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Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

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Page 1: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Algebra 1

Ch 2.4 – Solving Equations with Variables on Both Sides

Page 2: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Objective In the last lesson we solved multi-step

equations… In this lesson we will continue that

theme with another variation of equations…that is,

Students will solve equations with variables on both sides of the equation

Page 3: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Before we begin…

1. Distribute right away if necessary!!!2. Combine like terms on the same side of the

equation if necessary!!!3. Always start by getting the variable on the same

side!!!- Add or subtract the term with the variable from both sides of the equation.- When doing this you can only add or subtract like terms!!!

4. Use inverse operations to solve for the variable!!!5. Always multiply or divide as the last step!!!

Page 4: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

General Rule

It really doesn’t matter which side you collect the variable to….however, the general rule for solving equations with variables on both sides of the equals sign is to collect the variables to the side with the largest variable…

The reason we use the general rule is to minimize the amount of work with negative numbers…students are not always proficient with working with negative numbers….

Page 5: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Example 17x + 19 = - 2x + 55

Let’s analyze this equation first…

Deciding where to start….I see that on the left side is 7x and on the right side is – 2x.

In this instance I will use the general rule and collect the x’s to the left side of the equation…I do that by working on the right side of the equation and by adding 2x to both sides of the equation…which looks like this:

7x + 19 = - 2x + 55

+2x +2x

9x + 19 = + 55

The 2x’s on the right side cancel out leaving +55

7x + 2x = 9x

After collecting the variables to the left side I am left with a 2-step equation. Add/Subtract first and then multiply or divide…

Page 6: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Example #1 (continued)9x + 19 = + 55

In this 2-step equation to undo the +19, subtract 19 from both sides of the equation

9x + 19 = + 55

- 19 -19

9x = 36

To undo the multiplication, divide both sides by 9

9x = 36

9 9The 9’s cancel out leaving x x = 4

36 9 = 4

Page 7: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Example #290 – 9y = 6y

Let’s analyze this equation…

I see that there is a – 9y on the left and a +6y on the right…I’m going to make the decision to collect the variables to the right side of the equation.

To undo the – 9y, add 9y to both sides of the equation…it looks like this:

90 – 9y = 6y

+ 9y +9y

90 = 15y

The 9y’s on the left cancel out leaving 90

6y + 9y = 15y

You are left with a 1 step equation…

Page 8: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Example #2 (continued) 90 = 15y

To undo the multiplication of 15y…divide both sides by 15

90 = 15y

15 15

The 15’s cancel out leaving y

6 = y

90 15 = 6

The solution that will make the equation true is y = 6

Page 9: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Many or No Solutions In this lesson we will also look at equations that

have many solution or no solutions… When working with linear equations you will

come across situations in which there are many or no solutions to the equation…

Later on in the course when we plot systems of equations you will find that the many or no solution equations mean something…more about that later in the course….

Let’s look at some examples…

Page 10: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Equations with many solutions3(x + 2) = 3x + 6

First, we have to analyze this equation and decide what we want to do…

Notice, that the distributive property is illustrated on the left side of the equation…we must take care of that before we do anything else…

3(x + 2) = 3x + 6

Distribute the 3 on the left side of the equation to get:

3x + 6 = 3x + 6

Now you can collect the x’s to either side of the equation…I choose the left

3x + 6 = 3x + 6

- 3x -3x

6 = 6

The x’s on the left cancel each

other out

The x’s on the right cancel

each other out also

Page 11: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Equations with many solutions

6 = 6

What you are left with is a true statement 6 does equal 6

What does the solution mean?

This type of linear equation is called an identity

Any value of x will make this a true statement

If you choose any number and substitute it for x you will always get a true statement

Page 12: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Equations with no solutionsx + 2 = x + 4

Again, we should analyze the equation first…

I notice that there is an x on both sides of the equation…It doesn’t matter which side you collect the variables to….I will collect them to the left side like this:

x + 2 = x + 4

-x - x

2 = 4

I am left with the statement 2 = 4. This is not a true statement. To write this correctly you would use the symbol ≠ as follows:

2 ≠ 4

Page 13: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Equations with no solutions

2 ≠ 4

What does it mean?

In this equation, no value of x will make this a true statement

The solution is written as no solution

Page 14: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Solving Complex Equations

Ok…you have all the information you need to solve any equation…

The key is to analyze the equation first and make some decisions…

It’s ok if you don’t get it right the first time…go back and problem solve to see where you made your error…

I will walk you through a more complex equation so that you can see my thought process..

Page 15: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Example #34(1 –x) + 3x = -2(x – 1)

Ok…I notice right off that I have the distributive property on both the left and right side of the equation….

I must take care of the distributive property first before I do anything else

Don’t forget…at the level you are required to be able to recognize and know how to work with the distributive property….

4(1 –x) + 3x = -2(x – 1)

4 – 4x + 3x = -2x + 2

Don’t forget that -2 times -1 = +2

Ok…now I see that I have multiple x’s on the left side and -2x on the right…I’m going to make the decision to combine the x’s on the left before I collect all the x’s to the same side… My new equation will look like this…..

Page 16: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Example #3 (continued)

4 – 4x + 3x = -2x + 2

4 – x = -2x + 2

-4x + 3x = -1x. We write -1x as

just -xOk…now I have x’s on both sides of the equation…

I need to get the x’s on the same side…I am going to use the general rule and collect the x’s to the left side….that’s because –x is bigger than -2x and I don’t want to have to work with negative numbers…to undo the -2x add 2x to both sides…

4 – x = -2x + 2

+2x +2x

4 + x = + 2

The 2x’s cancel out leaving 2

-x + 2x = +1x. This is written

as just +x

Page 17: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Example #3 (continued)

4 + x = + 2

Ok…now all I have to do is isolate the x on the left and I have the solution….to undo the +4, subtract 4 from both sides of the equation….like this…

4 + x = + 2

-4 -4The 4’s on the left cancel out leaving

just x x = - 2

+ 2 – 4 = - 2

The solution that makes the equation true is x = -2

Reminder: Your answer must always be stated as a positive variable, If you have a negative variable you have to do something to make it positive!

Page 18: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

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…

Page 19: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

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 Equations with Variables on Both Sides. 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…

Page 20: Algebra 1 Ch 2.4 – Solving Equations with Variables on Both Sides

Homework: 2.4 B

1. 4x + 27 = 3x

2. -2m = 16m – 9

3. 12c – 4 = 12c

4. 12p – 7 = - 3p + 8

5. -7 + 4m = 6m - 56. 24 – 6r = (4 – r)

7. -4(x – 3) = -x

8. -2(6 – 10n) = 10(2n – 6)

9. ¼ (60 + 16s) = 15 + 4s

10. ¾ (24 – 8b) = 2(5b + 1)