Transcript Elementary Algebra - Seminole State College

Elementary Algebra
Exam 2 Material
Equations
• Equation – a statement that two expressions
are equal
– Equations always contain an equal sign, but an
expression does not have an equal sign
• Like a statement in English, an equation may be
true or false
• Examples:
5  9  14
47 5
T or F?
T or F?
True
False
.
Equations
• Most equations contain one or more
variables and the truthfulness of the
equation depends on the numbers that
replace the variables
• Example: x  4  9
• What value of x makes this true? x  5
• A number that can replace a variable to
make an equation true is called a solution
5 is a solution ot theequation
Types of Equations
• In Algebra you will study many different
types of equations
– Learn the names of each type
– Learn method for solving each type
• The simplest type of equation is called a
“linear equation”
Linear Equations
• Linear equation – an equation where, after
parentheses are gone, every term is either a
constant, or of the form: cx where c is a
constant and x is a variable with exponent1
Linear equations never have a variable in a
denominator or under a radical (square root
sign)
• Examples of Linear Equations:
.
4 x  5  13
3x  7  x  1
3
2 x  3  x
5
1
.72 x  6   x  83  x 
2
Identifying Linear Equations
• Identify linear equations:
x2 x
2
x7 1
  x  8
3
2
x5
7

2
x 3
4x 1  x  9
No
(exponent2 on variable)
Yes
(each termis c or cx)
No
(variablein denominator)
No
(variableunder radical)
Solving Linear Equations
• Any equation that is a true statement has
both sides with equal values. It is
“balanced on both sides of the equal sign.”
• In trying to find solutions to an equation we
try to do things that will keep both sides of
an equation balanced while progressing
towards a goal of ending up with the
variable alone on one side of the equal
sign
Solving Linear Equations by
Keeping Both Sides Balanced
• If we assume that both sides of an
equation really are equal and we add or
subtract the same thing on both sides,
then both sides will still be balanced
• In the earlier example, x  4  9 ,what
could we have done on both sides of
original equation to get a new equation
with “x” isolated? Add - 4 (or Subtract 4)
x44 94
x5
Solving an Equation by Balancing
with Addition or Subtraction
• Solve the equation:
x  9  12
x  9  9  12  9
x  3
• The solution to the original equation is:
x  3
Solving More Complicated
Equations
• In solving an equation, make sure that the expression on
each side is simplified, before proceeding
– Get rid of parentheses
– Combine like terms
• Next, choose which side will keep the variable and add
or subtract terms on both sides, as necessary, to isolate
the variable
2x  3  5x  4x  2
2 x  6  5x  4 x  8
2x  6  x  8
2x  6  x  x  x  8
x  6  8
x  6  6 .8  6
x  2
English Sentences that Translate
to Equations
• If an English sentence indicates that two
numerical expressions are equal, it can be
translated to an equation
• The phrases: “the result is”, or “is equal
to”, “equals” or “is”, can usually be
translated into algebra as an equal sign, =
• Translate: The sum of a number and three
is 15.
x  3  15
Translating Sentences to
Equations and Solving
• The major reason for translating a
sentence to an equation is to help us find
the value of an unknown number
described in the sentence
• Find the unknown number in the previous
example: “The sum of a number and 3 is
15”:
x  3  15
x  3  3  15  3
x  12
The number described must be 12.
Other Examples of Translating
English Sentences to Equations
• Translate and Solve: The product of 3 and
a number is equal to twice the number
plus 7.
3x  2 x  7
3x  2 x  2 x  2 x  7
x7
The number described must be 7.
Other Examples of Translating
English Sentences to Equations
• Translate and Solve: Twice the sum of a
number and 5 is the same as 8 less than
the number. What is the number?
2x  5  x  8
2 x  10  x  8
2 x  x  10  x  x  8
x  10  8
x  10  10  8  10
x  18
The number described must be - 18.
Homework Problems
• Section:
• Page:
• Problems:
2.1
100
Odd: 5 – 43, 47 – 65,
All: 69 – 72
• MyMathLab Section 2.1 for practice
• MyMathLab Homework Quiz 2.1 is due for
a grade on the date of our next class
meeting
Another Way to Keep Equations
Balanced
• We have learned that equations that are
true statements can be kept balanced by
adding or subtracting the same thing on
both sides of equal sign
• We can also keep both sides balanced by
multiplying or dividing both sides by any
number that is not zero
• Examples follow that show how multiplying
or dividing can help solve an equation
Solving Equations Using
Multiplication or Division
• Solve:
1
x  5
3
• The equation will be solved when we have
1
x by itself , not x
3
• What could we multiply on both sides? 3
1 
3 x   3 5
3 
x  15
Solving Equations Using
Multiplication or Division
• Solve:  2 x  12
• The equation will be solved when we have
x by itself , not  2 x
• What could we divide on both sides?  2
 2 x 12

