Transcript Ch. 2.6 power point

Chapter 2
Section 6
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2.6
1
2
3
Ratios and Proportions
Write ratios.
Solve proportions.
Solve applied problems by using
proportions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Objective 1
Write ratios.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 3
Write ratios.
A ratio is a comparison of two quantities using a quotient.
The ratio of the number a to the number b (b ≠ 0) is written
atob,
a : b, or
a
.
b
The last way of writing a ratio is most common in algebra.
Percents are ratios in which the second number is always 100.
For example, 50% represents the ratio 50 to 100, 27% represents
the ratio 27 to 100, and so on.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 4
EXAMPLE 1
Writing Word Phrases as Ratios
Write a ratio for each word phrase.
3 days to 2 weeks
Solution:
2weeks  7days  14days
3days
3days

14days
weeks
3

14
12 hr to 4 days
Solution:
4days  24hours  96hours
12hours
1
hours


96hours
8
4days
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 5
EXAMPLE 2
Finding Price per Unit
The local supermarket charges the following prices for a
popular brand of pancake syrup. Which size is the best
buy? What is the unit cost for that size?
Solution:
The 36 oz. size is the best
buy. The unit price is
$0.108 per oz.
$3.89
 $0.108
36
$2.79
 $0.116
24
$1.89
 $0.158
12
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 6
Objective 2
Solve proportions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 7
Solve proportions.
A ratio is used to compare two numbers or amounts. A
proportion says that two ratios are equal, so it is a special type of
equation. For example,
3 15

4 20
is a proportion which says that the ratios
In the proportion
3
4
and
15
20
are equal.
a c
  b, d  0  ,
b d
a, b, c, and d are the terms of the proportion. The terms a and d
are called the extremes, and the terms b and c are called the
a c
means. We read the proportions b  d as “a is to b as c is to d.”
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 8
Solve proportions. (cont’d)
Beginning with this proportion and multiplying each side by
the common denominator, bd, gives
a c

b d
a c
 bd      bd 
b d
ad  bc.
We can also find the products ad and bc by multiplying
diagonally.
bc
a c

b d
ad
For this reason, ad and bc are called cross products.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 9
Solve proportions. (cont’d)
If
a c
 ,
b d
then the cross products ad and bc are equal.
a c
  b, d  0  .
b d
a
c

From this rule, if
then ad = bc; that is, the product of
b
d
Also, if ad  bc, then
the extremes equals the product of the means.
a
b
If  , then ad = cb, or ad = bc. This means that the two
c d
proportions are equivalent, and the proportion a  c can
b d
a
b
also be written as
  c, d  0  .
c
d
Sometimes one form is more convenient to work with than the other.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 10
EXAMPLE 3
Deciding whether Proportions
Are True
Decide whether the proportion is true or false.
Solution: False
15  62  930
21 62

15 45
13 91

17 119
21 45  945
17  91  1547 Solution: True
13 119  1547
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 11
EXAMPLE 4
Finding an Unknown in a
Proportion
x 35

Solve the proportion
.
6 42
Solution: x 42  6  35
42 x 210

42
42
x5
The solution set is {5}.
The cross-product method cannot be used directly if there is more
than one term on either side of the equals symbol.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 12
EXAMPLE 5
Solving an Equation by Using
Cross Products
a6 2
 .
Solve
2
5
Solution:  a  6   5  2  2
5a  30  30  4  30
5a 26

5
5
26
a
5
The solution set is  26  .

5
When you set cross products equal to each other, you are really
multiplying each ratio in the proportion by a common denominator.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 13
Objective 3
Solve applied problems with
proportions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 14
EXAMPLE 6
Applying Proportions
Twelve gal of diesel fuel costs $37.68. How much
would 16.5 gal of the same fuel cost?
Solution:
Let x = the price of 16.5 gal of fuel.
$37.68
x

12 gal 16.5 gal
12 x 621.72

12
12
x  51.81
16.5 gal of diesel fuel costs $51.81.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2.6 - 15