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A lesson on complex fractions and their simplification. It includes steps for simplifying complex fractions, examples of complex fraction simplification, and important concepts such as the negative exponent rule.
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Simplifying complex fractions is basically just a
combination of the concepts from the previous three
lessons.
The rational expressions in the numerator and/or
denominator of the complex fraction need to be added
or subtracted first (Lesson 8).
Then the complex fraction gets converted to two
rational expressions being divided, which we don’t
actually divide at all, we simply convert to
multiplication by taking the reciprocal of the divisor
(Lesson 8).
When two fractions are being multiplied, we write
numerator times numerator and denominator times
denominator, but we don’t actually multiply them
because we need to cancel common factors (Lesson
7). Once we factor completely, we can simplify the
rational expressions by canceling common factors
(Lesson 7). Remember that the numerator and
denominator of the rational expression need to be
factored completely in order to simplify, and
factoring we covered in Lessons 6 and 7.
The ability to synthesize all the information from the
previous three lessons is imperative to being able to
simplify complex fractions.
It is imperative that you understand how to simplify, multiply, divide,
add, and subtract rational expressions, as well as how to factor
polynomials, in order to simplify complex fractions. All of those
concepts are combined into one problem when simplifying complex
fractions, and the inability to perform any of those tasks will prevent
you from correctly answering these types of problems.
Example 1 : Perform the indicated operations and simplify the expressions
completely.
a.
1
𝑥
−
1
2
𝑥
2
− 4
b.
𝑥
𝑦
2
−
𝑦
𝑥
2
1
𝑦
2
−
1
𝑥
2
b.
e.
1
𝑥 + ℎ
−
1
𝑥
ℎ
f.
f. 𝑎
𝑥
2
𝑥
2
1
(𝑥+ℎ)
2
1
𝑥
2
(𝑥+ℎ)
2
(𝑥+ℎ)
2
𝑥
2
𝑥
2
(𝑥+ℎ)
2
(𝑥+ℎ)
2
𝑥
2
(𝑥+ℎ)
2
𝑥
2
− (𝑥+ℎ)
2
𝑥
2
(𝑥+ℎ)
2
𝑥
2
− (𝑥+ℎ)(𝑥+ℎ)
𝑥
2
(𝑥+ℎ)
2
𝑥
2
− (𝑥
2
2
)
𝑥
2
(𝑥+ℎ)
2
𝑥
2
−𝑥
2
− 2 𝑥ℎ−ℎ
2
𝑥
2
( 𝑥+ℎ
)
2
− 2 𝑥ℎ−ℎ
2
𝑥
2
(𝑥+ℎ)
2
2
2
2
÷ ℎ
2
2
2
∙
2
2
2
2
−𝟐𝒙−𝒉
𝒙
( 𝒙+𝒉
)
3
3
2
c.
− 2
− 2
2
d.
− 1
− 2
− 2
− 2
d. a
1
𝑥
2
1
( 𝑥− 1
)
2
1
(𝑥− 1 )
2
(𝑥− 1 )
2
(𝑥− 1 )
2
1
𝑥
2
1
(𝑥− 1 )
2
𝑥
2
𝑥
2
1
( 𝑥− 1
)
2
(𝑥− 1 )
2
−𝑥
2
𝑥
2
( 𝑥− 1
)
2
1
(𝑥− 1 )
2
(𝑥− 1 )(𝑥− 1 )−𝑥
2
𝑥
2
(𝑥− 1 )
2
1
(𝑥− 1 )
2
𝑥
2
− 2 𝑥+ 1 −𝑥
2
𝑥
2
(𝑥− 1 )
2
1
( 𝑥− 1
)
2
1 − 2 𝑥
𝑥
2
(𝑥− 1 )
2
1
(𝑥− 1 )
2
2
2
÷
2
2
2
∙
2
2
2
2
𝟏−𝟐𝒙
𝒙
Answers to Examples:
1a. −
; 1b.
2
2
; 1 c. −
; 1 d.
2
2
2
;
1 e. −
; 1 f. −
2
2
; 2a.
; 2b. −
; 2 c.
2
;
2 d.
;