# Simplifying Easy Algebraic Fractions

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This pdf includes the following topics:-
Cancelling common factors
Multiplication and division of algebraic fractions
Addition and subtraction of algebraic fractions
1. Arithmetic of  
Algebraic Fractions 1.4
 
Introduction
Just as one whole number divided by another is called a numerical fraction, so one algebraic expression
divided by another is known as an algebraic fraction. Examples are
x 3x + 2y x2 + 3x + 1
, , and
y x−y x−4
In this Section we explain how algebraic fractions can be simplified, added, subtracted, multiplied
and divided.
 
Prerequisites • be familiar with the arithmetic of numerical
fractions
Before starting this Section you should . . .
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
Learning Outcomes • add, subtract, multiply and divide algebraic
fractions
On completion you should be able to . . .
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62 HELM (2008):
Workbook 1: Basic Algebra
2. ®
1. Cancelling common factors
10
Consider the fraction . To simplify it we can factorise the numerator and the denominator and then
35
cancel any common factors. Common factors are those factors which occur in both the numerator
and the denominator. Thus
10 65×2 2
= =
35 7× 6 5 7
Note that the common factor 5 has been cancelled. It is important to remember that only common
10 2
factors can be cancelled. The fractions and have identical values - they are equivalent fractions
35 7
2 10
- but is in a simpler form than .
7 35
We apply the same process when simplifying algebraic fractions.
Example 49
Simplify, if possible,
yx x x
(a) , (b) , (c)
2x xy x+y
Solution
yx
(a) In the expression , x is a factor common to both numerator and denominator. This
2x
common factor can be cancelled to give
y6x y
=
26x 2
x 1x
(b) Note that can be written . The common factor of x can be cancelled to give
xy xy
16x 1
=
6 xy y
x
(c) In the expression notice that an x appears in both numerator and denominator.
x+y
However x is not a common factor. Recall that factors of an expression are multi-
plied together whereas in the denominator x is added to y. This expression cannot be
simplified.
HELM (2008): 63
Section 1.4: Arithmetic of Algebraic Fractions
abc 3ab
Simplify, if possible, (a), (b)
3ac b+a
When simplifying remember only common factors can be cancelled.
abc 3ab
(a) = (b) =
3ac b+a
b
(a) (b) This cannot be simplified.
3
21x3
Simplify ,
14x
Factorising and cancelling common factors gives:
21x3 6 7 × 3× 6 x × x2 3x2
= =
14x 6 7 × 2× 6 x 2
36x
Simplify
12x3
Factorising and cancelling common factors gives:
36x 12 × 3 × x 3
3
= 2
= 2
12x 12 × x × x x
64 HELM (2008):
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4. ®
Example 50
3x + 6
Simplify .
6x + 12
Solution
First we factorise the numerator and the denominator to see if there are any common factors.
3x + 6 3(x + 2) 3 1
= = =
6x + 12 6(x + 2) 6 2
The factors x + 2 and 3 have been cancelled.
12
Simplify .
2x + 8
12
=
2x + 8
6×2 6
Factorise the numerator and denominator, and cancel any common factors. =
2(x + 4) x+4
Example 51
3 3(x + 4)
Show that the algebraic fraction and 2 are equivalent.
x+1 x + 5x + 4
Solution
The denominator, x2 + 5x + 4, can be factorised as (x + 1)(x + 4) so that
3(x + 4) 3(x + 4)
=
x2+ 5x + 4 (x + 1)(x + 4)
Note that (x + 4) is a factor common to both the numerator and the denominator and can be
3 3 3(x + 4)
cancelled to leave . Thus and 2 are equivalent fractions.
x+1 x+1 x + 5x + 4
HELM (2008): 65
Section 1.4: Arithmetic of Algebraic Fractions
x−1 1
Show that is equivalent to .
x2 − 3x + 2 x−2
First factorise the denominator:
x2 − 3x + 2 =
(x − 1)(x − 2)
Now identify the factor common to both numerator and denominator and cancel this common factor:
x−1
=
(x − 1)(x − 2)
1
. Hence the two given fractions are equivalent.
