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The Mathematics 11
Competency Test
Simplifying Complex Fractions
Complex fractions are fractions whose numerator and denominator themselves contain
fractions. The goal of simplifying a complex fraction is to rewrite it as an equivalent fraction in
which the numerator and denominator do not contain any other fractions. When we’re
considering numerical fractions, this means that the numerator and the denominator of the final
form are each just single numbers.
The most effective procedure consists of several steps involving the operations we have already
described in detail:
1. simplify the expression in the numerator of the complex fraction to a single fraction in
simplest form
2. simplify the expression in the denominator of the complex fraction to a single fraction in
simplest form
3. the original complex fraction now looks like one fraction divided by another fraction.
Carry out the division and simplify the result.
Example: Simplify
1 2
+
2 3 .
2 3
−
3 5
solution:
This is a complex fraction because the numerator is a sum of two fractions and the denominator
is a difference of two fractions. You might be tempted to immediately do some cancellations
(since, for example, the numerator and denominator both contain 2
3
, etc.), but none of the
common “things” in any of the fractions present are factors of either the numerator or the
denominator, and so cancellations to attempt to simplify things here would be an error. Instead,
we follow the systematic approach described above.
(1) simplify the expression in the numerator of the complex fraction to a single fraction in simplest
form:
1 2 1× 3 2 × 2 3 4 7
+ =
+
= + =
2 3 2×3 3× 2 6 6 6
(2) simplify the expression in the denominator of the complex fraction to a single fraction in
simplest form:
2 3 2 × 5 3 × 3 10 9
1
− =
−
=
−
=
3 5 3 × 5 5 × 3 15 15 15
(3) now:
David W. Sabo (2003)
Simplifying Complex Fractions
Page 1 of 3
1 2
+
2 3 =
2 3
−
3 5
7
6
1
15
=
7 15 7 × 15 7 × 3 × 5 35
=
×
=
=
6 1
6
2
2× 3
This last sequence includes whatever simplification appears possible. Thus, the final answer is
1 2
+
2 3
2 3
−
3 5
=
35
2
5
12
3 7
+
4 18
4−
Example: Simplify
.
solution: In steps
4−
5 4 × 12 5 48 − 5 43
=
−
=
=
12 1× 12 12
12
12
3 7 3 × 9 7 × 2 27 14 41
+
=
+
=
+
=
4 18 4 × 9 18 × 2 36 36 36
So
5
12
3 7
+
4 18
4−
43
12
41
36
=
=
=
43 36 43 × 36
×
=
12 41 12 × 41
43 × 2 × 2 × 3 × 3 43 × 3 129
=
=
41
41
2 × 2 × 3 × 41
as the final answer.
David W. Sabo (2003)
Simplifying Complex Fractions
Page 2 of 3
The method described here also covers the case where either a whole number is divided by a
fraction, or a fraction is divided by a whole number.
Examples:
5
3
7
3
7
5
=
5
1
3
7
=
5 7 5 × 7 35
× =
=
1 3 1× 3
3
=
3
7
5
1
=
3 1 3 ×1 3
× =
=
7 5 7 × 5 35
You can see that the general rule for this kind of problem is:
a
b
c
a
b
c
=
a×c
b
=
a
b×c
When a fraction is divided by a whole number, you just multiply that whole number into the
denominator of the fraction.
David W. Sabo (2003)
Simplifying Complex Fractions
Page 3 of 3