Transcript Chapter 1 - Equations, Inequalities, Modeling

Digital Lesson
Solving Linear Equations
in One Variable
Distinguish between expressions and equations.
Equations and inequalities compare algebraic expressions.
An equation is a statement that two algebraic expressions are equal.
An equation always contains an equals symbol, while an expression does
not.
3x – 7 = 2
Left
side
3x – 7
Right
side
Equation
Expression
(to solve)
(to simplify or evaluate)
Slide 2.1- 2
Simplify algebraic expressions, combine like terms
• Simplify means to combine as many terms as possible in an expression…
• Be sure to combine only like terms by addition or subtraction of the
coefficients (same variables with identical exponents)
• Example: Simplify 4m2 + 3m – 2m2 + 5m.
•
subtract coefficients of m with power 2
•
(4 - 2) (m2)
•
add coefficients of m with power 1
•
(3 + 5) (m)
2m2 + 8m
• Example: Simplify ( s – 3t) + (2t – 3s).
•
stack or line up coefficients of s and coefficients of t
•
s – 3t
•
-3s + 2t
•
add coefficients of s and t vertically
•
-2s + -t
A linear equation in one variable is an equation which can
be written in the form:
ax + b = c
for a, b, and c real numbers with a  0.
Linear equations in one variable:
2x + 3 = 11
2(x  1) = 8 can be rewritten 2x + (2) = 8.
1
2
x  5  x  7 can be rewritten  x + 5 =  7.
3
3
Not linear equations in one variable:
2x + 3y = 11
Two variables
(x 
1)2
=8
x is squared.
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2
5 x7
3x
Variable in the denominator
4
A solution of a linear equation in one variable is a real number
which, when substituted for the variable in the equation, makes
the equation true.
Example: Is 3 a solution of 2x + 3 = 11?
Original equation
2x + 3 = 11
Substitute 3 for x.
2(3) + 3 = 11
6 + 3 = 11
False equation
3 is not a solution of 2x + 3 = 11.
Example: Is 4 a solution of 2x + 3 = 11?
2x + 3 = 11
2(4) + 3 = 11
8 + 3 = 11
Original equation
Substitute 4 for x.
True equation
4 is a solution of 2x + 3 = 11.
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5
Addition and Subtraction Property of Equations
If a = b, then a + c = b + c and a  c = b  c.
That is, the same number can be added to or subtracted from
each side of an equation without changing the solution of the
equation.
Use these properties to solve linear equations.
Example: Solve x  5 = 12.
x  5 = 12
x  5 + 5 = 12 + 5
x = 17
17  5 = 12
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Original equation
The solution is preserved when 5 is
added to both sides of the equation.
17 is the solution.
Check the answer.
6
Multiplication and Division Property of Equations
a b
If a = b and c  0, then ac = bc and  .
c c
That is, an equation can be multiplied or divided by the same
nonzero real number without changing the solution of the
equation.
Example: Solve 2x + 7 = 19.
Original equation
2x + 7 = 19
The solution is preserved when 7 is
2x + 7  7 = 19  7
subtracted from both sides.
2x = 12
Simplify both sides.
1
1
( 2 x )  (12)
2
2
The solution is preserved when each side
is multiplied by 1 .
2
6 is the solution.
x=6
2(6) + 7 = 12 + 7 = 19 Check the answer.
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7
To solve a linear equation in one variable:
1. Simplify both sides of the equation.
2. Use the addition and subtraction properties to get all variable terms
on the left-hand side and all constant terms on the right-hand side.
3. Simplify both sides of the equation.
4. Divide both sides of the equation by the coefficient of the variable.
Example: Solve x + 1 = 3(x  5).
x + 1 = 3(x  5)
Original equation
x + 1 = 3x  15
Simplify right-hand side.
x = 3x  16
Subtract 1 from both sides.
Subtract 3x from both sides.
 2x =  16
Divide both sides by 2.
x=8
The solution is 8.
Check the solution: (8) + 1 = 3((8)  5)  9 = 3(3) True
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8
To solve a linear equation in one variable:
1. Simplify both sides of the equation.
2. Use the guess and check method by using a common value for both
variables and the substitution property.
3. Simplify both sides of the equation.
4. Check to see if solutions match; If not try values higher or lower.
Example: Solve x + 1 = 3(x  5).
x + 1 = 3(x  5)
Original equation
5 + 1 = 3(5  5)
Substitute a value; when x = 5.
5 + 1 = 6 : 3 (0) = 0
Check to see if solutions match.
Try another value for x = 8.
8 + 1 = 3(8  5)
Check – Solutions match
8 + 1 = 9 : 3 (3) = 9
The solution is 8.
Check the solution: (8) + 1 = 3((8)  5)  9 = 3(3) True
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9
To solve a linear equation in one variable:
1. Simplify both sides of the equation.
2. Count the number of constant value in the equation. Draw that many
boxes. Label the first box with your variable; last box is solution.
3. Use order of operations to construct flow chart from left to right.
4. Use inverse operation to construct flow chart from right to left.
Example: Solve x + 1 = 3(x  5).
x + 1 = 3(x  5)
Original equation
x + 1 = 3x  15
Simplify right-hand side.
 2x =  16
Subtract 1 from both sides.
x=
x=
x=8
-2
-2
-16
-16
-16
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Subtract 3x from both sides.
Construct two boxes ( -2 and -16)
Fill in arrows with operations; solve
The solution is 8.
10
Example: Solve 3(x + 5) + 4 = 1 – 2(x + 6).
3(x + 5) + 4 = 1 – 2(x + 6)
Original equation
3x + 15 + 4 = 1 – 2x – 12
Simplify.
3x + 19 = –2x – 11
Simplify.
3x = – 2x – 30
Subtract 19.
5x = – 30
Add 2x.
x = 6
The solution is  6.
3(– 6 + 5) + 4 = 1 – 2(– 6 + 6)
Divide by 5.
Check.
3(– 1) + 4 = 1 – 2(0)
3 + 4 = 1
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True
11
Equations with fractions can be simplified by multiplying both
sides by a common denominator.
1
2 1
x

