Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
4.6 Adding and Subtracting Mixed Numbers
We can use similar techniques from 4.5 to add and subtract mixed numbers.
Adding and Subtracting Mixed Numbers
1) Estimate the answer first.
 Round each mixed number to the nearest whole number.
 Perform the indicated operation to get an estimate of the answer.
2) Rewrite each mixed number in the problem as an improper fraction.
3) Add or subtract using methods from 4.4.
4) Write the answer in lowest terms if possible.
5) If possible, convert the answer to a mixed number.
6) Make sure the answer is reasonable. Compare it to the estimate. It should be
relatively close.
1  2
Example 1: For 4   9  , first round each number to estimate the answer. Then find the exact answer.
5  5
Solution:
1
2
Estimate: 4  4 and 9  9
5
5
1  2
So, 4   9   4   9 
5  5
 5
1  2
Actual: Convert each mixed number to an improper fraction. 4   9 
5  5

21  47 
 
5  5 

21  47 


5  5 
1
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers

Add as usual.


21   47 
5
26
26
or 
5
5
26
is improper.
5
Since the original problem started with mixed numbers, convert 

26
to a mixed number.
5
26
1
1
 5 . Check! 5 is relatively close to our estimate of 5 .
5
5
5
1  2
1
Therefore we are confident that 4   9   5 .
5  5
5
You Try It 1: First round each number to estimate the answer. Then find the exact answer.
3 
1 
3
1
a) 8   1 
b) 8  1
10  10 
10 10
1  5
Example 2: For 3   1  , first round each number to estimate the answer. Then find the exact answer.
4  6
Solution:
1
5
Estimate: 3  3 and 1  2
4
6
1  5
So, 3   1   3   2 
4  6
5
2
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
1  5
Actual: Convert each mixed number to an improper fraction. 3   1 
4  6

13  11 
 
4  6

13  11 


4  6 
Subtract as usual. Since the denominators are different, we find the LCD = 12
Rewrite each fraction as an equivalent fraction that has 12 as its new denominator.
13
3
needs to be multiplied by to get 12 as its denominator.
4
3
11
2
needs to be multiplied by
to get 12 as its denominator.
6
2
13  11  13 3  11 2 

 
   
4  6  4 3  6 2




39  22 


12  12 
39   22 
12
39   22 
12
61
12
61
is improper.
12
Since the original problem started with mixed numbers, convert
61
to a mixed number.
12
61
1
1
 5 . Check! 5
is relatively close to our estimate of 5 .
12
12
12
1  5
1
Therefore we are confident that 3   1   5 .
4  6
12
3
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
You Try It 2: First round each number to estimate the answer. Then find the exact answer.
1
5
1  5
a) 8   5 
b) 8  5
4
6
4  6
Example 3: For 8
7 
3 
  3  , first round each number to estimate the answer. Then find the exact answer.
15  10 
Solution:
Estimate: 8
7
3
 8 and 3  3
15
10
So, 8
7 
3
  3   8   3
15  10 
 11
Actual: Convert each mixed number to an improper fraction. 8


7 
3 
  3 
15  10 
127  33 
 
15  10 
127  33 


15
 10 
Add as usual. Since the denominators are different, we find the LCD = 30
Rewrite each fraction as an equivalent fraction that has 30 as its new denominator.
127
2
needs to be multiplied by
to get 30 as its denominator.
15
2
4
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
33
3
needs to be multiplied by to get 30 as its denominator.
3
10
127  33  127 2  33 3 

 
 

15
15 2  10 3 
 10 




254  99 


30
 30 
254   99 
30
353
353
or 
30
30
353
is improper.
30
Since the original problem started with mixed numbers, convert 

353
to a mixed number.
30
353
23
23
is relatively close to our estimate of 11 .
 11 . Check! 11
30
30
30
Therefore we are confident that 8
7 
3
23
  3   11 .
15  10 
30
You Try It 3: First round each number to estimate the answer. Then find the exact answer.
3  2
3  2
a) 18   21 
b) 18   21 
4  5
4  5
5
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
1
5
Example 4: For 8  5 , first round each number to estimate the answer. Then find the exact answer.
4
6
Solution:
1
5
Estimate: 8  8 and 5  6
4
6
1
5
So, 8  5  8  6
4
6
 14
1
5
Actual: Convert each mixed number to an improper fraction. 8  5
4
6


33 35

4
6
33 35

4
6
Subtract as usual. Since the denominators are different, we find the LCD = 12
Rewrite each fraction as an equivalent fraction that has 12 as its new denominator.
33
3
needs to be multiplied by to get 12 as its denominator.
3
4
35
2
needs to be multiplied by
to get 12 as its denominator.
2
6
33 35 33 3 35 2


  
4
6
4 3 6 2

99 70

12 12

99  70
12

99   70 
12
169
169
or 

12
12
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2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers

169
is improper.
12
Since the original problem started with mixed numbers, convert 

169
to a mixed number.
12
169
1
1
is relatively close to our estimate of 14 .
 14 . Check! 14
12
12
12
1
5
1
Therefore we are confident that 8  5  14 .
4
6
12
You Try It 4: First round each number to estimate the answer. Then find the exact answer.
1
2
1
5
a) 12  2
b) 8  5
4
5
4
6
Example 5: First round each number to estimate the answer. Then find the exact answer.
3
3
a) 12  4
b) 12  4
8
8
Solution:
3
a) Estimate: 12 is already an integer and 4  4
8
3
So, 12  4  12  4
8
 16
7
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
Actual: Convert any mixed number to an improper fraction. 12  4
 12 

