Transcript x - TonyReiter

11.1 Problem Solving Using
Ratios and Proportions
A ratio is the comparison of two numbers written as a fraction.
For example: Your school’s basketball team has won 7 games
and lost 3 games. What is the ratio of wins to
losses?
Because we are comparing wins to losses the first
number in our ratio should be the number of wins and
the second number is the number of losses.
7
The ratio is games won = 7 games
=
games lost
3 games
3
11.1 Problem Solving Using
Ratios and Proportions
In a ratio, if the numerator and denominator are measured in
different units then the ratio is called a rate.
A unit rate is a rate per one given unit, like 60 miles per 1 hour.
Example: You can travel 120 miles on 60 gallons of gas.
What is your fuel efficiency in miles per gallon?
120 miles
20 miles
Rate = 60 gallons =
1 gallon
Your fuel efficiency is 20 miles per gallon.
11.1 Problem Solving Using
Ratios and Proportions
An equation in which two ratios are equal is called a proportion.
A proportion can be written using colon notation like this
a:b::c:d
or as the more recognizable (and useable) equivalence of two fractions.
a c

b d
11.1 Problem Solving Using
Ratios and Proportions
When ratios are written in this order, a and d are the extremes, or
outside values, of the proportion, and b and c are the means, or
middle values, of the proportion.
a c

b d
a:b::c:d
Extremes
Means
11.1 Problem Solving Using
Ratios and Proportions
To solve problems which require the use of a proportion we can use one
of two properties.
The reciprocal property of proportions.
If two ratios are equal, then their reciprocals are equal.
The cross product property of proportions.
The product of the extremes equals the product of the means
11.1 Problem Solving Using
Ratios and Proportions
5 35

3 x
5x  3  35
5x  105
x  21
11.1 Problem Solving Using
Ratios and Proportions
2 6

x 9
92  6 x
18  6 x
3 x
x 2x 1

3
5
5 x  3(2 x  1)
5x  6 x  3
 x  3
x3
11.1 Problem Solving Using
Ratios and Proportions
Solve:
1
2

x x 1
1( x  1)  2 x
3
2

x  4 x 1
3( x  1)  2( x  4)
x – 1 = 2x
3x – 3 = 2x + 8
x = –1
x = 11
11.1 Problem Solving Using
Ratios and Proportions
x
1
Solve:

2x 1 x
x
1

2x  8 x
x  1(2 x  1)
x  1(2 x  8)
x2 = -2x - 1
x2 = 2x + 8
x2 + 2x + 1= 0
x2 - 2x – 8 = 0
(x + 1)(x + 1)= 0
(x + 2)(x - 4)= 0
(x + 1) = 0 or (x + 1) = 0
(x + 2) = 0 or (x - 4) = 0
2
x = -1
2
x = -2
x=4
11.2 Problem Solving Using
Percents


Percent means per hundred, or parts of 100
When solving percent problems, convert the
percents to decimals before performing the
arithmetic operations
Is means equals
Of means multiplication
11.2 Problem Solving Using
Percents
•
•
•
•
•
•
What is 20% of 50?
x = .20 * 50
x = 10
30 is what percent of 50?
30 = x * 50
x = 30/50 = .6 = 60%
11.2 Problem Solving Using
Percents
•
•
•
•
•
•
12 is 60% of what?
12 = .6x
x = 12/.6 = 20
40 is what percent of 300?
40 = x * 300
x = 40/300 = .133… = 13.33%
11.2 Problem Solving Using
Percents
•
•
•
•
•
•
10 is 30% of what?
10 = .3x
x = 10/.3 = 33.33
60 is what percent of 400?
60 = x * 400
x = 60/400 = .15 = 15%
11.2 Problem Solving Using
Percents

What percent of the region is shaded?
60
100 is what percent of 2400?
100 = x * 2400?
10
10
40
x = 100/2400
x = 4.17%
11.3 Direct and Inverse Variation
Direct Variation
The following statements are equivalent:



y varies directly as x.
y is directly proportional to x.
y = kx for some nonzero constant k.
k is the constant of variation or the constant of
proportionality
11.3 Direct and Inverse Variation
Inverse Variation
The following statements are equivalent:



y varies inversely as x.
y is inversely proportional to x.
y = k/x for some nonzero constant k.
11.3 Direct and Inverse Variation
If y varies directly as x, then y = kx.
If y = 10 when x = 2 ,
then what is the value of y when x = 8?
x and y go together.
Therefore, by substitution 10 = k(2).
What is the value of k? 10 = 2k
10 = 2k
5=k
11.3 Direct and Inverse Variation
k=5
Replacing k with 5 gives us y = 5x
What is y when x = 8 ?
y = 5(8)
y = 40
11.3 Direct and Inverse Variation
If y varies inversely as x, then xy = k.
If y = 6 when x = 4 , then what is the value of
y when x = 8?
x and y go together. Therefore, by substitution
(6)(4) = k.
What is the value of k?
24 = k
11.3 Direct and Inverse Variation
k = 24
Replacing k with 24 gives us xy = 24
What is y when x = 8 ?
8y = 24
y=3
11.3 Direct and Inverse Variation
Direct variation
15
y = kx
10
y = 2x
•
5
0
0
•
•
5
•
10
15
20
11.3 Direct and Inverse Variation
Inverse Variation
xy= k
•
15
10
•
xy= 16
•
5
•
0
0
5
•
10
15
20
11.4 Probability
The probability of an event P is the ration of
successful outcomes called successes, to the
outcome of the event, called possibilities.
number of successes
P(event) 
number of possibilit ies
11.4 Probability
Flipping a coin is an experiment and the possible
outcomes are heads (H) or tails (T).
One way to picture the outcomes of an experiment
is to draw a tree diagram. Each outcome is shown
on a separate branch. For example, the outcomes of
flipping a coin are
H
T
11.4 Probability
There are 4 possible outcomes when
tossing a coin twice.
First Toss
Second Toss
Outcomes
H
HH
T
H
HT
TH
T
TT
H
T
11.4 Probability
Drawing a card from a deck of 52.
P(heart)
13/52 = 1/4
P(red card)
26/52 = 1/2
P(face card)
12/52 = 3/13
P(queen)
4/52 = 1/13
11.4 Probability
Possible outcomes for two rolls of a die
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Find the following probabilities
1. Find the probability that the sum is a 2
2. Find the probability that the sum is a 3
3. Find the probability that the sum is a 4
4. Find the probability that the sum is a 5
5. Find the probability that the sum is a 6
6. Find the probability that the sum is a 7
7. Find the probability that the sum is a 8
8. Find the probability that the sum is a 9
9. Find the probability that the sum is a 10
10. Find the probability that the sum is a 11
11.Find the probability that the sum is a 12
•
•
•
•
•
•
•
•
•
•
•
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
11.5 Simplifying Rational
Expressions



Define a rational expression.
Determine the domain of a rational
function.
Simplify rational expressions.
11.5 Simplifying Rational
Expressions



Rational numbers are numbers that can be
written as fractions.
Rational expressions are algebraic fractions of
the form P(x) , where P(x) and Q(x)
Q(x)
are polynomials and Q(x) does not equal zero.
2
Example: 3x  2x  1
4x 1
11.5 Simplifying Rational
Expressions



P(x) ; Since division by zero is not
Q(x) possible, Q(x) cannot equal zero.
The domain of a function is all possible values of
x.
3x2  2x  1
For the example
, 4x + 1 ≠ 0
4x  1
so x ≠ -1/4.
11.5 Simplifying Rational
Expressions

The domain of
3 x 2  2 x is
1
4x  1
all real numbers except -1/4.
Domain = {x|x ≠ -1/4}
11.5 Simplifying Rational
Expressions

Find domain of
Solve:
2x  1
2
x  5x  6
x 2 –5x – 6 =0
(x – 6)(x + 1) = 0
The excluded values are x = 6, -1
Domain = {x | x ≠ -1, 6}
11.5 Simplifying Rational
Expressions

To simplify rational expressions, factor the
numerator and denominator completely.
Then reduce.

Simplify:
2 x  2 x  12
2
4 x  24 x  32
2
11.5 Simplifying Rational
Expressions
Factor:
Reduce:




2 x  2 x  12
2 x  x6

2
2
4 x  24 x  32 4 x  6 x  8
2
2
2 x  3 x  2  x  3 


2 4 x  4 x  2 2 x  4 
11.5 Simplifying Rational
Expressions
Simplify:
x2
2 x
Factor –1 out of the
denominator:
x2
 1 2  x 

x  2

 1 x  2
11.5 Simplifying Rational
Expressions
Reduce:

x  2

 1 x  2
1
 1
1
11.5 Simplifying Rational
Expressions


Multiply rational expressions.
Divide rational expressions
11.5 Simplifying Rational
Expressions




To multiply, factor each numerator and
denominator completely.
Reduce
Multiply the numerators and multiply the
denominators.
2
2
Multiply:
15 x
x  x  12
x  9 x  20
2

3 x  21
11.5 Simplifying Rational
Expressions
Factor:
Reduce:
15 x
x  x  12

2
2
x  9 x  20 3 x  21 x
2
2

x  4 x  3


 x  4 x  5 3 x x  7

5x
x  3


 x  5  x  7 
5 15 x
2
11.5 Multiplying and Dividing




5
x
x

3
5
x
x

3
Multiply:


 x  5  x  7   x  5 x  7 
5 x  15 x
 2
x  12 x  35
2
11.6 Multiplying and Dividing

To divide, change the problem to
multiplication by writing the reciprocal of
the divisor.