2 2
x  6
Other Examples of Translating
English Sentences to Equations
• Translate: The difference between 4 and
a number equals the number plus 5.
4 x  x5
4 x x  x x5
4  2x  5
 1  2 x 5 subtractedfrombothsides
1 2x

2
2
1
The number described must be - .
2
Other Examples of Translating
English Sentences to Equations
• Translate: If 11 times a number is
subtracted from 8 times the number, the
result is -9.
8 x  11x  9
 3x  9
 3x  9

3 3
x3
The number described must be 3.
Homework Problems
• Section: 2.2
• Page: 107
• Problems: Odd: 7 – 69,
All: 73 – 76
• MyMathLab Section 2.2 for practice
• MyMathLab Homework Quiz 2.2 is due for
a grade on the date of our next class
meeting
Solving Linear Equations
• Simplify each side separately
– Get rid of parentheses
– Multiply by LCD to get rid of fractions and decimals
– Combine like terms
• Get the variable by itself on one side by adding
or subtracting the same terms on both sides
• If the coefficient of the variable term is not 1,
then divide both sides by the coefficient
Determine if the equation is linear.
If it is, solve it: Is it linear? Yes
23x 1  4 x  8
6x  2  4x  8
6x  4x  2  4x  4x  8
2 x  2  8
2 x  2  2  8  2
2 x  6
2x  6

2
2
x  3
Determine if the equation is linear.
If it is, solve it: Is it linear? Yes
 8  6x  5  7  2x  4
 8  6 x  30  7  2 x  8
6 x  38  2 x  1
6 x  2 x  38  2 x  2 x  1
8 x  38  1
8 x  38  38  1  38
8 x  37
8 x 37
37

x
8
8
8
Determine if the equation is linear.
If it is, solve it: Is it linear? Yes
24  3x   3x  1  14
8  6 x  3x  3  14
8  6 x  3x  17
8  9 x  17
 9  9x
1  x
Determine if the equation is linear.
If it is, solve it: Is it linear? No
2 x  3  5x  x 2  4
• We can’t solve this equation yet. Later we
will learn its name, and how to solve it!
Determine if the equation is linear.
If it is, solve it: Is it linear? Yes
2
1
x  x  2  0.3x  1
3
4
2
1
1 3
x  x   x 1
3
4
2 10
1
1
2
3

60 x  x    60 x  1
4
2
3
 10

40 x  15 x  30  18 x  60
25 x  30  18 x  60
7 x  30  60
7 x  90
90
x
7
Example
• Solve:
• LCD:
5
3
x  x2
6
4
6  2 3
4  22
LCD  2  2  3  12
• Multiply both sides by LCD:
5 
3

12 x   12 x  2 
6 
4

10 x  9 x  24
x  24
Example
• Solve:
• LCD:
7
5
x  3  x 1
24
9
24  2  2  2  3
9  33
LCD  2  2  2  3  3  72
• Multiply both sides by LCD:
 7

