Download Module - Algebraic Fractions

Algebra
Module A35
Algebraic
Fractions
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED November, 2008
Algebraic Fractions
Statement of Prerequisite Skills
Complete all previous TLM modules before completing this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
An important component of solving algebraic problems is an understanding of algebraic
fractions. This module will help the student build the skills necessary to deal with
problems involving variables in the numerator, denominator, or both.
Learning Outcome
When you complete this module you will be able to…
Solve problems involving algebraic fractions.
Learning Objectives
1.
2.
3.
4.
5.
6.
Perform the operation of reducing fractions to lowest terms.
Identify the least common multiple for given terms.
Perform the operation of addition and subtraction of fractions.
Perform the operation of multiplication and division of fractions.
Perform the operation of simplifying complex fractions.
Solve linear equations involving algebraic fractions.
Connection Activity
Consider the problem
3
× 2 . You know from previous experience that this simplifies to 3
2
as the 2 in the numerator and the denominator divide and leave 1. The problem is
simplified before it is solved. The same principles apply to algebraic fractions.
a
× b = a.
b
The b’s cancel to simplify the problem. This module will introduce the student to more
advanced methods of simplifying and solving algebraic fractions.
1
Module A35 − Algebraic Fractions
OBJECTIVE ONE
When you complete this objective you will be able to…
Perform the operation of reducing fractions to lowest terms.
Exploration Activity
A fraction is defined as any number that can be written as a quotient of two integers. In a
similar way we define a fractional expression as any algebraic expression that can be
written with both a numerator and denominator.
When we deal with fractional expressions, we must be able to work effectively with
fractions. Since algebraic expressions are representations of numbers, the basic
operations of fractions will continue to form the basis of our algebraic operations.
NOTE:
One of the most important properties of fractions, often referred to as the Fundamental
Principle of Fractions, states, THE VALUE OF A FRACTION IS UNCHANGED IF
BOTH THE NUMERATOR AND DENOMINATOR ARE MULTIPLIED OR
DIVIDED BY THE SAME NUMBER PROVIDED THE NUMBER IS NOT ZERO.
EXAMPLE 1
Write 3 fractions equal to
a)
b)
c)
3
.
2
3 4 12
⋅ =
2 4 8
3 a 3a
⋅ =
2 a 2a
3 4 x 12 x
⋅
=
2 4 x 8x
Each of these examples obeys the above mentioned Fundamental Principle of Fractions.
2
Module A35 − Algebraic Fractions
EXAMPLE 2
Write a fraction equal to
a
with a numerator of a2 b.
3
In such a problem we make the numerator the desired expression, and the denominator is
changed accordingly since we always multiply the numerator and denominator by the
same value to make equivalent fractions.
Since a2 b = a(ab), we must multiply numerator and denominator of
a
by ab,
3
2
⎛ a ⎞ ab a b
=
⎜ ⎟
⎝ 3 ⎠ ab 3ab
Similarly we can use the Fundamental Theorem to reduce fractions to simpler forms.
EXAMPLE 3
14
21
Reduce
SOLUTION:
Since 7 is a common factor of the numerator and denominator, i.e.
be divided out so
14 7 ⋅ 2
=
the 7’s can
21 7 ⋅ 3
14 2
= .
21 3
EXAMPLE 4
Reduce the fraction
6x2 y3
to lowest terms.
12 x 4 y 2
SOLUTION:
This becomes:
6 ⋅ x2 ⋅ y2 ⋅ y
6 ⋅ 2 ⋅ x2 ⋅ x2 ⋅ y2
Divide out common factors 6, x2, y2 and get:
=
y
2x2
3
Module A35 − Algebraic Fractions
EXAMPLE 5
Reduce the fraction
a 2 − ab
2a − 2b
SOLUTION:
Step 1
Factor the numerator and denominator, and get,
a ( a − b)
2(a − b)
Step 2
Divide the numerator and denominator by (a − b), and get
a ( a − b) a
=
2(a − b) 2
EXAMPLE 6
Reduce the fraction
x2 − 9
2 ( x + 3)
SOLUTION:
Step 1
Factor the numerator and denominator, and get,
( x − 3)( x + 3)
2 ( x + 3)
Step 2
Divide the numerator and denominator by (x + 3), and get
( x − 3)
2
4
Module A35 − Algebraic Fractions
Experiential Activity One
1. Determine which of the following are true:
a.
b.
c.
3 1
=
12 4
20 2
=
30 3
−7 28
=
4
46
d.
6 − 12 x 2
=
7
14 x 2
e.
15a 2b − 15a 2
=
9ab
− 9a
2. Reduce to lowest terms:
a.
c.
e.
g.
8
12
17
68
150
400
222
74
b.
d.
f.
h.
