Math 6A, Lesson 7, Spring 2020, 4/5/2020

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Continue on Chapter 5.4 Forming Linear Equations to Solve Problems

1. Midterm Exam: all about equations

2. Constructing/Forming Linear Equations to Solve Problems

The steps involved in problem solving with linear equations are:

  • Step 1. Read the question carefully and identify the unknown quantity
  • Step 2. Use a letter to represent the unknown quantity (e.g. x)
  • Step 3. Express other quantities in terms of x
  • Step 4. Construct/Form an equation based on the given information
  • Step 5. Solve the equation
  • Step 6. Write down the answer statement

3. Homework:

  • Redo the problems you got wrong in last two homework assignments

Math 6A, Lesson 6, Spring 2020, 3/29/2020

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Continue on Chapter 5.4 Forming Linear Equations to Solve Problems

1. Quiz – and quiz go over

2. Constructing/Forming Linear Equations to Solve Problems

The steps involved in problem solving with linear equations are:

  • Step 1. Read the question carefully and identify the unknown quantity
  • Step 2. Use a letter to represent the unknown quantity (e.g. x)
  • Step 3. Express other quantities in terms of x
  • Step 4. Construct/Form an equation based on the given information
  • Step 5. Solve the equation
  • Step 6. Write down the answer statement

3. Homework:

  • Handout
    • Two pages
  • Workbook:
    • Page 31: 21, 22, 23, 24, 25
    • Page 32: 26, 27, 28, 29

Math 6A, Lesson 5, Spring 2020, 3/22/2020

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Chapter 5.3 Simple Fractional Equations

Chapter 5.4 Forming Linear Equations to Solve Problems

1. Fractional equation: when the variable of an equation is in the denominations of a term, the equation is called fractional equation

6 /(x – 2) = 3

1/(x + 3) = 2/x

Note: it is important to check the solutions, that they can’t be those values that make a denominator of the original equation zero.

2. Constructing/Forming Linear Equations to Solve Problems

Two pages handout of teaching material. The steps involved in problem solving with linear equations are:

  • Step 1. Read the question carefully and identify the unknown quantity
  • Step 2. Use a letter to represent the unknown quantity (e.g. x)
  • Step 3. Express other quantities in terms of x
  • Step 4. Construct/Form an equation based on the given information
  • Step 5. Solve the equation
  • Step 6. Write down the answer statement

3. Review basic concept of “open parenthesis”

  • 2(3x+y) – 5(0.2x-0.6y) =
  • -2(3x+y) + 5(0.2x-0.6y) =
  • -2(3x-y) + 5(-0.2x-0.6y) =
  • 2(3x-y) – 5(0.2x+0.6y) =

4.Homework:

  • Handout
    • Three pages
  • Workbook:
    • Page 30: 15, 16, 17, 18, 19, 20

Math 6A, Lesson 4, Spring 2020, 3/15/2020 — Online Class

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Chapter 5.2. Equations involving Parentheses

1. We apply the distributive law of multiplication over addition to help us solve equations involving parentheses.

Recall a(x + b) = ax + ab

Solve equation 9(x + 1) = 2(3x + 8)

2. When working with equations, always apply the same action to both sides of the equation.

Solve equation 5(2x – 9)/3 -8 = 2x

Solve equation (3x + 2)/5 = (4x – 7)/6

3. Rewriting equations

Giving the formula A = ½ * (a + b)h, find the value of a when b = 13, h = 9, and A = 90.

4. Homework:

  • Handout
    • Two pages
  • Workbook:
    • Page 27: 4
    • Page 28: 5
    • Page 29: 11, 12, 13, 14

Math 6A, Lesson 3, Spring 2020, 3/8/2020 — Online Class

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Chapter 5.1 Simple Linear Equations in One Variable

1. Quiz on  algebraic expression and manipulation: variables, constants, terms, like-terms, regrouping, combining like terms, simplifying expression, factoring.

2. Starting chapter 5. Solve simple linear equations in one variable:

  • Concepts: equation, variable, solution/root, linear equation (ax+b=c, where a,b,c are constant and a != 0), LHS, RHS, balancing.
  • Methods: subtract, add, divide or multiply to both sides by the same number.

3. Introduce concepts and methods of “isolate”, “move items to other side and change of sign”, “plug the answer/solution back in the equation”.

