Alg 1 - Ch 3.1 Solving Linear Eq w

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Algebra 1

Ch – 3.1 Solving Linear Equations

Using Addition and Subtraction

Objective

 Students will solve linear equations using addition and subtraction

Before we begin…

 Thus far we have discussed integers and written algebraic expressions and equations…

 Today we will look at how to solve linear equations using addition and subtraction…

 This is a fundamental algebraic skill…you are required to know how to do this…You will see this on tests….

Vocabulary

 Solve – When you solve an equation you are trying to find the value of the variable that makes the statement true .

 Solution – The value of the variable that makes the statement true. In an equality there is only 1 solution. In an inequality there are many solutions.

Solving Equations

 In order to solve equations you must understand what the goal is and the method to achieve the goal.

 When solving equations the goal is to isolate the variable …that means to get the variable all by itself on one side of the equation

 The method we use to accomplish that task is called the Balanced Equation Method . That means what you do to one side of the equation you must do to the other side.

 We use inverse operations to achieve the balanced equation method

Inverse Operations

 Inverse operations are the operations that

“undo” the four mathematical operations of add, subtract, multiply and divide

 Subtraction undoes addition

 Addition undoes subtraction

 Division undoes multiplication

 Multiplication undoes division

Equations

 Recall that equations have 2 parts – a left side and a right side separated by the equal sign.

Example

Separation of the two sides

12 + a = 20

Left Side

Right Side

Properties

 Before we can begin solving equations we must understand and prove that these properties are true.

 Today we will look at the following properties:

– The subtraction property of equality

– The addition property of equality .

Subtraction Property of Equality

 The subtraction property of equality states:

– If you subtract the same number from each side of an equation, the two sides remain equal

 Here is what it looks like

7 = 7

If I subtract 3 from both sides the equation both sides will remain equal

7 – 3 = 7 – 3

4 = 4

Note: It doesn’t matter what number you subtract from both sides the ending equation will still be equal

Addition Property of Equality

 After looking at the subtraction property of equality try to predict what the addition property of equality states…

Addition Property of Equality

 The addition property of equality states:

– If you add the same number to both sides of an equation, the two sides will remain equal

Example:

7 = 7

If I add 3 to both sides of the equation, both sides will remain equal

7 + 3 = 7 + 3

10 = 10

Again, it does not matter what number you add to both sides the ending equation will still be equal

Quick Review

 Before we go further…lets do a quick review of what we talked about

– What does it mean to solve an equation?

To find the value of the variable

– What’s the goal of solving an equation?

To isolate the variable

– How is the goal achieved?

Using the balanced equation method and inverse operations

– Where does the properties of equality fit into solving equations?

The properties of equality allow you to add or subtract from both sides of the equation thereby keeping the equation balanced

Comments

 At the 8 th grade level you are required to demonstrate your understanding of concepts.

 Therefore, I require that you show the steps used to solve these equations

 In many instances you can solve the equations mentally and that’s ok...However,

I will not give you credit for the work unless you show the steps…

 Lets look at some examples…

x + 5 = 3

- 5 -5 x = -2

Example #1

1. Write the Equation

2. Subtract 5 from both sides. This is the subtraction property of equality

3. After you subtract 5 from both sides you are left with the isolated variable and its value. In this case -2

Check

 You can check your answer by substituting the value of the variable in the original equation.

 If the answer creates a true statement than you have the correct solution

 If not…you did something wrong and must go back and problem solve

 Here is what the check looks like for the previous example…

Check

Original Problem x + 5 = 3

- 5 -5 x = -2

Check x + 5 = 3

(-2) + 5 = 3

3 = 3

1. Write the original equation

2. Substitute the value of the variable

3. Simplify

Algebraic Thinking…

 Solving these types of problems are all about how you think….

 The key is assessing what you see and knowing what to do to isolate the variable…

 Let’s look at how that works…

Example # 2

a + 4 = 10

1.

The first question you have to ask yourself is which side do you begin working on…keeping in mind that the goal is to isolate the variable…In this instance you begin working on the left side of the equation

2.

The second part is that I see that there is a + 4 on the left side of the equation…I need to know how to undo the

+ 4, and when I do that I must do it to both sides of the equation (Balanced Equation Method)

3.

Once I undo the + 4 I have the solution to the equation

4.

I can then check my solution by doing a check…

5.

Its that simple…

Example # 3

- 9 = 2 + y

- 2 -2

- 11 = y

1. Write the equation

2. Determine which side of the equation you will work on first…in this instance it’s the right side

3. To isolate the variable you have to use the subtraction property of equality. In this case subtract 2 from both sides

4. After you subtract 2 from both sides of the equation you will be left with the isolated variable and its value.

In this case -11

Comments

 On the next couple of slides are some practice problems…The answers are on the last slide…

 Do the practice and then check your answers…If you do not get the same answer you must question what you did…go back and problem solve to find the error…

 If you cannot find the error bring your work to me and I will help…

Your Turn

 Solve the equation

1.

x + 5 = 10

2.

11 = r – 4

3.

n – 5 = -9

4.

-3 + x = 7

5.

|-6| + y = 11

Your Turn

 Solve the equation

6.

|-8| + x = -3

7.

19 – (-y) = 25

8.

x + 4 – 3 = 6 ● 5

9.

4 = - b – 12

10.

-r – (-7) = 16

1.

5

2.

15

3.

-4

4.

10

5.

5

Your Turn Solutions

6.

-11

7.

6

8.

29

9.

-16

10.

-9

Summary

 A key tool in making learning effective is being able to summarize what you learned in a lesson in your own words…

 In this lesson we talked about Solving Linear

Equations using the addition and subtraction properties of equality … Therefore, in your own words summarize this lesson…be sure to include key concepts that the lesson covered as well as any points that are still not clear to you…

 I will give you credit for doing this lesson…please see the next slide…

Credit

 I will add 25 points as an assignment grade for you working on this lesson…

 To receive the full 25 points you must do the following:

– Have your name, date and period as well a lesson number as a heading.

– Do each of the your turn problems showing all work

– Have a 1 paragraph summary of the lesson in your own words

 Please be advised – I will not give any credit for work submitted:

– Without a complete heading

– Without showing work for the your turn problems

– Without a summary in your own words…

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