x−2
Example 52
6(4 − 8x)(x − 2)
Simplify
1 − 2x
Solution
The factor 4 − 8x can be factorised to 4(1 − 2x). Thus
6(4 − 8x)(x − 2) (6)(4)(1 − 2x)(x − 2)
= = 24(x − 2)
1 − 2x (1 − 2x)
x2 + 2x − 15
Simplify
2x2 − 5x − 3
First factorise the numerator and factorise the denominator:
x2 + 2x − 15
=
2x2 − 5x − 3
66 HELM (2008):
Workbook 1: Basic Algebra
6. ®
(x + 5)(x − 3)
(2x + 1)(x − 3)
Then cancel any common factors:
(x + 5)(x − 3)
=
(2x + 1)(x − 3)
x+5
2x + 1
Exercises
1. Simplify, if possible,
19 14 35 7 14
(a) , (b) , (c) , (d) , (e)
38 28 40 11 56
14 36 13 52
2. Simplify, if possible, (a) , (b) , (c) , (d)
21 96 52 13
5z 25z 5 5z
3. Simplify (a) , (b) , (c) 2
, (d)
z 5z 25z 25z 2
4. Simplify
4x 15x 4s 21x4
(a) , (b) , (c) , (d)
3x x2 s3 7x3
5. Simplify, if possible,
x+1 x+1 2(x + 1) 3x + 3 5x − 15 5x − 15
(a) , (b) , (c) , (d) , (e) , (f) .
2(x + 1) 2x + 2 x+1 x+1 5 x−3
6. Simplify, if possible,
5x + 15 5x + 15 5x + 15 5x + 15
(a) , (b) , (c) , (d)
25x + 5 25x 25 25x + 1
x2 + 10x + 9 x2 − 9 2x2 − x − 1
7. Simplify (a) , (b) , (c) ,
x2 + 8x − 9 x2 + 4x − 21 2x2 + 5x + 2
3x2 − 4x + 1 5z 2 − 20z
(d) , (e)
x2 − x 2z − 8
6 2x 3x2
8. Simplify (a) , (b) 2 , (c)
3x + 9 4x + 2x 15x3 + 10x2
x2 − 1 x2 + 5x + 6
9. Simplify (a) , (b) .
x2 + 5x + 4 x2 + x − 6
HELM (2008): 67
Section 1.4: Arithmetic of Algebraic Fractions
1 1 7 7 1
1. (a) , (b) , (c) , (d) , (e) .
2 2 8 11 4
2 3 1
2. (a) , (b) , (c) , (d) 4
3 8 4
1 1
3. (a) 5, (b) 5, (c) 2
, (d) .
5z 5z
4 15 4
4. (a) , (b) , (c) 2 , (d) 3x
3 x s
1 1
5. (a) , (b) , (c) 2, (d) 3, (e) x − 3, (f) 5
2 2
x+3 x+3 x+3 5(x + 3)
6. (a) , (b) , (c) , (d)
5x + 1 5x 5 25x + 1
x+1 x+3 x−1 3x − 1 5z
7. (a) , (b) , (c) , (d) , (e)
x−1 x+7 x+2 x 2
2 1 3
8. (a) , (b) , (c) .
x+3 2x + 1 5(3x + 2)
x−1 x+2
9. (a) , (b) .
x+4 x−2
2. Multiplication and division of algebraic fractions
To multiply together two fractions (numerical or algebraic) we multiply their numerators together
and then multiply their denominators together. That is
Key Point 19
Multiplication of fractions
a c ac
× =
b d bd
Any factors common to both numerator and denominator can be cancelled. This cancellation can be
performed before or after the multiplication.
To divide one fraction by another (numerical or algebraic) we invert the second fraction and then
68 HELM (2008):
Workbook 1: Basic Algebra
8. ®
Key Point 20
Division of fractions
a c a d ad
÷ = × = b 6= 0, c 6= 0, d 6= 0
b d b c bc
Example 53
2a 4 2a c 2a 4
Simplify (a) × , (b) × , (c) ÷
c c c 4 c c
Solution
2a 4 8a
(a) × = 2
c c c
2a c 2ac 2a a
(b) × = = =
c 4 4c 4 2
(c) Division is performed by inverting the second fraction and then multiplying.
2a 4 2a c a
÷ = × = (from the result in (b))
c c c 4 2
Example 54
1 1
Simplify (a) × 3x, (b) × x.