Key word: isolate, isolate, isolate. The key to solving many equations is to get the variables alone on one side of the equation. To solve a linear equation with one variable, we isolate the variable by following a few simple steps:

  • simplify both sides of the equation by combining like terms on each side;
  • move all the terms with the variable to one side and all the constants to the other using addition and subtraction, or just moving them to other side with change of signs;
  • after simplify the equation that results from the previous step, multiply by the reciprocal of the variable’s coefficient to solve for the variable.
  • you can always check your answer by plug the solution back to the variable in the equation, both sides should be equal. if not, go check your calculation.

4. Home Work

  • handout:
    • two pages
  • Workbook:
    • page 27: 1, 2, 3
    • page 28: 6

Math 6A, Lesson 2, Spring 2020, 3/1/2020

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Factorization by Extracting Common Factors, Factorization by grouping terms

1. The process of writing an algebraic expression as a product of its factors is called factorization or factoring:

  • 600 = 2x2x2x3x5x5
  • ax + ay = a(x+ y)
  • 15a + 20b = 5(3a) + 5(4b)=5(3a + 4b)
  • 24ax – 40ay + 8a = (8a)(3x) – (8a)(5y) + (8a)(1) = 8a(3x – 5y + 1)

2. Factorization by grouping

  • 12ax – 3ay + 8bx -2by

= (12ax -3ay) + (8bx -2by)

= 3a(4x-y) +2b(4x -y)

=(3a + 2b)(4x-y)

  • 49a + 42c -7ay -6cy

= (49a -7ay) + (42c – 6cy)

= 7a(7 – y) + 6c(7 – y)

= (7a + 6c)(7 – y)

3. Home Work:

  • Handout:
    • two pages
  • Workbook:
    • page 22: 7, 8
    • Page 24: 16, 17, 18, 20
    • Page 26: 26, 27, 28, 29

Math 6A, Lesson 1, Spring 2020, 1/19/2020

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Simplification of Linear Algebraic Expressions

1. The distributive law is applicable when removing parentheses in an algebraic expression. Examples in class:

  • 2(3x + 4y) =
  • -4(5a – 3b) + 7a
  • a(2x + 7y + 5z)
  • (2p + 3q -4r)(-6b)
  • 2[x- 5(3-x)]

2. Express each of the following as a single fraction in the simplest form. Recall LCM of denominators.

  • -p + p/3 +(3p)/5
  • (3p + 10)/4 -2
  • (3x – 4)/4 + (2x+5)/3
  • (1 – 2x)/3 + (3x + 1)/5 + (4x -3)/6

3. Home Work:

  • Handout:
    • two pages
  • Workbook: correct last week’s HW problem of
    • page 21: 3, 4, 5
    • Page 23: 11, 12, 13, 14, 15
    • Page 24: 21
    • Page 25: 22, 23
  • Hand out the report card for Fall 2019

Math 6A, Lesson 16, Fall 2019, 1/12/2020

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Distributive Law, Addition and Subtraction of Linear Algebraic Expressions

1. Use of parentheses and distributive law

  • a(x + y) = ax + ay
  • (x + y)a = a(x + y) = ax + ay = xa + xy
  • a(x – y) = a{x + (-y)] = ax + a(-y) = ax – ay
  • a(x + y + z) = ax + ay + az
  • x – (a – b) = a -a + b

2. Addition and subtraction of linear algebraic expressions: explained in class

  • (2a +3b) + (5a -4b) =
  • Find the sum of -2p + 3q – 4 and p + 5q – 3
  • (4x – 5) – (7x – 3)

3. Home Work:

  • Handout:
    • one pages
  • Workbook:
    • page 21: 3, 4, 5
    • Page 23: 11, 12, 13, 14, 15
    • Page 24: 21
    • Page 25: 22, 23

Math 6A, Lesson 15, Fall 2019, 1/5/2020

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Like Terms and Unlike Terms

1. Terms, coefficient, and constant terms

  • The expression 2x – 3y + 8 consists of three terms. They are 2x, -3y and 8. The numerical part, including the sign, of a term is called the coefficient of the variable.
  • Term 2x, the coefficient of x is 2
  • Term -3y, the coefficient of y is -3
  • Term 8, is called constant term

2. Classify the like terms and unlike terms: explained in class

3. Simplify the algebraic expression by combining (or collecting) like terms

  • 2x + 3x = 5x
  • 8y – 3y = 5y
  • 3a + 4b – 2a + 5b = (3a -2a) + (4b +5b) = a + 9b

4. Home Work:

  • Handout: two pages
  • Workbook:
    • page 21: 1, 2
    • Page 22: 6, 9, 10

Math 6A, Lesson 14, Fall 2019, 12/15/2019

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1. Exam on chapter 3 “Introduction to Algebra”

2. Home Work: 

  • Redo all the problems you got wrong in last three homework assignment.