jovan algebra | Page 8

Chapter 1
being large or by having variables in them, it usually easier to use the LCD to
add or subtract fractions. The solution requires less simplifying, too.
In the following practice problems one of the denominators will be the LCD;
you only need to rewrite one fraction before computing the sum or difference.
PRACTICE
Find the sum or difference.
1.
1
8 1
2
+=
2. 2
3 5
12
−=
3. 4
5 1
20
+=
4. 7
30 2
15
−=
5. 5
24 5
6
+=
SOLUTIONS
1. 1
8 1
2 1
8 1
2 4
4 1
8 4
8 5
8
+=
+⋅

 


=+
=
2. 2
3 5
12 2
3 4
4 5
12 8
12 5
12 3
12 1
4
−=
⋅


 


−
=−= =
3. 4
5 1
20 4
5 4
4 1
20 16
20 1
20 17
20
+=
⋅


 


+=
+=
4. 7
30 2
15 7
30 2
15 2
2 7
30 4
30 3
30 1
10
−=
−⋅

 


=−
==
5. 5
24 5
65
24 5
64
4 5
24 20
24 25
24
+=
+⋅

 


=+
=
 
Finding the LCD
We have a couple of ways for finding the LCD. Take, for example, 1
12 9
14
+. We
could list the multiples of 12 and 14—the first number that appears on each
list is the LCD: 12, 24, 36, 48, 60, 72, 84 and 14, 28, 42, 56, 70, 84.
PRACTICE
Find the sum or difference.
SOLUTIONS✔

Chapter 1 FraCtions 17
DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
Because 84 is the first number on each list, 84 is the LCD for 1
12 and 9
14. This
method works fine as long as the lists aren’t too long. But what if the denomi-
nators are 6 and 291 for example? The LCD for these denominators (which is
582) occurs 97th on the list of multiples of 6. We can use the prime factors of the denominators to find the LCD more
efficiently. The LCD consists of every prime factor in each denominator (at its
most frequent occurrence). To find the LCD for
1
12 and 9
14, we factor 12 and 14
into their prime factorizations: 12 = 2 ⋅ 2 ⋅ 3 and 14 = 2 ⋅ 7. There are two 2’s
and one 3 in the prime factorization of 12, so the LCD will have two 2’s and
one 3. There is one 2 in the prime factorization of 14, but this 2 is covered by
the 2’s from 12. There is one 7 in the prime factorization of 14, so the LCD
will also have a 7 as a factor. Once we have computed the LCD, we divide the
LCD by each denominator and then multiply the fractions by these numbers
over themselves.
LCD = 2⋅2 ⋅3 ⋅7 = 84
84 ÷ 12 = 7: multiply
1
12 by 7
7 84 ÷ 14 = 6: multiply 9
14 by 6
6
1
12 9
14 1
12 7
7 9
14 6
6 7
84 54
84 61
84
+=
⋅


 


+⋅ 

 


=+
=
EXAMPLE
Find the sum or difference after computing the LCD.

5
64
15
+
SOLUTION
We begin by factoring the denominators: 6 = 2
⋅ 3 and 15 = 3 ⋅ 5. The
LCD is 2 ⋅ 3 ⋅ 5 = 30. Dividing 30 by each denominator gives us 30 ÷ 6 = 5
and 30 ÷ 15 = 2. Once we multiply
5
6 by 5
5 and 4
15 by 2
2, we can add the
fractions.

5
6 4
15 5
65
5 4
15 2
2 25
30 8
30 33
30 1
+= ⋅


 


+⋅ 

 


=+
==11
10
EXAMPLE
Find the sum or difference after computing the LCD.
SOLUTION
We begin by factoring the denominators: 6 = 2
✔

18 alGebra D e myst i fieD DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
EXAMPLE
Find the sum or difference after computing the LCD.
17
24
5
36
+
SOLUTION
24 = 2
⋅ 2 ⋅ 2 ⋅ 3 and 36 = 2 ⋅ 2 ⋅ 3 ⋅ 3
The LCD = 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 = 72; 72 ÷ 24 = 3; and 72 ÷ 36 = 2. We multiply
17
24 by
3
3 and 5
36 by 2
2.
17
24 5
36 17
243
3 5
36 2
2 51
72 10
72 6
+= ⋅


 


+⋅ 

 


=+
=11
72
PRACTICE
Find the sum or difference after computing the LCD.
1.
11
12 5
18
−=
2. 7
15 9
20
+=
3. 23
24 7
16
+=
4. 3
8 7
20
+=
5. 1
6 4
15
+=
6. 8
75 3
10
+=
7. 35
54 7
48
−=
8. 15
88 3
28
+=
9. 119
180 17
210
+=
EXAMPLE
Find