Download Final Project-ED784.2

Wenying Chen
9th Grade
Educ 784.2
The outline of
the presentation
Lesson One
• Video-Distributive
Property
• Form-homework
Lesson Two
• Podcast-giving definitions
• Jing-showing the solutions
• Word-the vocabulary list
• Form-the assessment
questions
Lesson Three
• Blog-the review
questions
(websites and video are
included)
Lesson One
Aim: How do we simplify
algebraic expressions?
Do Now:
Simplify each expression
+
1.
=
2.
5
=6
+
+
+4
+2
-5
3X³Y
coefficient
term
variables
Like terms are terms that contain the same variables
raised to the same exponent.
X and 3X
2X² and 2X
2Y² and -5Y²
Are Unlike Terms
r³ and t³
3X²Y and ½X²Y
3rt³ and -0.5rt³ Are Like Terms
6 and 8

Are Like Terms
Look at these 10 terms. Find all the
like terms that can be combined:
Construct a like term that could
combine with each of these terms.
 7a³b²
 ¾mn³
4
Combine Like Terms
Example 1

7x² - 4x²
= (7-4) x²
= 3x²
Notice that we can combine like terms by adding or
subtracting the coefficients and keeping the
variables and exponents the same.
Practice
Simplify each expression by combing like terms.
A.
B.
C.
D.
12x + 30x
= 42x
6.8y² - 1y²
= 5.8y²
-4n + 11n²
-4n and 11n² are not like terms. Do not combine them
-20t – 8.5t
= -28.5t
Simplify the following expression by
combining like terms:
6xy + 3x²y + 1xy
xy + 6x²y + 10xy
+
+
17xy
+
+
+
9x²y
Distributive Property
n
n
n
n
n
n
11
11
1 1
1 1
11
1 1
3(n+2)
3n + 6
Distributive Property
What is the Distributive Property?
How to use the Distributive Property in
simplifying algebraic expressions?
Watch video Distributive Property Basics
Example 2
Simplify 2(x+6) + 3x.
Procedure
Justification
1. 2(x+6)+3x
2. 2(x)+2(6)+3x
Distributive Property
3. 2x + 12 + 3x
Multiply
4. 2x + 3x + 12
Commutative Property
5. 5x + 12
Combine like terms
Try this:
1) 6(x - 4) + 9
2) -12x - 5x + 3a + x
Closure



Answer the Aim
Discuss how the Commutative, Associative,
and Distributive Property to simplify
expressions.
Assign homework
Lesson Two
Aim: How do we solve equations ?

What is an equation?

What is a solution of an equation?

How do we find the solutions of an
equation?
Answer
An equation is like a balance scale.
What are the rules for keeping
an equation balanced?
What ever you do to
One side of the equal
Sign must be done to
The other side too
Use opposite math to
isolate the variable
on one side of the
equal sign
Motivation
How many
are in one
?
2 bags + 4 blocks = 3 bags + 2 blocks
- 2 bags
-2 bags
4 blocks = 1 bag + 2 blocks
- 2 blocks
- 2 blocks
2 blocks = 1 bag
Check
?
=
and
Since
and
=
=
8 blocks
8 blocks
Example 1
Solve 3x – 8 = 7. Check your answer.
+8 +8
3x = 15
3
3
x=5
Check 3x – 8 = 7
3(5) – 8 7
15 – 8 7
7 7
Example 2
Solve 4(x – 2) + 2x = 40
4x – 8 + 2x = 40
6x – 8 = 40
+8 +8
Distributive Property
combine like terms
add 8 on both sides of the
equation
6x = 48
6
6
divide 6 on both sides of
the equation
x=8
Solving Algebraic Equations
1.Use the distributive property to get rid of any parenthesis
2.Combine like terms
3.Move all of the variables to one side of the equal sign (make sure
it is positive!)
4.Get the variable by itself by doing opposite math to both sides of
the equal sign
5.Check your answer by substituting it into the original equation
Practice
1.
2.
3.
-4 + 7x = 3
2a + 3 – 8a = 8
9 = 6 – (x + 2)
Click here, if you need help.
Closure



Go over the Vocabulary List
Finish the Assessment Questions Sheet
Homework: finish Equations WebQuest
Lesson Three
Aim: How do we solve systems of
linear equation in two variables
by elimination?
Do Now:
A farmer has ducks and cows. There are 8
heads and 22 feet. How many ducks and
cows does he have?
Guess and Check
has __
has __
2 feet ; a
4 feet
a
The # of
Ducks
1
The # of
Cows
1
Total # of
Heads
2
Total # of
Feet
6=1*2+2*2
2
2
4
12
3
3
6
18
4
4
8
24
4
3
7
20
5
3
8
22
Method one:
Use one variable to set up an equation.
Let x = the amount of ducks, then 8-x = the amount of cows
the # of ducks’ feet * the # of ducks + the # of cows’ feet * the # of cows =total # of feet
2 * x
+
4 * (8–x)
2x + 32 – 4x = 22
-2x + 32 = 22
-2x = -10
x=5
the amount of cows = 8 – x = 8 – 5 = 3
So, there are 5 ducks and 3 cows.
= 22
The Idea of Elimination
+
1.
2.
+5
6
+4
+
+2
-
-5
Development
Example 1 elimination using addition
x – 2y = -19 (1) by elimination
5x + 2y = 1 (2)
Step 1 Add (1) and (2) to eliminate the y-terms.
6x = -18
Step 2 Simplify and solve for x.
x = -3
Step 3 Write one of the original equations.
x – 2y = -19
Step 4 Substitute -3 for x.
-3 – 2y = -19
Step 5 Simplify and solve for y
-2y = -16
y=8
Solve
Development
Example 2 elimination using subtraction
Solve
3x + 4y = 18 (1) by elimination
-2x + 4y = 8 (2)
Step 1 Subtract (1) and (2) to eliminate the y-terms.
5x = 10
Step 2 Simplify and solve for x.
x=2
Step 3 Write one of the original equations.
3x + 4y = 18
Step 4 Substitute 2 for x.
3(2) + 4y = 18
Step 5 Simplify and solve for y
6 + 4y = 18
4y = 12
y=3
Go Back to the Do Now
Let d = the # of ducks and c = the # of cows
d + c = 8 (1)
2d + 4c = 22 (2)
In some case, we will first need to multiply one or both of the
equations by a number so that one variable has opposite
coefficients. This will be the new step 1.
(1) * 2  2(d+c) = 8*2
2d + 2c = 16 (3)
(2) – (3)  2c = 6 c = 3 cows
Substitute c=3 into (1) or (2), but (1) would be easier
d + c = 8  d + 3 = 8  d = 5 ducks
Therefore, the farmer has 3 cows and 5 ducks.
Closure



Answer the Aim
Summarize the elimination method in
solving systems of linear equations in two
variables.
Homework: finish the review sheet