28
35
23
69
180
600
7y
28 z
6a 2
3a
2x + 2 y
x+ y
i.
a 2c
a 3c 2
j.
k.
32 x 2 y 2 z
48 x 2 y 3 z 2
l.
m.
2a 2 + 4a
6a 2
n.
cy
c 2 y 2 − cy
o.
a 2 + ab
2ab
p.
y 2 − 10 y
y2 − 5y
q.
x2 + 6x
x 2 + 3x
r.
28a 7b 9 cd 2
42a 9b 7 cd
Show Me
5
Module A35 − Algebraic Fractions
Experiential Activity One Answers
1.
a.
b.
c.
d.
e.
True
True
False
False
True
2.
a.
c.
e.
2
3
1
4
3
8
g. 3
1
ac
2
k.
3 yz
a+2
m.
3a
a+b
o.
2b
x+6
q.
x+3
i.
b.
d.
f.
h.
4
5
1
3
3
10
y
4z
j.
2a
l.
2
n.
p.
r.
1
cy − 1
y − 10
y −5
2b 2 d
3a 2
6
Module A35 − Algebraic Fractions
OBJECTIVE TWO
When you complete this objective you will be able to…
Identify the least common multiple for given terms.
Exploration Activity
Definition
The least common multiple (L.C.M. ) of two or more numbers is the smallest number
which is evenly divisible by each of the numbers.
EXAMPLE 1
What is the L.C.M. of the numbers 6, 15, and 30?
The L.C.M. must have the factors of 6, 15 and 30. The First number that has all of these
factors of these numbers is 30.
EXAMPLE 2
What is the L.C.M. of 6a2, 3ab3, and 4a2b?
The required multiple must be divisible by each of the given numbers. Hence it must
have each of these numbers as a factor. We first write each of the given numbers in
factored form:
6a2 = 2 · 3 · a · a
3ab3 = 3 · a · b · b · b
4a2b = 2 · 2 · a · a · b
We now see that in the L.C.M. the factors 2, 3, a, and a must appear in order that 6a2 be a
divisor. Similarly, we need the factors 3, a, b, b, and b in order that 3ab3 be a divisor; and
we need the factors, 2, 2, a, a, and b in order that 4a2b be a divisor. Thus in the L.C.M.
we need the factors
2,3,a,a,b,b,b,2 or ordering them
2, 2, 3, a, a, b, b, and b
that is, the L.C.M. is 12a2b3
Often the necessary factors can be determined by inspection. Then it is NOT necessary to
write the factorizations of all the numbers.
Complete exercise 2A now.
When combining fractions by addition or subtraction we need to find a common
denominator. The lowest common denominator is the lowest common multiple of the
denominators of the fractions.
The practice you just received in finding the L.C.M. will be helpful now in finding the
lowest common denominator (L.C.D.) for a number of fractions
7
Module A35 − Algebraic Fractions
EXAMPLE 3
Combine:
1 2 3
+ +
2 3 4
SOLUTION:
Here the L.C.D. for 2, 3, and 4 is 12. Therefore, each fraction must have its denominator
made into a 12.
For example:
1
6
becomes
by multiplying both the denominator and the numerator by 6.
2
12
2
8
becomes
here we multiply both denominator and numerator by 4.
3
12
3
9
becomes
here we multiply both denominator and numerator by 3.
4
12
so
1 2 3
6
8
9
+ + becomes
+ +
. Here each fraction has the same denominator.
2 3 4
12 12 12
So
6 8 9 23
+ + =
12 12 12 12
EXAMPLE 4
Find the L.C.D. for
3 2b 3a
+
+
ab a 2 b 2
and rewrite each fraction with the L.C.D.
ab has factors a and b
a2 has factors a and a
b2 has factors b and b
L.C.D. is a2b2 since a occurs twice in a2, and b occurs twice in b2.
8
Module A35 − Algebraic Fractions
Again, note that a2 b2 contains ab, a2, and b2 as factors.
Now,
3 ab 3ab
3
becomes
⋅
=
ab ab a 2b 2
ab
2b
2b b 2
2b 3
⋅
=
becomes
a2
a 2 b 2 a 2b 2
3a
3a a 2
3a 3
⋅
=
becomes
b2
b 2 a 2 a 2b 2
i.e., each of the fractions had its denominator multiplied by whatever was necessary to
convert it to the L.C.D. The numerator gets multiplied by the same quantity to keep the
fraction “balanced” or equivalent to the initial fraction.
So:
3 2b 3a
3ab
2b 3
3a 3
+ 2 + 2 becomes 2 2 + 2 2 + 2 2 which is an equivalent expression.
ab a
b
a b
a b
a b
Complete exercise 2B now.
