The Mathematics 11
Competency Test
Multiplication and Division With
Two Fractions
Multiplying one algebraic fraction by another algebraic fraction, or dividing one algebraic fraction
by another algebraic fraction, brings together three topics with which you are already familiar:
(i) the basic formalism:
a c ac
 
b d bd
and
a c
 
b d
a
b
c
d

a d ad
 
b c bc
(ii) the methods of simplifying fractions by cancellation of factors common to the numerator and
denominator, which we illustrated with many examples in the preceding note in this series.
and
(iii) using the laws of exponents when the factors being multiplied or divided are powers:
an am  an m
an
 a n m
m
a
a 
n
m
 an m
These laws of exponents imply, among other things, that we can move factors from the
numerator to the denominator or the reverse, as long as we change the sign of the exponent at
the same time. This property can make cancellations involving powers of a specific symbol very
easy. For instance, we can do
x9 x3 x6

 x3
6
x6
x
or
x9
 x 96  x 3
6
x
or even
x9
9  6
 x 9 x 6  x    x 96  x 3
x6
David W. Sabo (2003)
Multiplication and Division with Two Fractions
Page 1 of 11
The rest of this document consists of a collection of examples illustrating how these three sets of
properties can be used to carry out multiplication and division with algebraic fractions, ensuring
that the final result is in simplest possible form.
Example 1: Simplify:
 3 x 2   25 y 2 
.
 3 
4 
 5y   9 x 
solution:
Writing the two sets of brackets next to each other in this fashion indicates that the two fractions
are to be multiplied. So
 
 


2
2
 3 x 2   25y 2  3 x 25y

 3 
4 
5y 3 9 x 4
 5y   9 x 
 3   5   x y 

 5   3   x y 
2
2
5
3
2
2
4
3
5 x 2 y 2 5 24 23 5 2 1
 4 3  x y  x y
3x y
3
3
This result is technically correct, but many people view negative exponents as unattractive and
somewhat conducive to misunderstanding or misinterpretation. Most practitioners would
recommend writing the final answer here in the form of a proper fraction containing no negative
exponents. Thus, the best way to state the final answer is
5
3x 2 y
Example 2: Multiply
24a2b
49c 3
by
.
7c 2
8ab 2
solution:
This is just another way of posing the problem:
Simplify:
 24a2b  49c 3 

2 
2 
 7c  8ab 
Proceeding as in the previous example, we get
David W. Sabo (2003)
Multiplication and Division with Two Fractions
Page 2 of 11
 24a2b  49c 3 


2 
2 
 7c  8ab 
 24a b  49c 
7c 8ab 
2
3
2
2
 3   8   72   a2   b   c 3 


7

 7   8  a   b c 
2
 3  7   a2
 
 b  c3 c 







b
a  b 2  c 2



a
 
 
=
2

21ac
b
as the final solution. We could have done this last step as
21a 2bc 3
 21a 21b12c 32
2 2
ab c
21ac
b
 21a1b 1c 1 
to get the same final result, once factors are repositioned between the numerator and
denominator to get rid of negative exponents.
Example 3: Simplify
 14 x 2 y 3  12a5b 
.

3 2 
4 2 
 3a b  35 x y 
solution:
This is very similar to the expressions handled in the first two examples. Proceeding in the same
fashion, we get
 14 x 2 y 3  12a5b 


3 2 
4 2 
 3a b  35 x y 
14x y 12a b 
3a b 35x y 
2
3
3
5
2
4

 2 7   22  3   x 2 y 3a5b 
 3  5  7   x 4 y 2a3b2 

 2   7   22   3 
 3  5  7 
David W. Sabo (2003)
x
2 4
y
32
2
(This is what the actual multiplication of
the two fractions amounts to. Now this
result must be simplified.)
(Here we expand the numerical factors
into products of prime factors, and we
also sort out the various literal factors.)
a
5 3
1 2
b
(Cancel the common numerical factors
and combine powers of the literal
symbols.)
Multiplication and Division with Two Fractions
Page 3 of 11

8 2 2  1
x ya b
5

8a 2 y
5bx 2
(This is a fully simplified form, but it
contains negative exponents.)
as the final result with negative exponents eliminated.
Example 4: Simplify
12x 2 18 x 3
.