5

72 x  3   72 x  1
 24

9

21x  216  40 x  72
 216  19 x  72
 288  19 x
288

x
19
Example
• Solve: .3x  7  .52 x  .13
• LCD:
10 
100 
LCD 
2 5
2 255
2  2  5  5  100
T heLCD of multiplesof 10
is always thelargest multiple
• Multiply both sides by LCD:
100.3x  7  100.52x  .13
30 x  700  52 x  13
700  22 x  13
713  22 x
713
x
22
Example with Fractions & Decimals
• Solve:
• LCD:
6
4
10 
1
3
x   .3 x  5
6
4
2 3
22
2 5
LCD  2  2  3  5  60
• Multiply both sides by LCD:
3
1
60 x    60.3x  5
4
6
10 x  45  18 x  300
45  8 x  300
345  8 x
345
x
8
Linear Equations with No Solution
or All Real Numbers as Solutions
• Many linear equations only have one number as
a solution, but some have no solution and others
have all numbers as solutions
• In trying to solve a linear equation, if the variable
disappears (same variable & coefficient on both
sides) and the constants that are left make a
statement that is:
– false, the equation has “no solution” (no number can
replace the variable to make a true statement)
– true, the equation has “all real numbers” as solutions
(every real number can replace the variable to make
a true statement)
Solve the Linear Equation
2x  x  3  x  7
2x  x  3  x  7
x3 x7
x x3 x x7
37
False!
Equationhas no solution
Solve the Linear Equation
x  2  7 x   2 x  21  3x 
x  2  7x  2x  2  6x
8x  2  8x  2
8x  8x  2  8x  8x  2
 2  2
True!
All real numbers are solutions
Homework Problems
•
•
•
•
•
Section:
Page:
Problems:
Page:
Problems:
2.3
115
Odd: 7 – 45
117
Odd: 1 – 29
• MyMathLab Section 2.3 for practice
• MyMathLab Homework Quiz 2.3 is due for a
grade on the date of our next class meeting
Solving Application Problems
• Some problems may involve more than
one sentence and more than one unknown
• Such problems may seem as
overwhelming as trying to eat an elephant
• How do you eat an elephant?
One bite at a time!
• Application problems are solved easily if
you memorize the steps and do them one
at a time!
Steps in Solving
Application Problems
• Read the problem carefully trying to understand what the
unknowns are (take notes, draw pictures, don’t try to
write equation until all other steps below are done )
• Make word list that describes each unknown
• Assign a variable name to the unknown you know the
least about (the most basic unknown)
• Write expressions containing the variable for all the other
unknowns
• Read the problem one last time to see what information
hasn’t been used, and write an equation about that
• Solve the equation (make sure that your answer makes
sense, and specifically state the answer)
Example of Solving an
Application Problem
• Three less than 5 times a number is equal to 9
less than twice the number. What is the
number?
• List of unknowns
T hisis theonlyunknownso we call it :
– A number
x
• What else does the problem tell us that we
haven’t used?
Three less than 5 times a number is equal to 9
less than twice the number.
• What equation says this?
5x  3  2 x  9
Example Continued
5x  3  2 x  9
5x  2 x  3  2 x  2 x  9
3x  3  9
3 x  3  3  9  3
3x  6
x  2
• Solve the equation:
• Answer to question?
The number described is:
2
Example of Solving an Application
Problem With Multiple Unknowns
• A mother’s age is 4 years more than twice her
daughter’s age. The sum of their ages is 76.
What is the mother’s age?
• List of unknowns
– Mother’s age
– Daughter’s age
2x  4
x
Which do we know least about? Daughter's age
• What else does the problem tell us that we
haven’t used?
Sum of their ages is 76
x  2 x  4  76
• What equation says this?
Example Continued
• Solve the equation:
x  2 x  4  76
3x  4  76
3x  4  4  76  4
3x  72
x  24
• Answer to question?
Mother’s age is 2x + 4:
224  4  52
Solve the Application Problem
•
A 31 inch pipe needs to be cut into three pieces
in such a way that the second piece is 5 inches
longer than the first piece and the third piece is
twice as long as the second piece. How long
should the third piece be?
1. Read the problem carefully taking notes,
drawing pictures, thinking about formulas that
apply, making charts, etc.
Perhaps draw a picture of a pipe that is labeled
as 31 inches with two cut marks dividing it into
3 pieces labeled first, second and third
1st
2nd
3rd
31
Example Continued
2. Read problem again to make a “word list” of
everything that is unknown
What things are unknown in this problem?
The length of all three pieces (even though the
problem only asked for the length of the third).