3
8
35
8
12 35

1
8
Add as usual. Since the denominators are different, we find the LCD = 8
Rewrite each fraction as an equivalent fraction that has 8 as its new denominator.
12
8
needs to be multiplied by to get 8 as its denominator.
1
8
35
already has 8 as its denominator.
8
12 35 12 8 35

  
1 8
1 8 8

96 35

8
8

96  35
8

131
8
131
is improper.
8
Since the original problem started with mixed numbers, convert
131
to a mixed number.
8
131
3
3
 16 . Check! 16 is relatively close to our estimate of 16 .
8
8
8
3
3
Therefore we are confident that 12  4  16 .
8
8
8
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
3
b) Estimate: 12 is already an integer and 4  4
8
3
So, 12  4  12  4
8
8
Actual: Convert any mixed number to an improper fraction. 12  4
 12 

3
8
35
8
12 35

1
8
Subtract as usual. Since the denominators are different, we find the LCD = 8
Rewrite each fraction as an equivalent fraction that has 8 as its new denominator.
12
8
needs to be multiplied by to get 8 as its denominator.
1
8
35
already has 8 as its denominator.
8
12 35 12 8 35

  
1 8
1 8 8

96 35

8
8

96  35
8

61
8
61
is improper.
8
Since the original problem started with mixed numbers, convert
61
to a mixed number.
8
9
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
61
5
5
 7 . Check! 7 is relatively close to our estimate of 8 .
8
8
8
3
5
Therefore we are confident that 12  4  7 .
8
8
You Try It 5: First round each number to estimate the answer. Then find the exact answer.
1
1
a) 14  8
b) 14  8
6
6
Note: You may feel that the procedure to add or subtract mixed numbers seems long and tedious. Sometimes it
may very well be. Alternative methods to add or subtract mixed numbers exist and can make some of
the problems slightly less tedious.
However, sometimes these other methods lead to other complications such as borrowing and regrouping
as well as possible errors involving the signs of the final answer or fraction part.
If you have learned these alternative methods from a previous math course, we welcome you to use them
but please consider our warning above.
If you are interested in learning alternative methods to add or subtract mixed numbers, we welcome you
to ask a tutor or your instructor.
10
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
Example 6: Find the perimeter.
11
7 ft
3
ft
4
8 ft
9
1
ft
2
Solution:
To find the perimeter of the triangle we find the sum of all three sides.
We do not need the 7 ft measurement to find the perimeter.
P  8 ft  9
1
3
ft  11 ft
2
4
Before we calculate the exact perimeter, it is helpful to estimate the answer by first rounding
any mixed number to the nearest integer:
P  8 ft  9
1
3
ft  11 ft  8  10  12
2
4
 30 ft
(This way we know that our final answer should be relatively close to 30 ft .)
Continuing with the actual calculation:
P 89
1
3
 11
2
4
ft
Perimeter is measured in just feet ( or ft ).
P 89
1
3
 11
2
4
ft
Write any mixed numbers as improper fractions.
P 8
19 47

2
4
8 4 19 2 47
P    
1 4 2 2 4
P
32 38 47
 
4
4
4
P
32  38  47
4
Simplify.
ft
ft
LCD = 4
ft
ft
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2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
P
117
ft
4
117
is improper!
4
Convert
P  29
117
to a mixed number for a better understanding of the measurement.
4
1
ft
4
Check! Is this close to our original estimate of 30 ft ? Yes!
Now we can be confident that we have calculated the correct actual perimeter.
The perimeter of the triangle is 29
1
ft .
4
You Try It 6: Find the perimeter.
3
9 in
8
1
9 in
4
5 in
7 in
12
2015 Carreon
Math 40
Prealgebra
Section 4.6 – Adding and Subtracting Mixed Numbers
Example 7: First, round any mixed numbers to estimate the answer. Then find the exact answer.
7
Elise has a metal rod of length 10 inches. For a project, she needs to cut a 2 inch piece from the
16
metal rod. What is the length of the remaining piece?
Solution:
First round the mixed number to the nearest integer: 2
7
 2 inches.
16
Reread the problem using the rounded number to get a better understanding of the problem so you
can select the correct operation to use:
Elise has a metal rod of length 10 inches. For a project, she needs to cut a 2 inch piece from the
metal rod. What is the length of the remaining piece?
Cutting a 2 inch piece from the 10 inch rod indicates subtraction: 10  2  8 inch piece remaining.
7 10 39
 
16 1 16
10 16 39
  
1 16 16
160  39

16
9
9
Check! Is 7
relatively close to 8? Yes!
7
16
16
9
State the final answer: The length of the remaining piece is 7
inches.
16
Now calculate the actual answer: 10  2
You Try It 7: First, round any mixed numbers to estimate the answer. Then find the exact answer.
James has a piece of curtain material that measures 12 feet. He cuts a piece measuring 6
2
3
feet from the curtain material. What is the length of the remaining piece?
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