EXAMPLE 4
Combine:
1 2 3
+ −
x y x2
SOLUTION:
The L.C.D. is x 2 y so rewrite each fraction with x 2 y as the denominator.
=
2 xy 3x 2 3 y
+
−
x2 y x2 y x2 y
=
2 xy + 3x 2 − 3 y
x2 y
9
Module A35 − Algebraic Fractions
EXAMPLE 5
Combine:
2
2
+
x+ y x− y
SOLUTION:
The L.C.D. is ( x + y )( x − y ) so rewrite each fraction with ( x + y )( x − y ) as the
denominator.
=
=
=
=
2( x − y)
( x + y )( x − y )
2x − 2 y
+
2( x + y)
( x − y )( x + y )
+
2x + 2 y
( x + y )( x − y ) ( x − y )( x + y )
2x − 2 y + 2x + 2 y
( x + y )( x − y )
4x
( x + y )( x − y )
10
Module A35 − Algebraic Fractions
Experiential Activity Two
Exercise 2A:
Find a L.C.M. for each of the following:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
3,7, 9
2, 5, 7, 15
30, 65, 39
3x, 2y, z
2xy, y2, z2
6ab3, ab2, b3, 3a
24xy2, 18x2y, 54x3
6(a − b), 12(a + b)
2(x + y), 3(x − y), 5(x + y)
5(x − 7), 4(x − y)2, 2x
Show Me.
Exercise 2B:
In each of the following write the L.C.D. and then rewrite each fraction with the L.C.D.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
1 1 3
+ +
3 4 5
2 3 x
+ +
x y y
2x y z
+ −
7 5 R
1 1
1
+ +
8 12 18
1
2
−
2 x 3x
3 1+ x
+
7 x 14 x 2
a−b
b
+ 2
a
a b
x y z
3
+ + +
y z a y2z
1
2
1
−
+ 2
2
ab b
a
x
1
1+ + 2
x x z
1
1−
1− x
4
3
−
Show Me.
a +b a −b
1 + 3x 1 − 3x
−
1 − 3x 1 + 3x
x
x+2
1
Hint: Factor x2 – 2x
−
+ 2
x−2
x
x − 2x
11
Module A35 − Algebraic Fractions
Experiential Activity Two Answers
Exercise 2A Answers:
1. 63
2. 210
3. 390
4. 6xyz
5. 2xy2z2
6. 6ab3
7. 216x3y2
8. 12(a + b)(a − b)
9. 30(x + y)(x − y)
10. 20(x)(x − 7)(x − y)2
Exercise 2B Answers:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
60;
71
60
x 2 + 3x + 2 y
xy
10 Rx + 7 Ry − 35 z
35R;
35R
19
72;
72
−1
6 x;
6x
2 7x + 1
14 x ;
14 x 2
a 2 b − ab 2 + b
a 2 − ab + 1
a 2 b;
which
reduces
to
a 2b
a2
3
2 2
3a + axyz + ay + y z
ay 2 z;
ay 2 z
xy;
a 2 − 2ab + b 2
a 2b 2
x 2 + xz + x
xz + z + 1
x 2 z;
which reduces to
xz
x2 z
−x
1 − x;
1− x
a − 7b
(a + b)(a − b);
(a + b)(a − b)
12 x
(1 − 3 x)(1 + 3 x);
1 − 9x2
5
x( x − 2); 2
x − 2x
a 2b 2 ;
12
Module A35 − Algebraic Fractions
OBJECTIVE THREE
When you complete this objective you will be able to…
Perform the operation of addition and subtraction of fractions.
Exploration Activity
From arithmetic we recall that the sum of a set of fractions which all have the same
denominator is the sum of the numerators divided by the common denominator. Since
algebraic expressions represent numbers, this fact is also true in algebra.
To Add or Subtract Fractions:
1. Reduce each fraction to lowest terms.
2. Factor each denominator.
3. Find the L.C.D.
4. Rewrite each fraction with the L.C.D.
5. Add or subtract the numerators.
6. Reduce the resulting fractions to lowest terms.
EXAMPLE 1
Add
2 3
+ .
x y
SOLUTION:
Here the L.C.D. is xy.
Therefore
y
2
3
x
is multiplied by , and
is multiplied by
x
y
y
x
2 y 3 x
⋅ + ⋅
x y y x
2 y 3x
=
+
xy xy
2 y + 3x
=
xy
=
Remember: The L.C.D. must be divisible by each of the original denominators. Use this
fact as a check from time to time.
