The Mathematics 11
Competency Test
Simplification of Fractions
An indispensable aspect of any work with algebraic fractions is simplification of the result. We will
demonstrate techniques for simplification of fractions throughout these notes where the need
arises, but in this specific document, we’ll set the stage by working through a number of
examples, focussing primarily on the strategies for simplification themselves.
Despite the differences in appearance of the examples to follow, there really is just one approach
available for simplifying fractions:
(i) factor the numerator and denominator as completely as possible, and then
(ii) cancel any factors the numerator and denominator have in common.
If you take care to always do step (i) as a separate explicit step of work, you will rarely go wrong
in simplifying fractions.
What follows in this document is a series of quite a few worked out examples. They tend to be
quite repetitive because there is just the one simple strategy just described, which must be
applied to each one. Don’t just read through the rest of this document quickly to notice how
easily we’ve worked out the solutions (and perhaps to notice how boring this sort of stuff can
become). Instead, after studying the first few examples carefully, use the remaining examples as
practice problems: cover up the solutions and try to work them out on your own. When you are
done in each case, compare your method and final result with the one we’ve given.
Example 1: Simplify
10 xy
.
2y
solution:
The numerator and denominator of this fraction are already nearly fully factored. Using brackets
to make these factors explicit, we get
10 xy  2 5  x  y 

2y
 2 y 

 2   5  x   y 
 2  y 
 5x
as the final result.
David W. Sabo (2003)
Simplification of Fractions
Page 1 of 8
Example 2: Simplify
12 x 3 y 2
3x 2y 4
.
solution:
Writing the numerator and denominator with the factors explicitly separated gives
12x 3 y 2

3x 2 y 4
 2  3  x  y 
 3   x  y 
2
3
2
2
4
 2   3   x
2

 3  x 
2
 
 y2


2
 4y 
y




3
x

22 x 4 x
 2
y2
y
60 x 2  45 x 3
Example 3: Simplify
.
3x
solution:
The denominator is factored, but the numerator is not factored. Using the stepwise approach for
factoring illustrated earlier in these notes, we get
60x2 – 45x3 = 3[ 20x2 – 15x3 ]
= (3)(5) [ 4x2 – 3x3 ]
= (3)(5)(x2)[4 – 3x]
Thus
  5  x   4  3x 
 3 x
2
3
60 x 2  45 x 3

3x
 5   x 2   4  3 x 
x



x
 
 5x  4  3x 
as the final result. There is no need to multiply this last form out to remove the brackets, unless
there had been some specific instruction to do so.
David W. Sabo (2003)
Simplification of Fractions
Page 2 of 8
Note that the first step of factoring is absolutely essential here in order to get a correct result – a
simpler mathematical expression which is mathematically equivalent to the original expression. A
common error is to just focus on one of the terms in the numerator:
60 x 2  45 x 3
3x

 3   20   x   x   45x
 3  x 
3

20 x  45 x 3
This “simplified” expression is not equivalent to the original fraction, and so an error has been
made. You can easily demonstrate this by doing a test calculation:
60 x 2  45 x 3
3x
x 1

60 1  45 1
2
3 1
3

60  45
5
3
but
20x  45x 3
x 1
 20 1  45 1  20  45  25
3
Since for this one value of x, the expression 20x – 45x3 does not give the same value that the
original fraction gives, we know that 20x – 45x3 cannot be a correct simplification of the original
fraction. The error came from the fact that we cancelled the 3x of the denominator into just one of
the terms in the numerator. Instead, a valid cancellation must be against factors of the entire
numerator, as was done in the original solution of this example.
Example 4: Simplify
12x 4 y 2  15 x 2 y 4
.
3x 2 y 3
solution:
Since the numerator is the sum of two terms, we must begin by factoring it as much as possible:
(both terms in the original expression
share a factor of 3)
12 x 4 y 2  15 x 2 y 4  3  4 x 4 y 2  5 x 2 y 4 
 3 x 2  4 x 2 y 2  5 y 4 
(both terms in the square brackets of
the preceding line share a factor of x2)
 3 x 2 y 2  4 x 2  5 y 2 
(both terms in the square brackets of
the preceding line share a factor of y2)
Thus
   x   y   4x
 3   x   y
3
12x 4 y 2  15 x 2 y 4

3x 2y 3
David W. Sabo (2003)
2
2
2
3
2
y
 5y 2




Simplification of Fractions
Page 3 of 8

4 x 2  5y 2
y
as the final answer.
The initial factoring step is absolutely essential. If instead you decided to begin by cancelling
between the denominator and the first term of the numerator, along the lines of
4
12 x 4
x2
y 2  15 x 2 y 4
3 x2 y 3
y

