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Warm-Up Equations: Old & New Technology on Wheels Forms Returned Khan Academy Topics for This Week Class Work Today: Sept. 29, 2014

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Page 1: Sept. 29, 2014

Warm-Up Equations: Old & New Technology on Wheels Forms Returned Khan Academy Topics for This Week Class Work

Today:

Sept. 29,

2014

Page 2: Sept. 29, 2014

Show Me Pictures

Another Great Definition of Learning

*A Life Lesson for All of Us

~ A great graph and praise to this student for reasons that have nothing to do with algebra. ~ It’s all about not giving up; notquitting when things get tough. Stay with it and work harder, not less. Not giving up is the ONLY way to achieve whatever it is that you’d like to achieve. In this case....

“It's not that I'm so smart, it's just that I stay with problems longer.”

- Albert Einstein

Page 3: Sept. 29, 2014

Due on or before October 5, 7:00 pm

The root word of LINEar is?

Page 4: Sept. 29, 2014

Test #2: Class Averages

Page 5: Sept. 29, 2014

Equations.....truly are the foundation on which Algebra is

built upon. We cannot just “test it and move on.”

What’s the point of moving on if we can’t ultimately close out a problem by solving the equation?

Equations

y = (x + 1)2 – 2= 0

Even if we can turn this

into this,if we can’t solve for x and y, we’ll not

be able to solve the problem.

We are still learning new equation types, (literal, identity, absolute value). While learning & practicing the new equations, we’re going to have additional practice on those types of equations we have covered so far. Focus your efforts on understanding the rules

common to all equations. (Inverse Operations, Positive variable, etc.)

Page 6: Sept. 29, 2014

1. The width of a rectangle is 2 more than one-half the length. The perimeter is 89 yards. Find the length & width.

P = 89

Warm Up

9 – 6x = 2x + 1

Page 7: Sept. 29, 2014

* The same rules to solving equations apply, there are just more steps to the solution.

1. 10y - (4y + 8) = -20

Multi-Step Equations

2. −8z −(2+3z) = 5(z + 3)

Equations: 1-Step

1. x + 4 = 11

Equations: 2-Step

2. 3x - 5 = 10

2. x - 5 = 81. 2x + 4 = 18

Review

Page 8: Sept. 29, 2014

Class Notes &

Practice Problem

Section of

Notebook, Pls.

Page 9: Sept. 29, 2014

Solving Equations with Variables on Both Sides

When an equation has variables on both sides, three different results are possible.

Write & solve this equation: 2x + 2 = 4 + x The equation results in a solution of a single

number.In this case, x = 2

Page 10: Sept. 29, 2014

A 2nd Possibility

• 3x - 9 = 3x + 10

• A linear equation has NO SOLUTION when the solved equation has no variable and the equation is false.

3x - 3x = no variable, -9 = 10

Page 11: Sept. 29, 2014

A 3rd Possibility

• When an equation has variables on both sides, something else may occur. Solve the equation:

4(x – 5) = 4x - 204x -20 = 4x -20; Subtract 4x from the left side and -20 = -20This is called “identity”. It means ALL REAL NUMBERS make the sentence true.

A linear equation is an identity when the solved equation has NO VARIABLE and the solution is TRUE!In other words, no matter what number we substitute for x, the equation will be true.

Page 12: Sept. 29, 2014

Therefore…

• When solving equations with variables on both sides, 3 different things can happen.

1) solution is a single number (x = 4)

2) solution is NO SOLUTION (empty set)

3) solution is ALL REAL Numbers ( IDENTITY)

Page 13: Sept. 29, 2014

Class Notes:

Tips for Solving Literal & Other Equations:

1. If the variable you are solving for is inside the parenthesis, you must distribute first.

9p = 3(k + 5) for k

2. If the variable you are solving for is outside the parenthesis, simply perform the opposite operation.9 = k(p + 5) for k

3. When clearing fractions, if the denominator being multiplied has two or more terms, it must be distributed on the other side.

5k( 2p + 4) = 9 = for p

Literal Equations:

We need to separate the ‘k’ from the 5 in this case. Solve the equation.

Page 14: Sept. 29, 2014

Class Notes:

Tips for Solving Literal & Other Equations:

4. Always check to see if every term has something in common. You can then factor the common part out.3kp – 7km – 9k – 1 = -7p +

10 for k first

3kp – 7km – 9k = -7p + 11

Then,

k(3p – 7m – 9) = -7p + 11

Then What?

k= -7p + 11 (3p – 7m – 9)

Class Notes:

Page 15: Sept. 29, 2014

Tips for Solving Literal & Other Equations:

5. If an entire term can be factored out, replace it inside the parenthesis with a ‘1’

2p + 8px = ?

2p(1 + 4x) check by distributing the 2p again.

6. When solving fractional equations, you may have to clear fractions more than once.

Class Notes:

4 – 2x = 𝟔−𝟑 𝒙𝟑 20 – 10x =5

Complete the first step.Now we distribute the....

Solve for x.

Page 16: Sept. 29, 2014

Tips for Solving Literal & Other Equations:

7. Two equal fractions is called a proportion. Proportions can be solved by cross multiplying.

The following is not a proportion.

How must this problem be solved?

Finish It

Class Notes:

= solve

Page 17: Sept. 29, 2014

Class Work

Page 18: Sept. 29, 2014

Post-Test: 4 Most Missed

The first stage of a rocket burns 28 seconds longer than the second stage. If the total burning time for both stages is 182 seconds, how many seconds does the first stage burn for?

V. 3, Q 9

34%

A) 124 B) 77 C) 98 D) 90 E) 105

V. 1, Q 8

32%

- 6 = x - (3 – x) A) 1 B) -3 C) D) E) -2

Page 19: Sept. 29, 2014

Post-Test: 4 Most Missed

The Most Difficult Question was.....

V. 2, Q 5

30%

4(2x – 2) = -8(x – 4)

A) 2 B) 2 C) -2 D) E) No Solution

= .....

V. 3, Q10

26%

The sum of three consecutive even numbers is 132.

What is the largest of the three numbers?A) 33 B) 44 C) 42 D) 46 E) 43