13
Module A35 − Algebraic Fractions
EXAMPLE 2
Simplify:
4 xy
15 y
4x
+
+
2 2
2
24 x y
36 xy
60 x 2 y
SOLUTION:
Reduce fractions to lowest terms to obtain:
1
5
1
+
+
6 xy 12 xy 15 xy
Factor each denominator
1
5
1
(1)
+
+
2 ⋅ 3xy 2 23 xy 3 ⋅ 5 xy
Identify the L.C.D. as 22 · 3 · 5xy = 60xy and (1) becomes:
1(10) + 5(5) + 1(4)
=
60 xy
39
=
60 xy
13(3)
=
← divide out the 3' s to get :
20 xy (3)
13
=
20 xy
EXAMPLE 3
2
4
5
+
−
2
3
3S
3R
RS
Combine:
SOLUTION:
Since the L.C.D. must be factorable (divisible) by each of the original denominators, our
L.C.D.= 3R2S3.
Adjust each fraction according to what the denominator needs to become equal to the
L.C.D.
=
( S 3 )2
3(4) R R 2 S 2 (5)
+
−
( S 3 )3R 2 3R 2 S 3 R 2 S 2 (3S )
2S 3 + 12 R − 5 R 2 S 2
3R 2 S 3
− 5R 2 S 2 + 12 R + 3S 3
TLM answer =
3R 2 S 3
=
14
Module A35 − Algebraic Fractions
EXAMPLE 4
Perform the indicated operations and simplify:
t
1
2t
+ −
4t − 12 6 5t − 15
SOLUTION:
Step 1:
Factor the denominators to obtain:
t
1
2t
+ −
4(t − 3) 6 5(t − 3)
Step 2:
Identify a common denominator as 12 · 5(t − 3), which equals 60(t − 3).
Step 3:
Rewrite each term with the new common denominator to get:
15t + 1(10)(t − 3) − 24t
60(t − 3)
Step 4:
Simplify the numerator:
15t + 10t − 30 − 24t
60(t − 3)
t − 30
=
60(t − 3)
Before we go on to Objective 4 it is appropriate at this time to show a technique that is
useful in certain types of situations involving fractions.
If we are given the factor (x − 1) and the factor (1 − x) you shall see a relationship
between them. It is that the signs of one factor are the reverse of the signs of the other. In
other words if we factor −1 from either (x − 1) or (1 − x) we will get the following:
(x – 1) = −1(1 − x) or
(1 − x) = −1(−1 + x) = −1(x − 1)
That is, each factor, (1 − x) and (x − 1) is the negative of the other. This is helpful in some
situations.
15
Module A35 − Algebraic Fractions
EXAMPLE 5
The operation
2
3
can be performed very simply by using the above information:
+
1− x x −1
Factor −1 from the first denominator to get:
=
2
3
+
− 1( x − 1) x − 1
Put the − sign into the numerator so both denominators are the same
3
−2
+
( x − 1) x − 1
1
−2 + 3
=
=
x −1 x −1
=
This method is short and handy to use.
We will try this again with 2 more fractions:
5
10
−
Å same terms in the denominators, just opposite signs
y−x x− y
−5
10
Å remove the − sign and place it in the numerator
=
−
x− y x− y
−15
=
x− y
*Remember: This technique works only when two factors have identical terms and
opposite signs.
EXAMPLE 6
Perform the indicated operations and simplify:
x −1 3 − x
−
2x
xy
=
=
=
y ( x − 1)
2 xy
−
2 (3 − x )
2 xy
xy − y 6 − 2 x
−
2 xy
2 xy
( xy − y ) − (6 − 2 x)
2 xy
Å Make sure you use brackets and distribute the negative sign!!!
16
Module A35 − Algebraic Fractions
Experiential Activity Three
1.
3 6
+
5 5
2.
2
6
+
13 13
3.
1 7
+
x x
4.
2 3
+
a a
5.
1 3
+
2 4
6.
5 1
−
9 3
7.
3 7a
+
4x 4
8.
t −3 t
−
a
2a
9.
a b
−
x x2
10.
2 3
+
s2 5
11.
6
a
+
3
25 x
5x
12.
2 3
+
s2 s
13.
2 1 a
+ −
5a a 10
14.
2 6 9
− −
a b c
15.
x +1 x − 3 2 − x
−
−
x
y
xy
16. 5 +
17.
3
1
+
2x − 1 4x − 2
18.
5
a
−
6y + 3 8y + 4
19.
4
3
−
x( x + 1) 2 x
20.
3
1
−
ax + ay a 2
21.
s
1
3s
+ −
2 s − 6 4 4s − 12
1− x 3 + x
−
2
4
Show Me.
17
Module A35 − Algebraic Fractions
Experiential Activity Three Answers
1.
9
5
2.
8
13
3.
8
x
4.
5
a
5.
5
4
6.
2
9
7.
7ax + 3
4x
8.
t −6
2a
9.
ax − b
x2
10.
