jovan algebra | Page 6

11 11
112
11 2
11
=
⋅
⋅ =⋅
=
5. 39
123 31
3
34 13
313
41 13
41
=
⋅
⋅ =⋅
=
SOLUTIONS✔
PRACTICE
Simplify the fraction.

Chapter 1 FraCtions 9
DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
6. 18
4 23
3
22 2
233
2 9
2
=
⋅⋅
⋅ =⋅
⋅
=
7. 7
210 7
23 57 71
7235 7
71
23 51
30
=
⋅⋅⋅ =
⋅
⋅⋅⋅ =⋅
⋅⋅=
8. 240
165 222235
35 11 35 2222
35 11
=
⋅⋅⋅⋅⋅
⋅⋅ =
⋅
⋅⋅⋅⋅
⋅⋅ =
()
()3
35
35 2222
11
15
15 16
11 16
11 ⋅
⋅
⋅
⋅⋅⋅
=⋅ =
9. 55
3351
1
31 111
5
11 3 11
11 5
3 5
3
=
⋅
⋅ =
⋅
⋅=⋅
=
10. 150
30 2
355
23 5 23
55
23 51 23
5
23
=
⋅⋅⋅
⋅⋅ =
⋅⋅
⋅
⋅⋅ ⋅=
⋅⋅
⋅⋅
()
()
55 5
1 30
30
55
⋅= ⋅=
The Greatest Common Divisor
Fortunately there is a less tedious method for writing a fraction in its lowest
terms. We find the largest number that divides both the numerator and the
denominator. This number is called the greatest common divisor (GCD). We
factor the GCD from the numerator and denominator and then we rewrite
the fraction in the form:
GC D
GC D Ot
hern um erator fact ors
Ot herd enomin ator
⋅
f
act ors
In the previous practice problems, the product of the common primes was
the GCD.
EXAMPLES
Identify the GCD for the numerator and denominator and write the fraction
in lowest terms.
32
48 16
2
16 316
16 2
3 1
2
3 2
3
=
⋅
⋅=⋅
=⋅=
45
60 15
3
15 4 15
15 3
4 1
3
4 3
4
=
⋅
⋅ =⋅
=⋅=
EXAMPLES
Identify the GCD for the numerator and denominator and write the fraction

10 alGebra D e myst i fieD DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
PRACTICE
Identify the GCD for the numerator and denominator and write the fraction
in lowest terms.
1.
12
38 =
2. 12
54 =
3. 16
52 =
4. 56
21 =
5. 45
100 =
6. 48
56 =
7. 28
18 =
8. 24
32 =
9. 36
60 =
10. 12
42 =
SOLUTIONS
1. 12
38 26
21 9 2
26
19 6
19
=
⋅
⋅ =⋅
=
2. 12
54 62
69 6
62
9 2
9
=
⋅
⋅ =⋅
=
3. 16
52 44
41 3 4
44
13 4
13
=
⋅
⋅ =⋅
=
4. 56
21 78
73 7
78
3 8
3
=
⋅
⋅ =⋅
=
5. 45
100 59
52 0 5
59
20 9
20
=
⋅
⋅ =⋅
=
SOLUTIONS✔
PRACTICE
Identify the GCD for the numerator and denominator and write the fraction

Chapter 1 FraCtions 11
DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
6. 48
56 86
87 8
86
7 6
7
=
⋅
⋅ =⋅
=
7. 28
18 21
4
29 2
214
9 14
9
=
⋅
⋅ =⋅
=
8. 24
32 83
84 8
83
4 3
4
=
⋅
⋅ =⋅
=
9. 36
60 12
3
12 512
12 3
53
5
=
⋅
⋅=⋅
=
10. 12
42 62
67 6
62
7 2
7
=
⋅
⋅=⋅
=
Sometimes the greatest common divisor is not obvious. In these cases we
might want to simplify the fraction in multiple steps.
EXAMPLES
Write the fraction in lowest terms.
3990
6762 6
665
6 1127 665
1127 79
5
7 161 95
161
=
⋅
⋅ ==
⋅
⋅ =
644
2842 2
322
2 1421 322
1421 74
6
7 203 46
203
=
⋅
⋅ ==
⋅
⋅ =
PRACTICE
Write the fraction in lowest terms.
1.
600
1280 =
2. 68
578 =
3. 168
216 =
4. 72
120 =
5. 768
288=
EXAMPLES
Write the fraction in lowest terms.
PRACTICE
Write the fraction in lowest terms.

12 alGebra D e myst i fieD DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
SOLUTIONS
1. 600
1280 10 60
10 128 60
128 41
5
43 2 15
32
=
⋅
⋅ ==
⋅
⋅ =
2. 68
578 23
4
2 289 34
289 17
2
17 17 2
17
=
⋅
⋅ ==
⋅
⋅ =
3. 168
216 62
8
63 6 28
36 47
49 7
9
=
⋅
⋅ ==
⋅
⋅ =
4. 72
120 12
6
12 10 6
10 23
25 3
5
=
⋅
⋅ ==
⋅
⋅=
5. 768
288 4
192
47 2 192
72 29
6
23 6 96
36 42
4
49 24
= ⋅
⋅ ==
⋅
⋅==
⋅
⋅ =
99 38
33 8
3
=
⋅
⋅ =
For the rest of the book, we will write fractions in lowest terms.
Adding and Subtracting Fractions with Like Denominators
If we want to add or subtract two fractions having the
same denominators, we only need to add or subtract their
numerators. The rule is
a
b c
b ac
b
+=
+ and a
b c
b ac
b
−=
−.
Let us examine the sum
1
6 2
6
+ with a pie chart.
Adding 1 one-sixth segment to 2 one-sixth segment
gives us a total of 3 one-six segments, which agrees with
the formula:
1
6 2
6 12
6 3
6 1
2
+=
+
==.
EXAMPLES
Perform the addition or subtraction.
7
9 2
9 72
9 5
9
−=
−
=
8
15 2
15 82
15 10
15 52
53 2
3
+=
+
== ⋅
⋅=
PRACTICE
Perform the addition or subtraction.
1.
4
7 1
7
−=
2. 1
5 3
5
+=
1
61
61
6
FIGURE 1-7
SOLUTIONS✔
EXAMPLES
Perform the addition or subtraction.
PRACTICE
Perform the addition or subtraction.

Chapter 1 FraCtions 13
DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
3. 1
6 1
6
+=
4. 5
12 1
12
−=
5. 2
11 9
11
+=
SOLUTIONS
1. 4
7 1
7 41
7 3
7
−=
−
=
2. 1
5 3
5 13
5 4
5
+=
+
=
3. 1
6 1
6 11
6 2
6 1
3
+=
+
==
4. 5
12 1
12 51
12 4
12 1
3
−=
−
==
5. 2
11 9
11 29
11