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EVALUATING FRACTIONS
Two or more fractions can
evaluated by addition, subtraction, multiplication, division or a combination
of two or more of the above operations.
Fractions often appear easy to evaluate from first sight but a careful
attention often indicate the contrary. This was revealed by a survey I carried
out in a college form two class some two months back. Most of the students
acknowledged that fractions are very easy to evaluate but were not able to
explain why some still got some of the exercises wrong. This is one of the
reasons why I have decided to treat fractions and many other topics in Number Theory
here.
This means before solving any fraction, it must be well understood in
order to apply the correct method. Some students give wrong solutions to
questions involving fractions because they often assume that it is easy hence
begin solving without interpreting the question.
In this write-up, you will learn how to add,
subtract, multiply, divide fractions that: have the same denominators, have
different denominators, contain mixed fractions. You will be given a detail
guide on what to do concerning the evaluation of the different fractions above.
Each explanation ends with some practice exercises for you to solve and
evaluate our level of understanding. There will also be short VIDEOS to explain
some of the exercises. This is to ensure the concepts are well understood. You
can make request for further explanation if a concept is not understood. Use
the comment section for this.
Without any waste of time, let
us get into business right away. Follow the explanations with attention and
post your questions in the comments sections of this blog and be patient for
your answer.
A) ADDING OR SUBTRACTING PROPER AND IMPROPER
FRACTIONS.
A proper fraction is one in
which the numerator is less that the denominator example 5 .
7
An improper fraction is one in which the
numerator is greater than the denominator. For example, 11
3
. A mixed fraction is made up of
a whole number and a fraction. Example 5 2 .
3
To add or subtract fractions,
check if the denominators are same or different. If they are same, add or
subtract ONLY the numerators and maintain the denominator.
EXAMPLE 1: Evaluate each of the
following simplifying your result as far as possible
i)
1 + 3
2 2
ii)
4 + 2
7 7
iii)
5 – 3
8 8
iv)
7 –
4
11 11
SOLUTION
To simplify your answer means you write
it to a level where it can no longer be worked on.
i)
1 + 3 =? ( add only the numerators)
2 2
1
+ 3 = (1+3)
2
2 2
1
+ 3 = 4
2 2 2
1
+ 3 = 2 Ans.
2 2
ii)
4 + 2 =? ( add only numerators)
7
7
4 + 2
= (4+2)
7
7 7
4
+ 2 = 6
7
7 7 answer
5 _ 3 =?
( subtract only numerators)
8
8
5 _
3 = (5-3)
8 8 8
5 - 3 = 2
8 8
8
5 - 3 = 1
8 8 4 answer.
7 - 4 = ?
(Subtract only numerators)
11 11
7 - 4 = (7-4)
11 11
11
7 - 4 = 3
11 11 11
answer
EXERCISE FOR PRACTICE
Evaluate each of
the following simplifying your answer as far as possible.
i)
4 + 3 ii)
5 + 4 iii) 8 - 5 iv)
14 - 6 .
11
11 , 9
9 , 13
13 , 17
17
ANSWERS.
i) 7 . ii)
1.
iii) 3 .
iv) 8 .
11 13 17
B)
If the denominators are different, use their LOWEST COMMON MULTIPLE to simplify
them. The lowest common multiple denoted LCM is the product of the denominators
present.
EXAMPLE: Evaluate each of the following fractions
simplifying your result as far as possible.
i)
1 + 2 ii) 4 + 3
iii) 8 – 1 iv) 5
-2
3 5 7 2 11
2 9 3
SOLUTION
The denominators of each of the fractions above are
different. Use the lowest common multiple of each to evaluate it.
i)
1 + 2 = ?+ ? ( Where (3)(5) is the lowest common
multiple)
3 5 (3)(5)
Divide the lowest common multiple (LCM) by each of the
denominators and multiply your result by the corresponding numerator.
That is, (3)(5)÷3
=5. Multiply 5 by 1 which is the numerator of the fraction 1 to give
5(1).
3
Similarly,
(3)(5)÷5=3. Multiply 3 by 2 the numerator of the fraction 2 to give
3(2).
5
Substituting 5(1) and 3(2) respectively in place of”?” On
the right hand side of (i) above gives;
1 + 2 = 5(1) + 3(2)
3 5 (3)(5)
1 + 2 =
5 + 6
3 5 (3)(5)
1 + 2 =
11
3 5 15 answer.
ii)
4 + 3 = ? + ?
( where (7)(2) is the LCM )
7 2 (7)(2)
Divide the lowest common multiple (LCM) by each of the
denominators and multiply your result by the corresponding numerator.
That is, (7)(2) ÷7=2 . Multiply 2 by 4 the numerator of the
fraction 4 to obtain 2(4).
7
Similarly, (7)(2) ÷2=7. Multiply 7 by 3 the numerator of
the fraction 3 to obtain 7(3).