3s 2 + 10
5s 2
11.
ax 2 + 30
25 x 3
12.
3s + 2
s2
13.
− a 2 + 14
10a
14.
−9ab − 6ac + 2bc
abc
15.
− x 2 + 4 x + xy + y − 2
xy
16.
−3x + 19
4
17.
7
4x − 2
18.
−3a + 20
12(2 y + 1)
19.
−3 x + 5
2 x( x + 1)
20.
3a − x − y
a 2 ( x + y)
21.
−3
4( s − 3)
18
Module A35 − Algebraic Fractions
OBJECTIVE FOUR
When you complete this objective you will be able to…
Perform the operation of multiplication and division of fractions.
Exploration Activity
From arithmetic we recall that the product of two fractions is a fraction whose numerator
is the product of the numerators and whose denominator is the product of the
denominators of the given fractions. Also, we recall that we can find the quotient of two
fractions by inverting the divisor and proceeding as in multiplication.
Symbolically, multiplication is indicated by
a c ac
⋅ =
b d bd
and division is indicated by
a
b = a ⋅ d = ad
c b c bc
d
To Multiply Fractions
1. Factor each numerator and each denominator.
2. Reduce each fraction to lowest terms.
3. (a)
Multiply the resulting numerators to obtain the new numerator. Leave this
product in factored form.
(b)
Multiply the resulting denominators to obtain the new denominator. Leave
this product in factored form.
4. Reduce the resulting fraction to lowest terms, when possible, by dividing out like
factors.
NOTE:
In many cases reducing the fractions before doing the multiplication can shorten the
work.
See the next example for an explanation.
19
Module A35 − Algebraic Fractions
EXAMPLE 1
Simplify:
20 35a 2
⋅
7a 4b
SOLUTION:
SOLUTION I
SOLUTION II
2
20 35a
⋅
7a 4b
4⋅5 7⋅5⋅a ⋅a
=
⋅
7⋅a
4⋅b
25a
=
b
20 35a 2
⋅
7 a 4b
5 5a
⋅
← divide out like terms
1 b
25a
=
← collect what's left
b
In Solution I each term was factored completely prior to any dividing. In Solution II the
fractions were reduced first.
In general, choose the procedure which works better for you.
EXAMPLE 2
Simplify:
9ab 3 10a 2
⋅
4a 3b 15ab
We shall follow the Solution I method as in the previous example.
9ab 3 10a 2 3 ⋅ 3 ⋅ a ⋅ b ⋅ b ⋅ b 5 ⋅ 2 ⋅ a ⋅ a
⋅
=
⋅
4a 3b 15ab 2 ⋅ 2 ⋅ a ⋅ a ⋅ a ⋅ b 5 ⋅ 3 ⋅ a ⋅ b
3⋅3⋅b ⋅b 2⋅ a
=
⋅
2⋅ 2⋅ a ⋅ a 3⋅b
3⋅3⋅ 2⋅ a ⋅b ⋅b
=
2⋅ 2⋅3⋅ a ⋅ a ⋅b
Divide factors common to both the numerator and denominator before multiplying.
=
3b
2a
20
Module A35 − Algebraic Fractions
EXAMPLE 3
Find the quotient of:
5a 2
a4
÷
2b 10b 3
5a 2
a4
5a 2 10b 3
÷
=
⋅
Now divide out like terms to get :
3
2b 10b
2b a 4
25b 2
= 2
a
EXAMPLE 4
Simplify:
3x − 3 y 14 y − 14 x
÷
7z
15 z 2
3x − 3 y
15 z 2
=
⋅
7z
14 y − 14 x
=
3( x − y )
15 z 2
⋅
7z
−14 ( x − y )
=−
In this line we factored out a “−14” in order to
have a common factor that can be divided out.
45z
98
21
Module A35 − Algebraic Fractions
Experiential Activity Four
1.
3.
2a a
⋅
9 3
a 4a
⋅
5b b
2.
4.
a 2 27
⋅
9b a
2a 3ab 4b
7.
⋅
⋅
7b 2 8 9a 3
7a + 14 15a
9.
⋅
5
6a + 12
10a b
11.
÷
b
3a
6.
5.
13.
2 5a 2 7 a 3
⋅
⋅
15a 2 14 10b
21a 3a − 9
10.
⋅
5a − 15
7
125ab 5b 3
12.
÷
a
12
144a 3b 2 28a 5
15.
÷
39c 3
13bc 2
4 2
8 a2
⋅
ab 24
8.
11a 3
÷ 121a 2b
b
4 3
3a a
⋅
8 4
11a a
⋅
b 6b
14.
3x 6 x 2
÷
5 y 15 y
16.