5y 2
35 y
solution:
This is one fraction divided by another. Following the pattern given at the beginning of this
document, we know that the first step here is to rewrite the expression as the first fraction
multiplied by the reciprocal of the second fraction:
12x 2 18 x 3  12x 2   35y 



5y 2
35y  5y 2   18 x 3 
Now the remainder of the work is to simplify this multiplication, exactly as we dealt with the first
three examples. So
 12x 2   35y 


2 
3 
 5y   18 x 
12x  35y 
5y 18x 
2
2


3
12 35   x 2   y 
 5 18   x 3  y 2 
   5  7  x   y 
 5   2   3   x  y 
 22 2  3




2
2

14 23 1 2
x y
3

14 1 1
14
x y 
3
3 xy
3
3
2
as the final answer.
A very common error here is to start by correctly rewriting the original division problem as a
multiplication, but then doing the multiply step in a totally bizarre way – numerators with
denominators, as in
David W. Sabo (2003)
Multiplication and Division with Two Fractions
Page 4 of 11



12x 2 18 x 3
35y

18 x 3
5y 2  35y 
12x 2 18 x 3 12x 2
35y
12x 2




5y 2
35y
5y 2
18 x 3
5y 2


But this is totally wrong! When we multiply two fractions, it is always numerator times numerator
and denominator times denominator, regardless of where the multiplication problem originally
came from. If you examine the eventual result that would be obtained here, you’ll see that it
amounts to what we would get if we had multiplied the two original fractions together, rather than
dividing the first fraction by the second one. In other words, what has effectively been done by
using this erroneous method is to change the original division symbol to a multiplication symbol.
This must be an error. So, always remember: dividing by a fraction is equivalent to multiplying
by its reciprocal, and multiplying is always done the same way, regardless of from where the
original fractions were obtained.
Example 5: Divide
8ab2
16a 3 b
by
and simplify the result.
9a2c
27ac 2
solution:
When we rewrite this problem using the divide symbol, you’ll see that it is very similar to the
previous example:
8ab2 16a3b 8ab 2 27ac 2



9a2c 27ac 2 9a2c 16a3b
16a 3 b
27ac 2
That is, dividing by
is the same thing as multiplying by
. So, proceeding
27ac 2
16a 3 b
8ab2 16a3b 8ab 2 27ac 2



9a2c 27ac 2 9a2c 16a3b

8ab  27ac 
9a c 16a b 

 2   3  a   b  a  c 
 3   2  a  c  a   b 
2
David W. Sabo (2003)
2
2
3
3
3
2
4
3
2

3 1123 21 21
a
b c
2

3 3
3bc
a bc 
2
2a3
2
2
2
3
Multiplication and Division with Two Fractions
Page 5 of 11
as the final answer.
 4 x   5a  5b 
.

2 
 a  b   14 x 
Example 6: Simplify 
solution:
Suddenly addition symbols appear in a multiplication problem! This means we must apply the
basic rules of multiplication a bit more carefully. The strategy is still to multiply numerator by
numerator and denominator by denominator, and then simplify the result. So
 4x  5a  5b 
 a  b  14x 2 
 4 x  5a  5b 
 a  b  14 x 2  




 2   x 5a  b 
 a  b  2 7   x 

  5  a  b 
 a  b   2  7   x 
2
2
 22 2  x




2
x

10
7x
as the final answer. In the second step here, we factored wherever possible. Then in the third
step, we cancelled common factors between numerator and denominator. When multiplication
involves factors with more than one term, very careful use of necessary brackets is mandatory.
Example 7: Divide
z
2z 2  2z
by
.
xy
3 x  3y
solution:
This example brings together several operations illustrated in previous examples. Start by writing
the problem completely in mathematical notation:
z
2z 2  2z
z
3 x  3y


