Transcript Document

3.1
Order of Operations
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 1
Definitions




Algebra: a generalized form of arithmetic.
Variables: letters used to represent numbers
Constant: symbol that represents a specific
quantity
Algebraic expression: a collection of variables,
numbers, parentheses, and operation symbols.
Examples:
4x  2
x, x  4, 4(3 y  5),
, y 2  8y  2
3x  5
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 2
Order of Operations
1. First, perform all operations within
parentheses or other grouping symbols
(according to the following order).
2. Next, perform all exponential operations (that
is, raising to powers or finding roots).
3. Next, perform all multiplication and divisions
from left to right.
4. Finally, perform all additions and subtractions
from left to right.
Remember as:
PEMDAS or Please Excuse My Dear Aunt Sally
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 3
Example: Substituting for Two Variables

Evaluate 4 x 2  3 xy  5y 2 when x = 3 and y = 4.

Solution:
Copyright © 2009 Pearson Education, Inc.
-4x2 + 3xy – 5y2
-4(3)2 + 3(3)(4) – 5(4)2
-4(9) + 36 – 5(16)
-36 + 36 – 80
0-80
-80
Slide 6 - 4
3.2
Linear Equations in One Variable
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 5
Definitions




Terms are parts that are added or subtracted in
an algebraic expression.
Coefficient is the numerical part of a term.
Like terms are terms that have the same
variables with the same exponents on the
variables.
2x, 7x
 5x 2 , 8x 2
Unlike terms have different variables or different
exponents on the variables.
3
2
2x, 7
 5x , 6x
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 6
Properties of the Real Numbers
a(b + c) = ab + ac
a+b=b+a
Distributive property
Commutative property
of addition
ab = ba
Commutative property
of multiplication
(a + b) + c = a + (b + c) Associative property of
addition
(ab)c = a(bc)
Associative property of
multiplication
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 7
Example: Combine Like Terms

8x + 4x
=
=
Copyright © 2009 Pearson Education, Inc.

3x + 2 + 6y - 4 + 7x
=
=
Slide 6 - 8
Think of a balanced scale…
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 9
Solving Equations
Addition Property of Equality
If a = b, then a + c = b + c for all real numbers a,
b, and c.

Find the solution to the equation
x - 9 = 24.
Check:
=
=
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 10
Solving Equations continued
Subtraction Property of Equality
If a = b, then a - c = b - c for all real numbers
a, b, and c.

Find the solution to the equation
x + 12 = 31.
x + 12 - 12 = 31 - 12
x = 19
Check: x + 12 = 31
19 + 12 = 31 ?
31 = 31 true
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 11
Solving Equations continued
Multiplication Property of Equality
If a = b, then a • c = b • c for all real numbers a,
b, and c.

x
 9.
Find the solution to the equation
7
x
1
7x
9
 63
7
1
7
 x
7     7(9)
x  63
 7
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 12
Solving Equations continued
Division Property of Equality
If a = b, then a  b for all real numbers a, b,
c c
and c,
.

Find the solution to the equation 4x = 48.
4 x  48
4 x 48

4
4
x  12
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 13
General Procedure for Solving Linear
Equations



If the equation contains fractions, multiply both
sides of the equation by the lowest common
denominator (or least common multiple). This
step will eliminate all fractions from the
equation.
Use the distributive property to remove
parentheses when necessary.
Combine like terms on the same side of the
equal sign when possible.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 14
General Procedure for Solving Linear
Equations continued


Use the addition or subtraction property to
collect all terms with a variable on one side of
the equal sign and all constants on the other
side of the equal sign. It may be necessary to
use the addition or subtraction property more
than once. This process will eventually result in
an equation of the form ax = b, where a and b
are real numbers.
Solve for the variable using the division or
multiplication property. The result will be an
answer in the form x = c, where c is a real
number.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 15
Example: Solving Equations

Solve 3x - 4 = 17.
3x - 4 = 17.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 16
Example: Solving Equations

Solve 8x + 3 = 3(2x + 7).
8x + 3 = 3(2x + 7)
8 x  3  6 x  21
8 x  3  3  6 x  21  3
8 x  6 x  18
8 x  6 x  6 x  6 x  18
2 x  18
2 x 18

2
2
x 9
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 17
Example: Solving Equations

Solve 6(x  2)  2x  3  4(2x  3)  2.
6( x  2)  2 x  3  4(2 x  3)  2
6 x  12  2 x  3  8 x  12  2
8 x  9  8 x  10
8 x  8 x  9  8 x  8 x  10
9  10

False, the equation has no solution. The
equation is inconsistent.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 18
Example: Solving Equations

Solve 4(x  1)  6(x  2)  2(x  4).
4( x  1)  6( x  2)  2( x  4)
4 x  4  6 x  12  2 x  8
2 x  8  2 x  8
2 x  2 x  8  2 x  2 x  8
8  8
8  8  8  8
00

True, 0 = 0 the solution is all real numbers.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 19
Proportions

A proportion is a statement of equality between
two ratios.
a c

b d

Cross Multiplication
a c
If  , then ad = bc.
b d
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 20
To Solve Application Problems Using
Proportions


