jovan algebra | Page 5

method outlined above.
5
2
3 52
3 10
3
⋅=
⋅
=
or
52
3 5
1 2
3 52
13 10
3
⋅=⋅=
⋅
⋅ =
PRACTICE
Find the product with either of the two methods outlined above.
1.
6
7 9
⋅=
2. 8 1
6
⋅=
3. 4
2
5
⋅=
4. 3
14 2
⋅=
5. 12 2
15
⋅=
1
91
9
FIGURE 1-5
1
9
1
91
91
9
1
9
1
91
91
9
FIGURE 1-6
EXAMPLE
Find the product
PRACTICE
Find the product with either of the two methods outlined above.

Chapter 1 FraCtions 5
DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
SOLUTIONS
1. 6
7 9
69
7 54
7
⋅=
⋅
= or 6
7 9
1 69
71 54
7
⋅=
⋅
⋅ =
2. 8
1
6 81
6 8
6
⋅=
⋅
= or 8
1 1
6 81
16 8
6
⋅=
⋅
⋅ =
3. 4
2
5
⋅=42
5 8
5
⋅
= or 4
1 2
5 42
15 8
5
⋅=
⋅
⋅ =
4. 3
14 2
⋅=32
14 6
14
⋅
= or 3
14 2
1 32
14 1 6
14
⋅=
⋅
⋅ =
5.
12 2
15
⋅=12
2
15 24
15
⋅
= or 12
1 2
15 12
2
11 5 24
15
⋅=
⋅
⋅ =
Fraction Division
Fraction division is almost as easy as fraction multiplication. The rule for fraction
division is
a
b c
d a
b d
c ad
bc
÷=
⋅=, that is, we invert (switch the numerator and deno-
minator) the second fraction, and the fraction division problem becomes a fraction
multiplication problem. To see why this might be so, consider the computation
for
3
1 1
2
÷. We can think of this division problem as asking the question, “How
many halves go into 3?” Of course, the answer is 6, which agrees with the
formula:
3
1 1
2 32
6
÷= ⋅=.
EXAMPLES
Perform the division.
2
3 4
5 2
3 5
4 10
12
÷=
⋅=
3
4 5
3
4 5
1 3
4 1
5 3
20
÷=
÷= ⋅=
EXAMPLES
Perform the division.
SOLUTIONS✔

6 alGebra D e myst i fieD DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
PRACTICE
Perform the division.
1.
7
6 1
4
÷=
2. 8
15 6
5
÷=
3. 5
3 9
10
÷=
4. 40
9 2
3
÷=
5. 3
7 30
4
÷=
6. 4
2
3
÷=
7. 10
21 3
÷=
SOLUTIONS
1. 7
6 1
4 7
6 4
1 28
6
÷=
⋅=
2. 8
15 6
58
15 5
640
90
÷=
⋅=
3. 5
39
10 5
310
9 50
27
÷=
⋅=
4. 40
9 2
3 40
9 3
2 120
18
÷=
⋅=
5. 3
7 30
4 3
7 4
30 12
210
÷=
⋅=
6. 42
3 4
1 2
3 4
1 3
2 12
2
÷= ÷=
⋅=
7. 10
21 3
10
21 3
1 10
21 1
3 10
63
÷=
÷= ⋅=
SOLUTIONS✔
PRACTICE
Perform the division.

Chapter 1 FraCtions 7
DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
Simplifying Fractions
When working with fractions, we are usually asked to “reduce the fraction to
lowest terms” or to “write the fraction in lowest terms” or to “simplify the fraction.”
These phrases mean that the numerator and denominator have no common
factors (other than 1). For example,
2
3 is in written in lowest terms but 4
6 is not
because 2 is a factor of both 4 and 6. Simplifying fractions is like fraction mul-
tiplication in reverse. For now, we will use the most basic approach to simplify-
ing fractions. In the next section, we will learn a quicker method. First write the numerator and denominator as a product of prime numbers.
(Refer to the Appendix if you need to review finding the prime factorization of
a number.) Next collect the prime numbers common to both the numerator and
denominator (if any) at beginning of each fraction. Split each fraction into two
fractions, the first with the common prime numbers. This puts the fraction in
the form of “1” times another fraction. This might seem like unnecessary work
(actually, it is), but it will drive home the point that the factors that are common
in the numerator and denominator form the number 1. Thinking of simplifying
fractions in this way can help you avoid common fraction errors later in algebra.
EXAMPLE
Simplify the fraction with the method outlined above.
6
18
SOLUTION
We begin by factoring 6 and 18.
6
18 23
23 323
1
23 3
=
⋅
⋅⋅ =
⋅⋅
⋅⋅
()
()
We now write the common factors as a separate fraction.
6
18 23
23 323
1
23 323
23 1
3 6
61
3
=
⋅
⋅⋅ =
⋅⋅
⋅⋅ =
⋅
⋅⋅=
⋅
()
()
Because 6
6 is 1, we see that 6
18 simplifies to 1
3.
6
18 23
23 323
1
23 323
23 1
3 6
61
3 1
3
=
⋅
⋅⋅ =
⋅⋅
⋅⋅ =
⋅
⋅⋅=⋅=
()
()
42
49 72
3
77 7
723
7 1
6
7 6
7
=
⋅⋅
⋅ =⋅
⋅
=⋅ =
SOLUTION
We begin by factoring 6 and 18.
✔
EXAMPLE
Simplify the fraction with the method outlined above.

8 alGebra D e myst i fieD DeMYstiFieD / algebra DeMYstiFieD / Gibilisco / 000-0 / Chapter 1
PRACTICE
Simplify the fraction.
1.
14
42 =
2. 5
35 =
3. 48
30 =
4. 22
121 =
5. 39
123 =
6. 18
4 =
7. 7
210 =
8. 240
165 =
9. 55
33=
10. 150
30 =
SOLUTIONS
1. 14
42 27
23 727
1
27 327
27 1
3 14
14 1
3
=
⋅
⋅⋅ =
⋅⋅
⋅⋅ =
⋅
⋅⋅= ⋅=
()
()11
3
2. 5
35 5
57 51
57 5
51
7 1
7
=
⋅=
⋅
⋅ =⋅
=
3. 48
30 2222
3
23 5 23
222
23 5 23
232
= ⋅⋅⋅⋅
⋅⋅ =
⋅
⋅⋅⋅
⋅⋅ =
⋅
⋅⋅
()
()⋅⋅⋅
=⋅ =
22
5 6
68
5 8
5
4. 22
121 21
1
11