 2
x  y 3 x  3y x  y 2z  2z
David W. Sabo (2003)

 z  3 x  3y 
 x  y   2z2  2z 

 z  3  x  y 
 x  y   2  z   z  1
Multiplication and Division with Two Fractions
Page 6 of 11
3
2  z  1

as the final answer.
Example 8: Multiply
5a 2b 2  10ab3
12a 2b  12ab 2
by
and simplify the result.
4a3 b  4a 2b 2
15a3 b  30a 2b 2
solution:
Although the expressions are rather more complicated here than in any previous examples, the
same old strategy still applies. We need to simplify the product of the two fractions:
5a b  10ab 12a b  12ab 
 4a b  4a b 15a b  30a b 
2
 5a2b2  10ab3   12a2b  12ab2 


3
2 2 
3
2 2 
4
a
b

4
a
b
15
a
b

30
a
b



2
3
3
2
2
2
3
2
2
2
Even though the numerator and denominator here contain products of binomials, there is a
temptation to multiply both of the products out separately. Here this would give us something like
60a 4 b3  60a3 b 4  120a3b 4  120a2b5
60a6b2  120a5b3  60a5a3  120a 4b 4
which can be simplified, but surely has to look like we’re going in the wrong direction. Since
simplifying fractions occurs via cancelling common factors between the numerator and the
denominator, it is always far better to attempt more factorization than to consider multiplying to
remove whatever factors are already present. So, we will abandon (at least for the present)
continuing with this last monstrosity just above, and attempt to factor the original product.
Considering existing factors one at a time, this turns out to be quite easy.
5a2b2  10ab3  5ab2 a  2b 
12a2b  12ab2  12ab a  b 
4a3b  4a2b2  4a2b a  b 
15a3b  30a2b2  15a2b a  2b 
Thus
5a b  10ab 12a b  12ab 
 4a b  4a b 15a b  30a b 
2
2
3
David W. Sabo (2003)
3
2
2
2
3
2
2
2
5ab  a  2b 12ab a  b 

 4a b  a  b  15a b  a  2b 
2
2
Multiplication and Division with Two Fractions
2
Page 7 of 11
 5   2 3  a b a  2b   a  b 
 2   3   5  a b  a  b  a  2b 
2

2
2

4
3
a2
b
2
b  a  2b 
a2  a  2b 
as our final answer. It is tempting to cancel the (a + 2b) in the numerator with the (a – 2b) in the
denominator, but these two expressions are not identical, so it would be an error to cancel them.
(If you enjoy a challenge, or if you are a glutton for punishment, you could attempt to simplify the
gross “multiplied out” expression above. If you follow the strategies for factoring, it’s not
impossible to do, and you will end up with the same final answer. Along the way, you may realize
that some of the work you’re having to do is simply undoing the initial expansion operation –
hence that initial step really was no progress at all!)
Example 9: Simplify the result of dividing
6 p 3 q 3  3 p 2q 4
5 p2q 2  15 pq 3
18 p2q  9 pq 2
.
7 p4q 2  21p3q 3
by
solution:
The overall work to complete this problem is very similar to what appeared in the previous
example, so you might try to do this one by yourself first, before looking at the solution to follow.
6 p 3 q 3  3 p 2q 4
18 p2q  9 pq 2
6 p 3 q 3  3 p 2q 4
7 p 4q 2  21p3q 3



5 p2q 2  15 pq 3
7 p4q 2  21p3q 3
5 p2q 2  15pq 3
18p2q  9pq 2
 6p q  3p q 7p q  21p q 

