprocal of the divisor
Concept Example:
Example:
1  2 = 1  3 = 3
2
3
2
2
4
x + 1

2
2x  x + 1
3x2  3
x2 3 x + 2
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Example:
Your Turn
Example:
4x2  16
x + 1
x2 + 8
x2  16


8x3  64
x2 + 9x + 8
x3  10
x2  9x + 20
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SS6.3 Adding and Subtracting Rational Expressions w/ Common
Denominators
When we have a common denominator, as with fractions, we simply add or
subtract the numerators. We do have to be cautious because when
subtracting it is the entire numerator that's subtracted, so we must use the
distributive property to subtract.
Concept Example: 7  3 = 4
15
15
15
Example:
2y
7n
+ 3 =
7n
Example:
2x + 5  x  5 =
y + 2
y + 2
Just as with fractions sometimes we will need to reduce our answer before
we are finished.
Concept Example: 2 +
15
Example:
4 = 6 = 2
15
15
5
y

2
=
y  2
y  2
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Your Turn
Example:
x2 + 1 +
3x
x + 2
x + 2
=
This leads us to the point that we always need to factor the numerator and
denominator in order to reduce and also for our next application of finding
a common denominator.
Finding the LCD of a rational expression is the same as finding the LCD of
fractions. We just need to remember that a polynomial must first be
factored. Each factor that isn't a constant is considered like a prime number.
Concept Example: Find the LCD of
2 , 7
15
36
Example:
Find the LCD of
1 , 2x + 5
x
x(x + 5)
Example:
Find the LCD of
x + 2
, x + 1
x + 3x + 2
x + 2
2
1) Factor x2 + 3x + 3
2) LCD
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Example:
Find LCD of
2
,
x + 1
2
4x + 13x + 3 x + 6x + 9
2
1) Factor 4x2 + 13x + 3
2) Factor x2 + 6x + 9
3) LCD
Your Turn
Example: Find LCD of
2x
2
x + 2x + 1
,
x2 + 1
x2  1
Next, we need to build a higher term so that we can add and subtract
rational expressions with unlike denominators.
Building A Higher Term
1) Find LCD
2) Divide LCD by denominator
3) Multiply numerator by quotient
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Concept Example: Build the higher term
2 , 7
15
36
1) LCD was 180
2) 180  15=
180  36 =
3) 2 
15
=
7 
36
Example:
Find the LCD of
180
=
180
1 , 2x + 5
x
x(x + 5)
1) LCD was x(x + 5)
2) x(x + 5) =
x
3) 1 
x
x(x + 5) =
x(x + 5)
=
2x + 5
x(x + 5)
x(x + 5)
Your Turn
Example: Find LCD of
2
,
x + 1
2
4x + 13x + 3 x + 6x + 9
2
1) LCD was (4x + 1)(x + 3)2
2)
(4x + 1)(x + 3)2 =
(4x + 1)(x + 3)
3)
2

(4x + 1)(x + 3)
x + 1
(x + 3)2

(4x + 1)(x + 3)2 =
(x + 3)2
=
(4x + 1)(x + 3)2=
=
(4x + 1)(x + 3)2
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SS6.4 Adding & Subtracting Rational Expressions w/ Unlike
Denominators
Addition and Subtraction of Rational Expressions w/ Unlike Denominators
1) Find LCD
2) Build Higher Term
3) Add or Subtract as normal
Example:
15a + 6b
b
5
1) LCD
2) Build Higher Term
3) Add
Example:
15
+
x
2
2x  4
x  4
1) LCD
2) Build Higher Term
3) Add
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Example:
x