4a 3b
14a 2b 2
Show
÷
55( x + y ) 11( x + y ) 2
Me.
3
17.
72 x y 21a x
81x
⋅
÷
2
2
105a
32 y
160a 4 y 3
18.
6a − 30 3a + 9 3a − 15
⋅
÷
7 a − 21 2a + 14 7a + 49
19.
12a 2b 3
24
÷
2
5c
5ac 2
20.
3rs
xy + y 2
⋅
x + xy 6r 2 s 3
21.
3x 2 − x
10
x−2
⋅ 2
⋅
5
3
2x − 8x x − 1
22.
6a 10bx
⋅
5b 9my
2
22
Module A35 − Algebraic Fractions
Experiential Activity Four Answers
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
2a 2
27
4a 2
5b 2
3a
b
1
21a
7a
2
30a 2
b2
a
11b 2
12b 3
7a 2c
8a 6 x 3 y 4
9
3 3
a b
2
x−2
x−4
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
22.
3a 2
32
11a 2
6b 2
a
3b
a3
30b
9a
5
25a 2
12b 2
3
2x
2a ( x + y )
35b
3(a + 3)
a −3
y
2rs 2 x
4ax
3my
23
Module A35 − Algebraic Fractions
OBJECTIVE FIVE
When you complete this objective you will be able to…
Perform the operation of simplifying complex fractions.
Exploration Activity
A complex fraction is any fraction in which either the numerator or the denominator or
both are themselves fractions.
Some examples are:
x
x
y
a)
Here the denominator is
x
y
3
a
4
a2
b)
⎛3⎞
⎝a⎠
⎛ 4 ⎞
.
2⎟
⎝a ⎠
Here we have a fraction in both the numerator ⎜ ⎟ and the denominator ⎜
To Simplify a complex fraction means to reduce it to a simple fraction, i.e. one that does
not have a fraction in its numerator or denominator.
EXAMPLE 1
Simplify:
1
3
1
2
3
4
2
4
=
1 4 4
2
⋅ = which reduces to
2 3 6
3
24
Module A35 − Algebraic Fractions
EXAMPLE 2
Simplify
1
a
1
a−
2
2+
Note: both numerator and denominator have L.C.D.'s.
1
2a 1
+
a = a a = 2a + 1 ⋅ 2 = 4a + 2
1
2a − 1
a
2a − 1 a (2a − 1)
a−
2
2
2+
EXAMPLE 3
The following is a more complicated example of a complex fraction requiring
simplification:
a+
1+
4
1
a+b
2
a+b
a −b
Step 1 As a first step, simplify the numerator.
4
1
a+b
⎛ a+b⎞
a + 4⎜
⎟
⎝ 1 ⎠
= a + 4a + 4b
= 5a + 4b
a+
This is the simplified numerator.
25
Module A35 − Algebraic Fractions
Step 2 Now simplify the denominator.
2
a+b
a −b
⎛ a −b ⎞
= 1+ 2⎜
⎟
⎝ a+b⎠
2a − 2b
= 1+
a+b
a + b + 2a − 2b
=
a+b
3a − b
=
a+b
1+
This is the simplified denominator.
Step 3 Now put the pieces back together and the original fraction becomes:
5a + 4b
3a − b
a+b
⎛ a+b ⎞
= (5a + 4b) ⎜
⎟
⎝ 3a − b ⎠
There is nothing here that can be cancelled by factoring, so we leave the answer in this
form.
(a + b)(5a + 4b)
3a − b
26
Module A35 − Algebraic Fractions
Experiential Activity Five
Simplify each of the following:
1.
3.
5.
7.
9.
11.
13.
1 1
−
2 x
1 1
+
2 2
a
a+
b
c
c+
b
1 1
−
2 3
2.
1 1
+
4 9
1
a−
2
4.
1
−2
a
1
−b
a
6.
1
a−
b
2a
7b
8.
5a
14b
1
x+
y
10.
x
x2 +
y
1
12.
1
1−
a
1−
b
1
1 1
−
x y
14.
1
x
−1
y
m
1− m
−1
1+ m
x 3
−
3 x
1 1
−
3 x
x x
−
6 9
x x
−
3 18
1 1
+
2 3
1
1−
6
nx − ny
C 2 − CF
Cx − Cy
nC + nF
2
1 1
+
x y
2
y−
1 1
+
x y
Show Me.
x−
15.
Show Me.
27
Module A35 − Algebraic Fractions
Experiential Activity Five Answers
1.
3.
5.
x−2
2x
a
c
m + 1 −m − 1
or
−2
2
7. x + 3
9.
1
5
11. 1
n 2 (C + F )
C 2 (C − F )
−x
15.
y
13.
6
13
−a
4.
2
−b
6.
a
4
8.