Represent the unknown quantity by a variable.
Set up the proportion by listing the given ratio
on the left-hand side of the equal sign and the
unknown and other given quantity on the righthand side of the equal sign. When setting up
the right-hand side of the proportion, the same
respective quantities should occupy the same
respective positions on the left and right. For
example, an acceptable proportion might be
miles miles

hour
hour
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 21
To Solve Application Problems Using
Proportions continued


Once the proportion is properly written, drop the
units and use cross multiplication to solve the
equation.
Answer the question or questions asked using
appropriate units.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 22
Example

A 50 pound bag of fertilizer will cover an area of
15,000 ft2. How many pounds are needed to
cover an area of 226,000 ft2?
50 pounds
x

2
15,000 ft
226,000 ft 2
(50)(226,000)  15,000 x
11,300,000  15,000 x
11,300,000 15,000 x

15,000
15,000
753.33  x
Copyright © 2009 Pearson Education, Inc.

754 pounds of fertilizer
would be needed.
Slide 6 - 23
3.3
Formulas
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 24
Definitions

A formula is an equation that typically has a
real-life application.

To evaluate a formula, substitute the given
value for their respective variables and then
evaluate using the order of operations.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 25
Perimeter

The formula for the perimeter of a rectangle is
Perimeter = 2 • length + 2 • width or P = 2l + 2w.

Use the formula to find the perimeter of a yard
when l = 150 feet and w = 100 feet.
P = 2l + 2w
P = 2(150) + 2(100)
P = 300 + 200
P = 500 feet
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 26
Example

The formula for the volume of a cylinder is
Use the formula to find the height of a cylinder
with a radius of 6 inches and a volume of
565.49 in3.
2
V  r h
565.49   (62 )h
565.49  36 h
565.49 36 h

36
36
5.000  h
The height of the cylinder is 5 inches.
Slide 6 - 27
Copyright © 2009 Pearson Education, Inc.
Exponential Equations: Carbon Dating


Carbon dating is used by scientists to find the
age of fossils, bones, and other items. The
 t / 5600
formula used in carbon dating is P  P0 2
.
If 15 mg of C14 is present in an animal bone
recently excavated, how many milligrams will be
present in 4000 years?
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 28
Exponential Equations: Carbon Dating
continued

P  P0 2
 t / 5600
P  15(2)4000 / 5600
P  15(2).71
P  15(0.61)
P  9.2 mg

In 4000 years, approximately 9.2 mg of the
original 15 mg of C14 will remain.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 29
Solving for a Variable in a Formula or
Equation

Solve the equation 3x + 8y  9 = 0 for y.
3 x  8y  9  0
3 x  8y  9  9  0  9
3 x  8y  9
3 x  3 x  8y  9  3 x
8y  9  3 x
8y 9  3 x

8
8
9  3x
y
8
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 30
Solve
h
A  (b1  b2 )
2
for b2.
h
A  (b1  b2 )
2
h

2   A   2   (b1  b2 ) 
2

2 A  h(b1  b2 )
2 A h(b1  b2 )

h
h
2A
 b1  b2
h
2A
 b1  b2
h
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 31
3.4
Applications of Linear
Equations in One Variable
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 32
Translating Words to Expressions
Phrase
Ten more than a number
Mathematical
Expression
x + 10
A number increased by 5
x+5
Four less than a number
x–4
A number decreased by
8
x–8
Twice a number
Copyright © 2009 Pearson Education, Inc.
2x
Slide 6 - 33
Translating Words to Expressions
Phrase
Four times a number
Mathematical
Expression
4x
2 decreased by a number
2–x
The difference between a
number and 6
x–6
Five less than 7 times a
number
7x – 5
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 34
Translating Words to Expressions
Phrase
Mathematical
Expression
Eleven more than twice a
number
2x + 11
The sum of 6 times a
number and 4
6x + 4
Nine times a number,
decreased by 5
9x – 5
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 35
Translating Words to Expressions
Phrase
Seven more than a
number is 12
Three less than a
number is 4
Twice a number,
decreased by 3 is 8.
A number decreased by
15 is 9 times the number
Copyright © 2009 Pearson Education, Inc.
Mathematical
Equation
x + 7 = 12
x–3=4
2x  3 = 8
x  15 = 9x
Slide 6 - 36
To Solve a Word Problem







Read the problem carefully at least twice to be
sure that you understand it.
If possible, draw a sketch to help visualize the
problem.
Determine which quantity you are being asked
to find. Choose a letter to represent this
unknown quantity. Write down exactly what this
letter represents.
Write the word problem as an equation.
Solve the equation for the unknown quantity.
Answer the question or questions asked.
Check the solution.
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 37
Example
The bill (parts and labor) for the repairs of a car
was $496.50. The cost of the parts was $339.
The cost of the labor was $45 per hour. How
many hours were billed?
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 38
Example

Sandra Cone wants to fence in a rectangular
region in her backyard for her lambs. She only
has 184 feet of fencing to use for the perimeter
of the region. What should the dimensions of
the region be if she wants the length to be 8 feet
greater than the width?
Copyright © 2009 Pearson Education, Inc.
Slide 6 - 39
continued, 184 feet of fencing, length 8
feet longer than width



Let x = width of region
Let x + 8 = length
P = 2l + 2w
Copyright © 2009 Pearson Education, Inc.
x
x+8
Slide 6 - 40