Word List of Unknowns:
Length of first
Length of second
Length of third
Example Continued
3. Give a variable name, such as “x” to the “most
basic unknown” in the list (the thing that you
know least about)
What is the most basic unknown in this list?
Length of first piece is most basic, because
problem describes second in terms of the first,
and third in terms of second
Give the name “x” to the length of first
Example Continued
4.
Give all other unknowns in the word list an algebraic
expression name that includes the variable, “x”
How would the length of the second be named?
x+5
How would the length of the third be named?
2(x + 5)
Word List of Unknowns:
Algebra Names:
Length of first
x
Length of second
x+5
Length of third
2(x + 5)
Example Continued
5.
Read the problem one last time to determine what
information has been given, or implied by the problem,
that has not been used in giving an algebra name to
the unknowns and use this information to write an
equation about the unknowns
What other information is given in the problem that has
not been used?
Total length of pipe is 31 inches
How do we say, by using the algebra names, that the
total length of the three pieces is 31?
x + (x + 5) + 2(x + 5) = 31
Example Continued
6.
Solve the equation and answer the original question
This is a linear equation so solve using the appropriate
steps:
x + (x + 5) + 2(x + 5) = 31
x + x + 5 + 2x + 10 = 31
4x + 15 = 31
4x = 16
x=4
Is this the answer to the original question?
No, this is the length of the first piece.
How do we find the length of the third piece?
The length of the third piece is 2(x + 5):
2(4 + 5) = (2)(9) = 18 inches = length of third piece
Solving Application Problems
Involving Angles
• Angles are measured in units called degrees
• A complete rotation of a ray from a starting position back to the
starting position has a measure of 360o
• Half of a rotation of a ray from a starting position to a position
pointing the opposite direction has a measure of 180o and the angle
is called a straight angle
• One quarter of a rotation of a ray has a measure of 90o and the
angle is called a right angle
• Two angles whose sum makes a right angle (whose sum is 90o) are
called complementary angles
• Two angles whose sum makes a straight angle (whose sum is 180o)
are called supplementary angles
• If x represents the measure of an angle,
– The measure of the complementary angle is:
– The measure of the supplementary angle is:
90  x
180  x
Example of Solving an Angle
Application Problem
• Find the measure of an angle such that the supplement
is 15o more than twice the complement
• List of unknowns
– Angle measure
– Complement measure
– Supplement measure
x
90  x
180  x
Which do we know least about? Measure of angle
• What else does the problem tell us that we haven’t
used?
Supplement is 15o more than twice the complement
• What equation says this?
180 x  290  x   15
Example Continued
• Solve the equation: 180 x  290  x   15
180  x  180  2 x  15
 x  2 x  15
x  15
• Answer to question?
o
The measure of the angle is: 15
Solving Application Problems
Involving Consecutive Integers
• If an application problem involves consecutive
integers, consecutive even integers, or
consecutive odd integers, remember that
consecutive integers differ by 1 and consecutive
even, as well as consecutive odd, integers differ
by 2
• If x represents an integer, the next integer is
x
+ 1, and the next is x + 2, etc
• If x represents an even integer, the next even
integer is x + 2, and the next is x + 4, etc
• If x represents an odd integer, the next odd
integer is x + 2, and the next is x + 4, etc
Example of Solving a Consecutive
Integer Application Problem
• Find two consecutive odd integers such that
three times the smaller is thirteen less than twice
the larger
• List of unknowns
– smaller odd integer
– next larger odd integer
x
x2
Which do we know least about? First odd integer
• What else does the problem tell us that we
haven’t used?
Three times smaller is 13 less than twice larger
• What equation says this?
3x  2x  2  13
Example Continued
• Solve the equation:
3x  2x  2  13
3 x  2 x  4  13
3x  2 x  9
x  9
• Answer to question?
The smaller odd integer is x =  9
7
The larger odd integer is x + 2 =
Homework Problems
•
•
•
•
•
Section:
2.4
Page:
126
Problems: Odd: 11 – 57
MyMathLab Section 2.4 for practice
MyMathLab Homework Quiz 2.4 is due for
a grade on the date of our next class
meeting