5x x
Solution
3x 1 1 3x 3x 3
(a) Note that 3x = . Then × 3x = × = =
1 5x 5x 1 5x 5
x 1 1 x x
(b) x can be written as . Then × x = × = = 1
1 x x 1 x
HELM (2008): 69
Section 1.4: Arithmetic of Algebraic Fractions
1 y
Simplify (a) × x, (b) × x.
y x
1 1 x x
(a) ×x= × =
y y 1 y
y y x yx
(b) ×x= × = =y
x x 1 x
Example 55
2x
y
Simplify
3x
2y
Solution
2x 3x
We can write the fraction as ÷ .
y 2y
Inverting the second fraction and multiplying we find
2x 2y 4xy 4
× = =
y 3x 3xy 3
70 HELM (2008):
Workbook 1: Basic Algebra
10. ®
Example 56
4x + 2 x+3
Simplify ×
x2 + 4x + 3 7x + 5
Solution
Factorising the numerator and denominator we find
4x + 2 x+3 2(2x + 1) x+3 2(2x + 1)(x + 3)
× = × =
x2 + 4x + 3 7x + 5 (x + 1)(x + 3) 7x + 5 (x + 1)(x + 3)(7x + 5)
2(2x + 1)
=
(x + 1)(7x + 5)
It is usually better to factorise first and cancel any common factors before multiplying. Don’t remove
any brackets unnecessarily otherwise common factors will be difficult to spot.
Simplify
15 3
÷
3x − 1 2x + 1
To divide we invert the second fraction and multiply:
15 3 15 2x + 1 (5)(3)(2x + 1) 5(2x + 1)
÷ = × = =
3x − 1 2x + 1 3x − 1 3 3(3x − 1) 3x − 1
HELM (2008): 71
Section 1.4: Arithmetic of Algebraic Fractions
11. Exercises
5 3 14 3 6 3 4 28
1. Simplify (a) × , (b) × , (c) × , (d) ×
9 2 3 9 11 4 7 3
5 3 14 3 6 3 4 28
2. Simplify (a) ÷ , (b) ÷ , (c) ÷ , (d) ÷
9 2 3 9 11 4 7 3
3. Simplify
x+y 1 2
(a) 2 × , (b) × 2(x + y), (c) × (x + y)
3 3 3
4. Simplify
x+4 1 3 x x+1 1 x2 + x
(a) 3 × , (b) × 3(x + 4), (c) × (x + 4), (d) × , (e) × ,
7 7 7 y y+1 y y+1
πd2 Q Q
(f) × 2, (g)
4 πd πd2 /4
6/7
5. Simplify
s+3
3 x
6. Simplify ÷
x + 2 2x + 4
5 x
7. Simplify ÷
2x + 1 3x − 1
5 14 9 16
1. (a) , (b) , (c) , (d)
6 9 22 3
10 8 3
2. (a) , (b) 14, (c) , (d)
27 11 49
2(x + y) 2(x + y) 2(x + y)
3. (a) , (b) , (c)
3 3 3
3(x + 4) 3(x + 4) 3(x + 4) x(x + 1) x(x + 1)
4. (a) , (b) , (c) , (d) , (e) , (f) Q/4,
7 7 7 y(y + 1) y(y + 1)
4Q
(g)
πd2
6
5.
7(s + 3)
6
6.
x
5(3x − 1)
7.
x(2x + 1)
72 HELM (2008):
Workbook 1: Basic Algebra
12. ®
3. Addition and subtraction of algebraic fractions
To add two algebraic fractions the lowest common denominator must be found first. This is the
simplest algebraic expression that has the given denominators as its factors. All fractions must be
written with this lowest common denominator. Their sum is found by adding the numerators and
dividing the result by the lowest common denominator.
To subtract two fractions the process is similar. The fractions are written with the lowest common
denominator. The difference is found by subtracting the numerators and dividing the result by the
lowest common denominator.
Example 57
State the simplest expression which has x + 1 and x + 4 as its factors.
Solution
The simplest expression is (x + 1)(x + 4). Note that both x + 1 and x + 4 are factors.
Example 58
State the simplest expression which has x − 1 and (x − 1)2 as its factors.
Solution
The simplest expression is (x − 1)2 . Clearly (x − 1)2 must be a factor of this expression. Also,
because we can write (x − 1)2 = (x − 1)(x − 1) it follows that x − 1 is a factor too.
HELM (2008): 73
Section 1.4: Arithmetic of Algebraic Fractions
13. Example 59
3 2
Express as a single fraction +
x+1 x+4
Solution
The simplest expression which has both denominators as its factors is (x + 1)(x + 4). This is the
lowest common denominator. Both fractions must be written using this denominator. Note that
3 3(x + 4) 2 2(x + 1)
is equivalent to and is equivalent to . Thus writing
x+1 (x + 1)(x + 4) x+4 (x + 1)(x + 4)
both fractions with the same denominator we have
3 2 3(x + 4) 2(x + 1)
+ = +
x+1 x+4 (x + 1)(x + 4) (x + 1)(x + 4)
The sum is found by adding the numerators and dividing the result by the lowest common denomi-
nator.
