jovan algebra | Page 7

11
11
1
+= +
==
Adding and Subtracting Fractions with Unlike Denominators
If we need to find the sum or difference of two fractions having different
denominators, then we must rewrite one or both fractions so that they have the
same denominator. Let us use the pie model to find the sum
1
4 1
3
+.
If we divide the pie into
34 12
×= equal pieces, we see that 1
4 is the same as
3
12 and 1 3 is the same as 4
12.
Now that we have these fractions written so that they have the same deno-
minator, we can add them:
1 4 1
3 3
12 4
12 34
12 7
12
+=
+= +
=.
1
4
1
3
FIGURE 1-8
4
12
3
12
FIGURE 1-9
SOLUTIONS✔

14 alGebra D e myst i fieD DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
To compute a
b c
d
+ or a
b c
d
−, we can “reverse” the simplification process
to rewrite the fractions so that they have the same denominator. This process
is called finding a common denominator. Multiplying
a
b by d
d (the second denom-
inator over itself ) and
c
d by b
b (the first denominator over itself ) gives us equiv-
alent fractions that have the same denominator. Once this is done, we can add
or subtract the numerators.

a b c
d a
b d
d c
d b
b ad
bd cb
bd
+=
⋅+⋅= +
Noww ecan ad dthe nu
um erators. 
=+
ad cb
bd
a b c
d a
b d
d c
d b
b ad
bd cb
bd
−=
⋅−⋅= −
Noww ecan subtractt
th en um erators. 
=−
ad cb
bd
Note that this is essentially what we did with the pie chart to find 1 4 1
3
+
when we divided the pie into 43
12
×= equal parts.
For now, we will use the formula
a b c
d ad
cb
bd
±=
± to add and subtract two
fractions. Later, we will learn a method for finding a common denominator
when the denominators have common factors.
EXAMPLES
Find the sum or difference.
1
2 3
7
8
15 1
2
+
−
SOLUTIONS
In this sum, the first denominator is 2 and the second denominator is 7. We
multiply the first numerator and denominator of the first fraction,
1
2, by 7
and the numerator and denominator of the second fraction,
3
7, by 2. This
gives us the sum of two fractions having 14 as their denominator.
1
2 3
7 1
2 7
7 3
7 2
2 7
14
  
                 
+= ⋅


 


+⋅ 

 


=  
      
+= 6
14 13
14
8
15 1
2 8
15 2
2 1
2 15
15
  
                
−= ⋅


 


−⋅ 

 


=  
     
16
30 15
30 1
30
−=
SOLUTIONS
In this sum, the first denominator is 2 and the second denominator is 7. We
✔
EXAMPLES
Find the sum or difference.

Chapter 1 FraCtions 15
DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
PRACTICE
Find the sum or difference.
1.
5
6 1
5
−=
2. 1
3 7
8
+=
3. 5
7 1
9
−=
4. 3
14 1
2
+=
5. 3
4 11
18
+=
SOLUTIONS
1. 5
6 1
5 5
65
5 1
5 6
6 25
30 6
30 19
30
−=
⋅


 


−⋅ 

 


=−
=
2. 1
3 7
8 1
3 8
8 7
8 3
3 8
24 21
24 29
24
+=
⋅


 


+⋅ 

 


=+
=
3. 5
7 1
9 5
7 9
9 1
9 7
7 45
63 7
63 38
63
−=
⋅


 


−⋅ 

 


=−
=
4. 3
14 1
2 3
14 2
2 1
2 14
14 6
28 14
28 20
28
+=
⋅


 


+⋅ 

 


=+
===5
7
5. 3
4 11
18 3
4 18
18 11
18 4
4 54
7244
72 9
+= ⋅


 


+⋅ 

 


=+
=88
72 49
36
=
The Least Common Denominator (LCD)
Our goal is to add/subtract two fractions having the same denominator. In the
previous example problems and practice problems, we found a common
denominator. Now we will find the
least common denominator (LCD). For
example in
1 3 1
6
+, we could compute 1
3 1
6 1
3 6
6
1
63
3 6
18 3
18 9
18 1
2
+=
⋅()+⋅()=+ ==.
But we really only need to rewrite
1 3: 13 1
6 1
3 2
2 1
62
6 1
6 3
6 1
2
+=
⋅()+=+= =.
While 18 is a common denominator in the above example, 6 is the smallest
common denominator. When denominators get more complicated, either by
SOLUTIONS✔
PRACTICE
Find the sum or difference.

16 alGebra D e myst i fieD DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 /