2
x  4
4  x2
2
1) LCD
2) Build Higher Term
3) Subtract
Example: y + 2
y + 3
 2
1) LCD
2) Build the higher term
3) Subtract
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Example:
x 
2
2
x  1 x  2x + 1
2
1) LCD
2) Build Higher Term
3) Subtract
Your Turn
Example:
2x
2
x + 2x + 1
+
x2 + 1
x2  1
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SS6.5 Simplifying Complex Fractions
A complex fraction is a fraction with an expression in the numerator and an
expression in the denominator.
Simplifying Complex Fractions Method #1
1) Solve or simplify the problem in the numerator
2) Solve or simplify the problem in the denominator
3) Divide the numerator by the denominator
4) Reduce
Example:
3
4
----------------2
3
Example:
6x  3
5x2
------------------------2x  1
10x
Example:
1 + 2
2
3
------------------------5  5
9
6
1) Solve numerator
1 + 2
2
3
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2) Solve denominator
5  5
9
6
3) Divide & Reduce
Example:
1  1
5
x
------------------------7 + 1
10
x2
1) Simplify numerator
1  1
5
x
2) Simplify denominator
7 + 1
10 x2
3) Divide numerator by denominator
4) Reduce
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Simplifying Complex Fractions Method #2
1) Find the LCD of the numerator and the denominator fractions
2) Multiply numerator and denominator by LCD
3) Simplify resulting fraction
Example:
3
4
----------------2
3
1) LCD of 3 and 4
2) Multiply
3
4
----------------2
3
3) Simplify
6x  3
5x2
------------------------2x  1
10x
2
1) LCD of 5x and 10x
Example:
2) Multiply
6x  3
5x2
------------------------2x  1
10x
3) Simplify
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Example:
1 + 2
2
3
------------------------5  5
9
6
1) Find LCD of
2) Multiply
2,3, 9, 6
1 + 2
2
3
------------------------5  5
9
6
3) Simplify
Example:
1  1
5
x
------------------------7 + 1
10
x2
1) Find LCD of 5, x, 10 and x2
2) Multiply
1  1
5
x
------------------------7 + 1
10
x2
3) Simplify
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SS6.6 Solving Equations Containing Rational Expressions
The only difference between solving rational expression equations and
regular equations is that all solutions must be checked to make sure that it
does not make the denominator of the original expression zero. If one of the
solutions makes the denominator zero, it is called an extraneous solution.
Steps
1) Find the LCD of the denominators in the equation
2) Multiply both sides by the LCD
3) Solve as usual
4) Check for extraneous solutions
Example:
9
= -2
2a  5
1) Find LCD
2) Multiply
3) Solve
4) Check for extraneous solutions
9
= -2
2a  5
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Example:
7 + 1 = x
x  2 x + 2
1) Find LCD
2) Multiply
3) Solve
4) Check for extraneous solutions
Examples:
y
+
2y + 2
2y  16 = 2y  3
4y + 4
y + 1
1) Find LCD
2) Multiply
3) Solve
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4) Check for extraneous solutions
2x  2 = x  8
x + 2
x  2
1) Find LCD
Example:
2) Multiply
3) Solve
4) Check for extraneous solutions
In the next examples, we will want to solve for a variable when there are two
variables in an equation.
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Example: Solve for b:
3a + 2 = -4
3b  2
2a
1) Find LCD
2) Multiply
3) Solve for b
Example:
Solve for n
m2  n = p
6
3
2
1) Find LCD
2) Multiply
3) Solve for n
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SS6.7 Ratios and Proportions
A ratio is a quotient of two numbers where the divisor isn't zero. A ratio is
stated as: a to b
a : b or a where a & b are whole numbers and b0
b
A proportion is a mathematical statement that two ratios are equal.
2 = 4
3
6
is a proportion
It is read as: 2 is to 3 as 4 is to 6
The numbers on the diagonal from left to right, 2 and 6, are called the
extremes
The numbers on the diagonal from right to left, 4 and 3, are called the
means
If we have a true proportion, then the product of the means equals the
product of the extremes.
These are also called the cross products and finding the product of the means
and extremes is called cross multiplying.
Example: Find the cross products of the following to show that this is a true
proportion
27 = 3
72
8
The idea that in a true proportion, the cross products are equal is used to
solve for unknowns!
This is strictly a review of material covered as a supplement to Chapter
1. The only problem type that was not covered at that time was the
following:
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Example:
x + 1 = 2
3
3x
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SS6.8 Rational Expressions and Problem Solving
This section is on word problems. In order to do well with word problems
you must:
1) Read the problem and understand what is being said and asked
2) Write down all the information that is important
3) Assign a variable to an unknown and put all other unknowns in
terms of this one
4) Translate any equations given in the problem
5) Create your final algebraic expression
6) Solve
7) Check
In this section, the only catch is that we are working in parts of things, the
distance formula where time is the key or with similar triangles. Therefore,
we will have rational expressions coming into our equation.
Example: Three times the reciprocal of a number equals 9 times the
reciprocal of 6. Find the number.
Example: In 2 minutes, a conveyor belt moves 300 pounds of tin from the
delivery truck to a storage area. A smaller belt moves the same amount the
same distance in 6 minutes. If both belts are used, find how long it takes to
move the cans to the storage area.
Minutes
Part of Job in 1 Minute
Large
Small
Together
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Example: A cyclist rode the first 20-mile portion of his workout at a
constant speed. For the 16-mile cooldown portion of his workout, he
reduced his speed by 2 miles per hour. Each portion of the workout took the
same time. Find the cyclist's speed during the first portion and find his
speed during the cooldown portion.
Rate 
Distance =
Time
Similar Triangles are triangles that have proportional side lengths and the
same measure in angles
A
D
B
E
F
C
AB : DE
as
BC : EF
If we don't know the length of a side of a triangle and we know that the
triangles are similar, then we can use proportions to solve the problem. The
same is true with any figures that are said to be similar.
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Example: See #13 on p. 396
Example: See #35 on p. 396
In this section there are four patterns in the problems as shown by the above
examples. Look for these patterns and solve all problems that follow the
pattern just as I have solved these problems.
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