 ( x  4). The lowest common denominator
Example: Solve
of all fractions in the equation is 6.
2
3 3
1
2
1


6  x    6  ( x  4)  Multiply by 6.
3
2
3

3x + 4 = 2x + 8
Simplify.
3x = 2x + 4
Subtract 4.
Subtract 2x.
x=4
1
2 1
(4)   ((4 )  4) Check.
2
3 3
2 1
2   (8)
3 3
8 8
True

3 3
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12
CLASSROOM
EXAMPLE 2
Using the Properties of Equality to Solve a Linear Equation
Solve.
4x + 8x = –9 + 17x – 1
Solve.
6 – (4 + x) = 8x – 2(3x + 5)
Solve.
0.02(60) + 0.04x = 0.03(50 + x)
Identify conditional equations, contradictions, and
identities.
Type of
Linear
Equation
Number of
Solutions
Indication when Solving
Conditional
One
Final line is x = a number.
Identity
Infinite;
solution set
{all real
numbers}
Final line is true, such as 0 = 0.
Contradiction None; solution Final line is false, such as
set 
–15 = –20 .
Slide 2.1- 14
CLASSROOM
EXAMPLE 6
Recognizing Conditional Equations, Identities, and Contradictions
Solve each equation. Decide whether it is a conditional equation, an
identity, or a contradiction.
5(x + 2) – 2(x + 1) = 3x + 1
x 1 2x
1

 x
3
3
3
5(3x + 1) = x + 5
Alice has a coin purse containing $5.40 in dimes and quarters.
There are 24 coins all together. How many dimes are in the
coin purse?
Let the number of dimes in the coin purse = d.
Then the number of quarters = 24  d.
10d + 25(24  d) = 540
Linear equation
10d + 600  25d = 540
Simplify left-hand side.
10d  25d =  60
15d =  60
d=4
Subtract 600.
Simplify right-hand side.
Divide by 15.
There are 4 dimes in Alice’s coin purse.
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16
The sum of three consecutive integers is 54. What are the
three integers?
Three consecutive integers can be represented as
n, n + 1, n + 2.
n + (n + 1) + (n + 2) = 54
3n + 3 = 54
3n = 51
n = 17
Linear equation
Simplify left-hand side.
Subtract 3.
Divide by 3.
The three consecutive integers are 17, 18, and 19.
17 + 18 + 19 = 54.
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Check.
17
Teaching Concept #1
Modeling Algebraic Expressions
Using operation
words:
+,,*,/,,,,
Total, Sum,
Difference, Between,
Product, per,
Quotient, average,
Split equally, each,
Give or take.
Examples:
8 less than the product of 5
times a number.
5n - 8
The quotient of 11 and the
sum of 7 and the value of x.
11 / (x + 7)
The total cost of notebooks if
each notebook costs $2.25
2.25n
Teaching Concept #2
Modeling Algebraic Expressions
Using Graphs:
A number greater than or equal
to 1
Positive numbers
Values between -2 and 2
The cost of x notebooks is each
notebook costs $2.25
Teaching Concept #5
Modeling Algebraic Expressions
Using Tables:
The cost of notebooks
purchased if each
notebook costs $2.25.
The data plan for your
cell phone with has a
flat rate of $30 for
10MB and $10 for
each additional 2MB .
0
1
2
3
4
5
0
2.25
4.50
6.75
9.00
11.25
0
5
10
12
14
20
30.00
30.00
30.00
40.00
50.00
130.00
What’s an inequality?
•
• Is a range of values,
rather than ONE set number
An algebraic relation showing that a
quantity is greater than or less than
another quantity.
Speed limit:
55  x  75
Symbols