4 x 2  15 x 2 y 4
y3
y

4 x 2  15 x 2 y 3
you would end up with a completely wrong result. We’re being very repetitive on this point here
because errors of the type demonstrated above, where the factoring step is skipped, are very,
very commonly made by people writing the BCIT Mathematics 11 Competency Test. This kind of
thing is not wrong so much because we say so, or because you didn’t use “our” method, the
method “we” told you to use. Rather, it is wrong because the result does not have the same
mathematical meaning, nor will it give the same mathematical results as the original expression,
and so it’s about the same sort of thing as deciding the change the meaning of words when you
speak (but pretending you haven’t!). If you substitute specific pair of values for x and y into this
proposed simplified expression, you will get different value from it than you would get when you
substitute the same values of x and y into the original expression. Simplification is intended to
give a mathematical expression which is easier to write down and work with than the original
expression, but which gives exactly the same results as the original expression in all
circumstances. If our apparent simplified expression does not do this, then we have made a
mistake.
Example 5: Simplify
5x 2
.
5 x 2  20 x
solution:
It is so tempting just to cancel the ‘5x2’ terms and be done:
5x 2
5x  20 x
2

1
1  20 x

1
20 x
or even
5x 2
5x 2  20 x
By now you’re screaming back, “No! No! Both of these results are wrong! Wrong! Wrong!” In
both cases, the 5x2 that is being cancelled is not a factor in the denominator (it is a “term” in the
denominator) and hence cannot be validly cancelled. In the second version above, another error
is made as well. When factors are cancelled, they always leave a result of “1” behind. If it is a
multiplication by 1, we need not write down the 1, as in
David W. Sabo (2003)
Simplification of Fractions
Page 4 of 8
5x y
3x y

5 11 5

3 11 3
But, if it is an “added” 1, you cannot drop it:
1 20x   20x
1   20x   20x
but
This latter is as incorrect as writing
5x 2

5x 2
This is supposed to look like a fraction with nothing written in the numerator or in the denominator
(that is, the ultimate in cancellation – perhaps we should consider this to be annihilation!). Of
course, you can easily see this is nonsense.
Anyway, back to the original example. Before any cancelling can be contemplated, both the
numerator and denominator must be factored completely. The numerator is already in factored
form. However, for the denominator we have
5x2 + 20x = 5 [ x2 + 4x ]
=5x(x + 4)
Thus
x
5x 2
5 x2
x


2
5 x  20 x 5 x  x  4  x  4
as the final correct answer.
Example 6: Simplify
3t 2  5ts  7s 2t  4st 2
.
t
solution: The denominator is obviously fully factored. For the numerator, only slight factoring is
possible:

3t 2  5ts  7s 2t  4st 2  t 3t  5s  7s 2  4st

Therefore

2
3t 2  5ts  7s 2t  4st 2 t 3t  5s  7s  4st )

t
t

 3t  5s  7s 2  4st
David W. Sabo (2003)
Simplification of Fractions
Page 5 of 8
as the final simplified form.
Example 7: Simplify
a 2  4a
.
ax  4 x
solution:
First, factoring
a2 + 4a = a(a + 4)
and
ax + 4x = x(a + 4)
Thus
a 2  4a a  a  4  a


ax  4 x x  a  4  x
as the final simplified result.
Example 8: Simplify
x 2  5x  6
.
x 3
solution:
There is no possibility of factoring the denominator further here, since neither term in the
denominator is a product. The numerator is a trinomial, and so possibly can be factored into a
product of two binomials:
x 2  5x  6
?

 x  a  x  b 
As explained earlier in these notes, for this to be possible, we would need to find numbers ‘a’ and
‘b’ such that
a + b = -5
and
ab = 6
You have probably already noticed that a = -2 and b = -3 will work. It is easy to confirm that
(x – 3)(x – 2) = x2 – 5x + 6.
So, our original fraction becomes
x 2  5 x  6  x  3   x  2

 x 2
x 3
 x  3
David W. Sabo (2003)
Simplification of Fractions
Page 6 of 8
in simplest form.
Example 9: Simplify
x 2  2x  15
.
5x
solution:
Proceeding as in the previous example, we quickly find that
x 2  2 x  15  x  5  x  3 

5x
5x
At first, it may not appear that any cancellation is possible because neither of the factors in the
numerator look the same as the denominator. However, recall that
(x – 5) = -(5 – x)
So, we can write our fraction here as
x 2  2 x  15  x  5  x  3 

5x
5x

 5  x   x  3
5  x 
   x  3
since now the presence of the common factor (5 – x) in the numerator and denominator is
obvious. Cancelling that common factor leads to the final result.
Example 10: Simplify
x 2  5x  6
.
x2  9
solution:
By now, you are very familiar with the process: factor the numerator and denominator and cancel
common factors. Here this gives
x 2  5x  6  x  2  x  3  x  2


x2  9
 x  3  x  3 x  3
as the final answer.
David W. Sabo (2003)
Simplification of Fractions
Page 7 of 8
20a 2b3c 2  15ab 2c 3
Example 11: Simplify
.
5abc
solution:
Factoring and taking into account for the minus sign in the denominator gives
2
2
20a2b3c 2  15ab2c 3 5 a b c  4ab  3c 

5abc
 1 5 a b c
b
c
 bc  4ab  3c 
as the final simplified result.
David W. Sabo (2003)
Simplification of Fractions
Page 8 of 8