5p q  15pq 18p q  9pq 
3
3
2
2
2
4
4
3
2
2
3
3
2
To avoid trying to do so many things at once that making errors becomes almost inevitable, we’ll
do the factoring of each bracketed expression first. It’s probably prudent to do these
factorizations in the step-wise fashion described earlier in these notes, since there is a fair
amount of detail to keep track of here. So
6p3q3 + 3p2q4 = 3(2p3q3 + p2q4)
= 3p2(2pq3 + q4)
= 3p2q3(2p + q)
To save space, we’ll just list the final results of the factoring of the remaining three expressions:
7p4q2 – 21p3q3 = 7p3q2(p – 3q)
David W. Sabo (2003)
Multiplication and Division with Two Fractions
Page 8 of 11
5p2q2 – 15pq3 = 5pq2(p – 3q)
18p2q – 9pq2 = 9pq(2p – q)
So,
 6p q  3p q 7p q  21p q   3p q   2p  q  7p q   p  3q 
5p q  15pq 18p q  9pq  5pq   p  3q 9pq  2p  q 
3
2
3
2
4
2
4
3
2
2

2

3
2
2
3
3
2
2
 3  7  p q   2p  q   p  3q 
 5   3   p  q   p  3q   2p  q 


5

3
 7   p5

15 
3
5
2
3
 5 q 
 q
  2p  q 


2
3
p
q  2p  q 
p3
2
  
7 p3q 2  2 p  q 
15  2p  q 
as the final result.
This problem, like the one in the previous example, looked horrible at a first glance. But you
should spend enough time studying it step-by-step to see that as bad as it may look as a whole,
the final result shows up almost automatically if you just carefully employ the basic strategy for
simplifying fractions. Anyone can do the individual steps one-at-a-time required for solving this
problem.
Example 10: Simplify
ab b
 .
a5 5
solution:
This looks too simple. We get
a  b b  a  b  5   a  b  5 
 

a  5 5  a  5 
 b   a  5  b 
Even though there appears to be a lot of similar things in the numerator and denominator, no
further cancellations are possible, and so this is the final answer. Probably most people would
write it as
5 a  b 
b a  5
David W. Sabo (2003)
.
Multiplication and Division with Two Fractions
Page 9 of 11
How do you know that, for example,
   5 b 1
5b
b  a  5
5 a b
is wrong? Because in the first step, the a’s that are cancelled are not factors in either the
numerator or the denominator. If what you’re cancelling are not common factors, then you’re
making a mistake. The same applies here to suggestions that we cancel the b’s or the 5’s. In
both cases, although one is a factor, the other is not, and so cancellation produces an invalid
result. This is an example where, once we’ve done the initial division of one fraction by another,
no further simplification of the result is possible.
Example 11: Simplify:
 18 x 3 y 2  50 x 4 y  15 xy 4 
.

5 
2 2 
5 3 
 75 xy  27 x y  16 x y 
solution:
Although we’ve only looked at products of two factors so far, it is obvious that products of three or
more fractions work the same way – just multiply the numerators together and multiply the
denominators together, and then simplify the results. So here, we get








18 x 3 y 2 50 x 4 y 15 xy 4
 18 x 3 y 2  50 x 4 y  15 xy 4 


5 
2 2 
5 3 
75 xy 5 27 x 2 y 2 16 x 5 y 3
 75 xy  27 x y  16 x y 

18  50 15   x 341  y 214 
 75  27 16   x125  y 523 

 2 3  2 5  3 5  x y
3 5 3  2  x y
2
2
2
8
3
4
8
7
10
 2114321135212 x 88 y 710
 223151 x 0 y 3 
5
5

3
2 3y
12y 3
2
There was a lot of bookkeeping of factors here, so we took a few extra steps to ensure no errors
were made. Recall that any number to the power zero just gives 1, and so
x0 = 1
in the last line above.
David W. Sabo (2003)
Multiplication and Division with Two Fractions
Page 10 of 11
In a similar fashion, the rule for division can be extended to situations involving more than two
fractions. However, in such cases, we would be dealing with some variation of a mixed
multiplication and division problem. You should be able to sort such problems out if you pay
careful attention to brackets or other symbols of grouping.
David W. Sabo (2003)
Multiplication and Division with Two Fractions
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