Less than
Greater than
Less than OR EQUAL TO
Greater than OR EQUAL TO
Solutions….
You can have a range of answers……
-5 -4 -3 -2 -1 0
1
2
All real numbers less than 2
x< 2
3
4
5
Solutions continued…
-5 -4 -3 -2 -1 0
1
2
All real numbers greater than -2
x > -2
3
4
5
Solutions continued….
-5 -4 -3 -2 -1 0
1
2
3
4
5
All real numbers less than or equal to 1
x 1
Solutions continued…
-5 -4 -3 -2 -1 0
1
2
3
4
5
All real numbers greater than or equal to -3
x  3
Did you notice,
Some of the dots were solid
and some were open?
x2
-5 -4 -3 -2 -1
0
1
2
3
4
5
-5 -4 -3 -2 -1
0
1
2
3
4
5
x 1
Why do you think that is?
If the symbol is > or < then dot is open because it can not be
equal.
If the symbol is  or  then the dot is solid, because it can be
that point too.
Write and Graph a Linear
Inequality
Sue ran a 2-K race in 8 minutes. Write an inequality
to describe the average speeds of runners who were
faster than Sue. Graph the inequality.
Faster average speed >
1
s
4
Distance
2
s
8
Sue’s Time
-5 -4 -3 -2 -1 0
1
2
3
4
5
Solving an Inequality
Solving a linear inequality in one variable is much like
solving a linear equation in one variable. Isolate the
variable on one side using inverse operations.
Solve using addition:
x–3<5
Add the same number to EACH side.
x 3  5
+3
+3
x<8
Solving Using Subtraction
Subtract the same number from EACH side.
x  6  10
-6
-6
x4
Using Subtraction…
x 5  3
Graph the solution.
-5 -4 -3 -2 -1 0
1
2
3
4
5
Using Addition…
2  n4
Graph the solution.
-5 -4 -3 -2 -1 0
1
2
3
4
5
THE TRAP…..
When you multiply or divide each side of
an inequality by a negative number, you
must reverse the inequality symbol to
maintain a true statement.
Solving using Multiplication
Multiply each side by the same positive number.
1
(2)
x  3 (2)
2
x6
Solving Using Division
Divide each side by the same positive number.
3x  9
3
3
x3
Solving by multiplication of a
negative #
Multiply each side by the same negative number
and REVERSE the inequality symbol.
(-1)
 x  4 (-1)
Multiply by (-1).
See the switch
x  4
Solving by dividing by a negative
#
Divide each side by the same negative
number and reverse the inequality symbol.
 2x  6
-2
-2
x3
Finding Absolute Value
Absolute value is used to describe how to operate with positive and
negative numbers.
Geometric Meaning of Absolute Value a ,
The absolute value of a real number a, denoted
is the distance from 0 to a on the number
line. This distance is always nonnegative.
5  5
3  3
The absolute value of 3 is +3 because 3 is 3 units
from 0 on the number line.
Exploration
• Determine the solution for each equation.
–
x 4
4, -4
–
n 9
9, -9
–
c  6
No Solution
What did you notice?
• Summarize what you noticed from the
previous solutions.
When a variable is inside an absolute value, there
are two solutions.
 When an absolute value is set equal to a negative
number, there is no solution. (this is important to
remember)
 Can you think of a situation where there would be
one solution? When the absolute value is equal to zero.

• Steps for solving
absolute value
equations.
1. Distribute
2. Combine Like Terms
3. Move Variable to One Side
4. Undo + or –
5. Undo × or ÷
**Need to isolate the absolute
value expression**
1) Undo addition or
subtraction outside of
absolute value.
2) Undo multiplication or
division outside of absolute
value.
3) Set expression inside
absolute value equal to the
given value and its opposite.
4) Solve for variable using
steps for solving equations.
Examples
• Solving basic
absolute value
equations
1.
x  5  12
x  5  12 and x  5  12
x  5  12 and x  5  12
5 5
x  17
5
5
x  7
Examples continued
2. 2 x  6  4
1, 5
1
3.
x4 8
2
-24, 8
More Examples
• Solving
absolute value
equations
when there are
terms outside
the symbols
x  1  4  12
1.
x  1  4  12
4 4
x  1  16
x 1  16 and x 1  16
x 1  16 and x 1  16
1
1
1
1
x  15 and x  17
Even More Examples
5.
3x  4  6  10
0, 8/3
6. 3 2 x  4  6  18
-2, 6