3(x + 4) 2(x + 1) 3(x + 4) + 2(x + 1) 5x + 14
+ = =
(x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4) (x + 1)(x + 4)
Key Point 21
Addition of two algebraic fractions
Step 1: Find the lowest common denominator
Step 2: Express each fraction with this denominator
Step 3: Add the numerators and divide the result by the lowest common denominator
Example 60
1 5
Express + as a single fraction.
x − 1 (x − 1)2
Solution
The simplest expression having both denominators as its factors is (x − 1)2 . We write both fractions
with this denominator.
1 5 x−1 5 x−1+5 x+4
+ 2
= 2
+ 2
= 2
=
x − 1 (x − 1) (x − 1) (x − 1) (x − 1) (x − 1)2
74 HELM (2008):
Workbook 1: Basic Algebra
14. ®
3 5
Express + as a single fraction.
x+7 x+2
First find the lowest common denominator:
(x + 7)(x + 2)
Re-write both fractions using this lowest common denominator:
3 5
+ =
x+7 x+2
3(x + 2) 5(x + 7)
+
(x + 7)(x + 2) (x + 7)(x + 2)
Finally, add the numerators and simplify:
3 5
+ =
x+7 x+2
8x + 41
(x + 7)(x + 2)
Example 61
5x 3x − 4
Express − as a single fraction.
7 2
Solution
In this example both denominators are simply numbers. The lowest common denominator is 14, and
both fractions are re-written with this denominator. Thus
5x 3x − 4 10x 7(3x − 4) 10x − 7(3x − 4) 28 − 11x
− = − = =
7 2 14 14 14 14
HELM (2008): 75
Section 1.4: Arithmetic of Algebraic Fractions
1 1
Express + as a single fraction.
x y
The simplest expression which has x and y as its factors is xy. This is the lowest common denom-
1 y 1 x
inator. Both fractions are written using this denominator. Noting that = and that =
x xy y xy
we find
1 1 y x y+x
+ = + =
x y xy xy xy
No cancellation is now possible because neither x nor y is a factor of the numerator.
Exercises
x x 2x x 2x 3x x 2 x+1 3
1. Simplify (a)+ , (b) + , (c) − , (d) − , (e) + ,
4 7 5 9 3 4 x+1 x+2 x x+2
2x + 1 x x+3 x x x
(f) − , (g) − , (h) −
3 2 2x + 1 3 4 5
2. Find
1 2 2 5 2 3 x+1 x+4
(a) + , (b) + , (c) − , (d) + ,
x+2 x+3 x+3 x+1 2x + 1 3x + 2 x+3 x+2
x−1 x−1
(e) + .
x − 3 (x − 3)2
5 4
3. Find + .
2x + 3 (2x + 3)2
1 11
4. Find s+
7 21
A B
5. Express + as a single fraction.
2x + 3 x + 1
A B C
6 Express + + as a single fraction.
2x + 5 (x − 1) (x − 1)2
A B
7 Express + as a single fraction.
x + 1 (x + 1)2
76 HELM (2008):
Workbook 1: Basic Algebra
16. ®
Ax + B C
8 Express + as a single fraction.
x2+ x + 10 x − 1
C
9 Express Ax + B + as a single fraction.
x+1
x1 x1 x2 x3
10 Show that is equal to .
1 1 x2 − x3

x3 x2
3x x x 3x  x x 
11 Find (a) − + , (b) − + .
4 5 3 4 5 3
11x 23x x x2 − 2 x2 + 6x + 2
1. (a) , (b) , (c) − , (d) , (e) ,
28 45 12 (x + 1)(x + 2) x(x + 2)
x+2 9 + 2x − 2x2 x
(f) , (g) , (h)
6 3(2x + 1) 20
3x + 7 7x + 17 1
2. (a) , (b) , (c) ,
(x + 2)(x + 3) (x + 3)(x + 1) (2x + 1)(3x + 2)
2x2 + 10x + 14 x2 − 3x + 2
(d) , (e)
(x + 3)(x + 2) (x − 3)2
10x + 19
3.
(2x + 3)2
3s + 11
4.
21
A(x + 1) + B(2x + 3)
5.
(2x + 3)(x + 1)
A(x − 1)2 + B(x − 1)(2x + 5) + C(2x + 5)
6.
(2x + 5)(x − 1)2
A(x + 1) + B
7.
(x + 1)2
(Ax + B)(x − 1) + C(x2 + x + 10)
8.
(x − 1)(x2 + x + 10)
(Ax + B)(x + 1) + C
9.
x+1
53x 13x
11. (a) , (b)
60 60
HELM (2008): 77
Section 1.4: Arithmetic of Algebraic Fractions