5
1
10.
x
a−b
12.
a
2.
14. −x
28
Module A35 − Algebraic Fractions
OBJECTIVE SIX
When you complete this objective you will be able to…
Solve linear equations involving algebraic fractions.
Exploration Activity
Example 1
Solve for x:
7 1
+ = −3
x 2
SOLUTION:
Multiply both sides by the L.C.D. which is 2x.
⎛7 1⎞
2 x ⎜ + ⎟ = 2 x ( −3)
⎝ x 2⎠
14 x 2 x
+
= −6 x
2
x
14 + x = −6 x
Isolate x.
14 = −7x
−2 = x
CHECK
NOTE:
We will now use a slightly different approach for checking our answers. LHS means left
hand side. RHS means right hand side. Substitute the solution into LHS, simplify, and
check to see if it is equal to the quantity on the RHS of the equal sign.
LHS
7 1
+
x 2
7
=
+
−2
−7
=
+
2
−6
=
2
= −3
RHS
−3
1
2
1
2
LHS = RHS
COMMENT: Following is shown a slightly different method for solving Example 1. Use
whichever method you prefer.
29
Module A35 − Algebraic Fractions
EXAMPLE 1 - repeated
Solve
7 1
+ = −3 for x.
x 2
SOLUTION:
Rather than multiply through the whole equation by the L.C.D., simply combine the left
hand side into a single fraction and then cross-multiply.
7 1
+ = −3
x 2
14 + x −3
=
2x
1
1(14 + x) = −3(2x)
14 + x = −6x
7x = −14
x = −2
If the right hand side of the equation were a combination of several fractions, then it
should be combined into a single fraction prior to cross-multiplying. This alternative
process can be used for all the examples presented in this objective.
EXAMPLE 2
Solve for x:
5
3
+
=2
2x + 4 x + 2
SOLUTION:
Factor the denominators if possible.
5
3
+
=2
2( x + 2) x + 2
Multiply every term by the L.C.D. and simplify
⎡ 5
3 ⎤
2( x + 2) ⎢
+
= 2( x + 2)(2)
+
+
2(
x
2)
x
2 ⎥⎦
⎣
2( x + 2)5 2( x + 2)3
+
= 2( x + 2)(2)
2( x + 2)
x+2
Å Divide out like factors
5 + 6 = 4 ( x + 2)
30
Module A35 − Algebraic Fractions
Isolate x.
5 + 6 = 4(x + 2)
11 = 4x + 8
3 = 4x
x = 0.75
See check that follows.
CHECK:
LHS
5
3
+
2x + 4 x + 2
5
3
=
+
2(0.75) + 4 0.75 + 2
5
3
=
+
1.5 + 4 2.75
5
6
=
+
5.5 5.5
11
=
5.5
=2
RHS
2
LHS = RHS
EXAMPLE 3
Solve for x:
1 3
2
+
=
x 2x x + 1
SOLUTION:
Multiply by the L.C.D. which is 2x(x + 1).
⎛1 3 ⎞
⎛ 2 ⎞
2 x( x + 1) ⎜ + ⎟ = 2 x( x + 1) ⎜
⎟
2
x
x
⎝
⎠
⎝ x +1⎠
2 x( x + 1) ( 3)( 2 x ) ( x + 1)
+
= 2x ( 2)
2x
x
2 ( x + 1) + 3 ( x + 1) = 4 x
2 x + 2 + 3x + 3 = 4 x
5x + 5 = 4x
x = −5
31
Module A35 − Algebraic Fractions
CHECK
NOTE: Here we must substitute into both sides of the equation and check for equality.
LHS
RHS
1 3
+
x 2x
1
3
=
+
− 5 2(−5)
1 3
=− −
5 10
2
3
=− −
10 10
−5
=
10
1
=−
2
2
x +1
2
=
− 5 +1
2
=
−4
1
=−
2
LHS = RHS
EXAMPLE 4
Solve for a:
10
7a + 5
4a
+
=
16 12a − 20 3a − 5
SOLUTION:
Reduce fractions if possible (factor).
5
7a + 5
4a
+
=
8 4(3a − 5) 3a − 5
Multiply by the L.C.D. which is 8(3a − 5).
⎛5
7a + 5 ⎞
⎛ 4a ⎞
8(3a − 5) ⎜ +
⎟ = 8(3a − 5) ⎜
⎟
8
4(3
a
−
5)
⎝ 3a − 5 ⎠
⎝
⎠
⎛ ( 5 )( 8 ) (3a − 5) ( 8 ) (3a − 5) ( 7 a + 5 ) ⎞ ⎛ ( 4a )( 8 ) (3a − 5) ⎞
+
⎜
⎟=⎜
⎟
8
4(3a − 5)
3a − 5
⎝
⎠ ⎝
⎠
5(3a − 5) + 2(7 a + 5) = 8(4a )
32
Module A35 − Algebraic Fractions
Isolate a.