2
Substituting 2(4) and 7(3) respectively in place of “?” on
the right hand side of (ii) above gives;
4 + 3 =
2(4) +7(3)
7 2 (7)(2)
4 + 3 =
8 + 21
7 2 14
4 + 3 = 29
7 2 14.
iii)
8 -
1 = ? +?
11 2 (11)(2)
As above,
(11)(2) ÷11 =2. Multiply 2 by 8 the numerator of the
fraction 8 to obtain (2)(8).
11
(11)(2) ÷2 = 11. Multiply 11 by 1 the numerator of the fraction
1 to obtain (11)(1).
2
Substituting the products (2)(8) and (11)(1) in place of
“?” on the right hand side of (iii) above gives;
8 - 1 = (2)(8) - (11)(1)
11 2 ( 11)(2)
8 - 1 =
16 -11
11 2 22
8 - 1 = 5
11 2 22
iv)
5 - 2 = ? + ?
9 3 (9)(3)
As above ;
(9)(3) ÷9 = 3. Multiply 3 by 5 the numerator of the
fraction 5 to obtain (3)(5).
9
(9)(3)÷3=9. Multiply
9 by 2 the numerator of the fraction 2 to obtain (9)(2).
3
Replacing “?” in iv above by (3)(5) and (9)(2) respectively gives;
5 - 2 =
(3)(5) - (9)(2)
9 3 (9)(3)
5 – 2 = 15
- 18
9 3 27
5 – 2
= -3 = -1
9 3 27 9 answer.
EXERCISE FOR PRACTICE
Evaluate each of the following
simplifying your results as far as possible.
i)
3 +
1 (ii) 3 + 4 (iii)
2 - 4 (iv) 3 - 8
5 4 4
5 3 5 7 9
Answers: i) 17
ii) 19 iii) -2 iv -13
20 20 15 35
EVALUATING MIXED FRACTIONS
A mixed fraction is any fraction that is made up of a whole
number and a fraction. For example 4 2
3
Where 4 is the
whole number and 2 is the
fraction.
3
To add or subtract mixed fractions, convert the mixed
fraction into an improper fraction and proceed as in the exercises treated so
far. If the denominators are the same, convert the mixed fraction into an
improper fraction and add or subtract ONLY the numerators and maintain the
denominator.
EXAMPLE 1: Evaluate and simplify each of the following
simplifying your answer as far as possible.
i)
31 + 41 ii) 4 2 -
51
2 2 3 3
Solution
i)
31 + 41 (convert each fraction to an improper fraction
and simplify)
2 2
To convert a
mixed fraction to an improper fraction, add the numerator to the product of the
whole number and denominator and maintain the denominator.
31 + 41 = 1+ (3×2) + 1+ (4×2)
2 2 2 2
31 + 41 = 1
+ 6 + 1 + 8
2 2 2 2
31 + 41 = 7 + 9 ( add only numerators)
2
2 2
2
31 + 41 = 2+9
2 2 2
31 + 41 = 11
2 2 2
31 + 41 = 51 Answer.
2 2 2
ii)
42 - 51 (convert
to improper fraction and simplify).
3 3
42 - 51 = 2+ ( 4×3) -- 1+( 5×3)
3 3 3 3
42 -- 51 = 2+ 12 -- 1+ 15
3 3 3 3
42 -- 51 = 14 - 16
3 3 3
42 -- 51 = -2 Answer.
3 3 3
If
the mixed fractions have different denominators, convert to improper fractions
and look for the LCM of the fractions .Proceed as above.
Example 2:
Evaluate each of the following simplifying your result as far as possible.
i)
31 + 42 ii) 42 –51
2 3 3 2
Solution
i)
31 + 42 ( Convert to improper fractions and
simplify)
2 3
31 + 42 = 1 +( 3×2) + 2 +(4× 3)
2 3 2 3
31 + 42 = 1+6
+ 2+ 12
2 3 2 3
31 + 42 = 7
+ 14 = ? + ?
2 3 2 3 (2)(3)
3 1 + 4 2 = 7 + 14 = 3(7) + 2(14)
2 3 2 3 (2)(3)
2 3 2 3 (2)(3)
31 + 42 = 21
+ 28
2 3 6
31 + 42 = 49 or 41
2 3 6 6 Answer.
ii)
73 -- 21 ( Convert to improper fractions)
4 5
73 -- 21 = 3+ (7×4) -- 1+
( 2×5)
4 5 4 5
73 -- 21 = 3+ 28 -- 1+10
4 5 4 5
73 --- 2 1 = 31 --
11 = ? --
?
4 5 4 5 ( 4)(5)
7 3 -- 21 = 31(5) --
11(4)
4
5 20
73 -- 21 = 155—44
4
5
20
73 - 2 1 = 111 OR 511
4 5 20 20 Answer.
EXERCISE FOR PRACTICE
Evaluate each of
the following, simplifying your answer as far as possible.
i)
31 + 2 1 ii)
31 + 23 (iii)
22 --- 11 . Answers:
i) 6 , ii) 14 ,
iii) 9
2
2 2 5 3 5
15 10
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