Mrs. Meehan 6th Grade Algebra - PowerPoint PPT Presentation

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Mrs. Meehan 6th Grade Algebra

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Title: Mrs. Meehan 6th Grade Algebra


1
Mrs. Meehan6th GradeAlgebra
2
Combining unlike terms
  • I can simplify algebraic expressions

3
Do NowSimplify each expression
  • -
  • 5 4 2 - 5
  • 6

4
term
  • 3X³Y
  • coefficient variables
  • Like terms are terms that contain the same
    variables raised to the same exponent.
  • X and 3X
  • 2Y² and -5Y²
  • 3X²Y and ½X²Y
  • 6 and 8

2X² and 2X r³ and t³ 3rt³ and -0.5rt³
Are Unlike Terms
Are Like Terms
Are Like Terms
5
Look at these 10 terms. Find all the like terms
that can be combined
6
Construct a like term that could combine with
each of these terms.
  • 7a³b²
  • ¾mn³
  • 4

7
Combine Like TermsExample 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.

8
PracticeSimplify each expression by combing like
terms.
  • 12x 30x
  • 42x
  • 6.8y² - y²
  • 5.8y²
  • -4n 11n²
  • -4n and 11n² are not like terms. Do not
    combine them
  • -20t 8.5t
  • -28.5t

1
9
Simplify the following expression by combining
like terms
  • 6xy 3x²y xy 6x²y 10xy

6xy
1xy
10xy
3x²y
6x²y
9x²y
17xy
10
Distributive Property
n
n
n
n
n
n
1
1
1
1
1
1
1
1
1
1
1
1
  • 3(n2) 3n
    6
  • 3(n 2) 3n 6

11
Distributive Property
  • What is the Distributive Property?
  • How to use the Distributive Property in
    simplifying algebraic expressions?
  • Watch video Distributive Property Basics

12
Example 2
  • Simplify 2(x6) 3x.
  • Try this
  • 1) 6(x - 4) 9 2) -12x
    - 5x 3a x

Procedure Justification
1. 2(x6)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
13
Solving equations
  • I can solve equations

14
  • What is an equation?
  • What is a solution of an equation?
  • How do we find the solutions of an equation?


Answer
15
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
16
Motivation
17
  • 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

18
Check
  • ?
  • and and
  • Since
  • 8 blocks 8 blocks

19
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

20
Example 2
  • Solve 4(x 2) 2x 40
  • 4x 8 2x 40
    Distributive Property
  • 6x 8 40 combine
    like terms
  • 8 8 add 8 on
    both sides of the

  • equation
  • 6x 48
  • 6 6
    divide 6 on both sides of

  • the equation
  • x 8

21
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

22
Practice
  • -4 7x 3
  • 2a 3 8a 8
  • 9 6 (x 2)
  • Click here, if you need help.

23
Two Variables
  • I can solve systems of linear equation in two
    variables by elimination

24
Do Now
  • A farmer has ducks and cows. There are 8 heads
    and 22 feet. How many ducks and cows does he have?

25
Guess and Checka has __ feet a
has __ feet
2
4
The of Ducks The of Cows Total of Heads Total of Feet
1 1 2 61222
2 2 4 12
3 3 6 18
4 4 8 24
4 3 7 20
5 3 8 22
26
Method oneUse 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 ) 22
  • 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.

27
The Idea of Elimination
  • -
  • 5 4 2 - 5
  • 6


-

-
28
DevelopmentExample 1 elimination using addition
(1)
  • Solve x 2y -19 by elimination
  • 5x 2y 1
  • 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

(2)
29
DevelopmentExample 2 elimination using
subtraction
(1)
  • Solve 3x 4y 18 by elimination
  • -2x 4y 8
  • 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

(2)
30
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(dc) 82
  • 2d 2c 16 (3)
  • (2) (3) ? 2c 6? c 3 cows
  • Substitute c3 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.

31
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