15a – 25 + 14a + 10 = 32a
29a −15 = 32a
−3a = 15
−5 = a
CHECK
LHS
RHS
10
7a + 5
+
16 12a − 20
5
7(−5) + 5
= +
8 12(−5) − 20
5 − 30
= +
8 − 80
50 30
=
+
80 80
80
=
80
=1
4a
3a − 5
4(−5)
=
3(−5) − 5
− 20
=
− 20
=1
LHS = RHS
EXAMPLE 5
Solve for y:
13
9
−9
−
= 2
y − 2 y − 3 y − 5y + 6
SOLUTION:
Factor to get:
13
9
−9
−
=
y − 2 y − 3 ( y − 2)( y − 3)
multiply by the L.C.D. which is (y − 2)(y − 3).
⎛ 13
⎛
⎞
−9
9 ⎞
−
( y − 2)( y − 3) ⎜
⎟ = ( y − 2)( y − 3) ⎜
⎟
y
−
y
−
y
−
y
−
2
3
(
2)(
3
)
⎝
⎠
⎝
⎠
⎛ 13( y − 2)( y − 3) 9( y − 2)( y − 3) ⎞ ⎛ −9( y − 2)( y − 3) ⎞
−
⎜
⎟=⎜
⎟
y−2
y −3
⎝
⎠ ⎝ ( y − 2)( y − 3) ⎠
13( y − 3) − 9( y − 2) = −9
13 y − 39 − 9 y + 18 = −9
4 y = 21 − 9
4 y = 12
y=3
33
Module A35 − Algebraic Fractions
CHECK
LHS
RHS
13
9
−
y −2 y −3
13
9
=
−
3−2 3−3
13 9
= −
1 0
−9
y2 − 5y + 6
−9
=
9 − 15 + 6
−9
=
0
There is no solution here since y = 3 gives division by zero which is undefined.
Some fractional equations look more ominous than others. However there are some
techniques that make the questions easier. The following example illustrates this.
EXAMPLE 6
Solve for x in the following equation:
1
(−43 x + 516)
7x
−
=
−6
6 x + 20
− 516 + x
SOLUTION:
Step 1
Clean up the left side of the equation by changing
1
−1
to
i.e. write the fraction with
−6
6
a positive denominator; and change (−43x + 516) to (43x − 516) by multiplying the
negative sign into the numerator; also change (−516 + x) to (x – 516).
Step 2
Rewrite the original equation incorporating the changes indicated in Step 1 to get:
−1 43 x − 516
7x
+
=
6
6 x + 20
x − 516
Step 3
Add the left hand side together using a L.C.D. of 6(6x + 20) to get:
−1(6 x + 20) 6(43x − 516)
+
6(6 x + 20)
6(6 x + 20)
−1(6 x + 20) + 6(43x − 516)
6(6 x + 20)
34
Module A35 − Algebraic Fractions
Step 4
Now simplify the numerator to obtain:
−6 x − 20 + 258 x − 3096
6(6 x + 20)
=
252 x − 3116
6(6 x + 20)
The left hand side is now simplified.
Put it back into the equation to solve
Step 5
Now cross multiply
252 x − 3116
7x
=
6(6 x + 20)
x − 516
(252 x − 3116)( x − 516) = 7 x(6)(6 x + 20)
252 x 2 − 3116 x − 130032 x + 1607856 = 252 x 2 + 840 x
The x2 terms cancel out, so we get:
−133988x = −1607856
x = 12
35
Module A35 − Algebraic Fractions
Experiential Activity Six
45 1
1
+ =−
x 11
4
17 1
1
+
=−
2.
x 12
5
16 1
3.
− =3
x 5
−6 1
4.
− = −1
x 3
7 1
5.
+ = −3
x 2
3
6. 4 − = 55
x
2 3
2
7.
−
=
x 2x x + 1
2a − 3
2a
=
8.
5a − 10 a − 2
1
2
−3
−
= 2
9.
x − 2 x − 3 x − 5x + 6
5a − 1
3
=
10.
10a − 2 5a − 1
1.
Show Me.
Experiential Activity Six Answers
1.
2.
3.
4.
5.
–132
–60
5
9
–2
6.
−
7.
8.
1
or −0.0588
17
1
or 0.3333
3
3
− or −0.375
8
9. 4
10.
7
or 1.4
5
Practical Application Activity
Complete the Algebraic Fractions assignment in TLM.
Summary
This module presented the student with the basic principles of dealing with algebraic
fractions.
36
Module A35 − Algebraic Fractions