(Change to multiplication and flip the second
fraction.)
Divide:
2x  x  3 2x  x  6

2
2
x  4x  5
 x  1
2
2
11.5 Multiplying and Dividing
2x  x  3 2x  x  6

2
2
x  4x  5
 x  1
2
Change to
multiplication:
Factor
completely:
2
2x  x  3 x  4x  5
 2
2
2x  x  6
 x  1
2
2
2 x  3 x  1   x  1 x  5
 x  1 x  1 2 x  3 x  2
11.5 Multiplying and Dividing
Reduce:
Multiply:
2 x  3 x  1   x  1 x  5
 x  1 x  1 2 x  3 x  2
x5

x2
11.7 Dividing Polynomials
Dividing a Polynomial by a Monomial
Let u, v, and w be real numbers, variables or
algebraic expressions such that w ≠ 0.
uv u v
1.
 
w
w w
u v u v
2.
 
w
w w
11.7 Dividing Polynomials
 12 x  6 x  9 x
2
 4 x  2 x  3
3x
3
2
18c  27c  45c
9c
4
2
 2c  3c  5
3
11.7 Dividing Polynomials
( x 2  4 x  12)  ( x  2)
Use Long Division
x + 6
x  2 x  4 x  12
2
x2 -2x
6x - 12
6x - 12
0
Note: (x + 6) (x – 2) =
x2 + 4x - 12
11.7 Dividing Polynomials
( x 2  4 x  1)  ( x  1)
Use Long Division
x +5
x 1 x  4x 1
x2 - x
2
5x - 1
5x - 5
4
4
x  5 r 4 or x  5 
x 1
11.7 Dividing Polynomials
( x3  2 x  1)  ( x  2)
Note: x2 term is missing
x2 - 2x + 6
x  2 x  0x  2x 1
3
2
x  2 x  6 r  13
2
x3 +- 2x2
 13
2
2
-2x + 2x
or x  2 x  6 
x

2
2
-2x – 4x
6x - 1
6x +- 12
-13
11.8 Solving Rational Equations
3 1 12
 
x 2 x
2 x * 3 2 x 2 x *12


x
2
x
6  x  24
LCD: 2x
Multiply each fraction
through by the LCD
3
1 12
 
 18 2  18
3  9  12
 x  18
x  18
Check your solution!
11.8 Solving Rational Equations
Solve.
5x
5
 4
x 1
x 1
5 x( x  1)
5( x  1)
 4( x  1) 
( x  1)
( x  1)
5x  4 x  4  5
5x  4 x  1
x  1
Check your solution!
LCD: ?
LCD: (x+1)
5(1)
5
 4
11
1 1
5
5
 4
0
0
?
No Solution!
11.8 Solving Rational Equations
Solve.
3x  2
6
 2
1
x2 x 4
Factor 1st!
3x  2
6

1
x  2 ( x  2)( x  2)
LCD: (x + 2)(x - 2)
(3x  2)( x  2)( x  2) 6( x  2)( x  2)

 ( x  2)( x  2)
( x  2)
( x  2)( x  2)
3x 2  6 x  2 x  4  6  x 2  2 x  2 x  4
3x 2  4 x  4  x 2  2
x  3 or x  1
2x  4x  6  0
2
x  2 x  3  0 x  3  0 or x 1  0
Check your solutions!
( x  3)( x  1)  0
2
11.8 Solving Rational Equations
Short Cut!
When there is only fraction on each side
of the =, just cross multiply as if you are
solving a proportion.
11.8 Solving Rational Equations
Example: Solve.
3
1

2
x  4x x  4
x  4 x  3( x  4)
2
x  4 x  3 x  12
2
x  x  12  0
2
( x  4)( x  3)  0
x  4  0 or x  3  0
x  4 or x  3
Check your solutions!
11.8 Solving Rational Equations
Solve.
6
x2

2
2x  2x x 1
6
x2

2 x( x  1) x  1
2 x( x  1)( x  2)  6( x  1)
2 x ( x  2)  6
x( x  2)  3
x  2x  3  0
2
( x  3)( x  1)  0
x  3  0 or x 1  0
x  3 or x  1
Check your solutions!