Solving Eq W-variables On Both Sides

Algebra 1 Ch 3.4 – Solving Equations with Variables on Both Sides

Objective 

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

Before we begin… 

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In the last lesson we solved multi-step equations… In this lesson we will continue that theme with another variation of equations…that is, equations that have variables on both sides of the equals sign… In this lesson we will also look at equations that have many solution or no solutions…

General Rule 

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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….

Example 1 7x + 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 + 2x = 9x

7x + 19 = - 2x + 55 +2x 9x + 19 =

+2x

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

+ 55

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

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

=

- 19 9x

+ 55 -19

=

36

To undo the multiplication, divide both sides by 9 9x The 9’s cancel out leaving x

=

9 x

36 9

=

4

36 ÷ 9 = 4

Example #2 90 – 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: The 9y’s on the left cancel out leaving 90

90 – 9y = 6y

6y + 9y = 15y

+ 9y +9y 90

= 15y

You are left with a 1 step equation…

Example #2 (continued) 90

= 15y

To undo the multiplication of 15y…divide both sides by 15 90 ÷ 15 = 6

90

= 15y

15

15

6

The 15’s cancel out leaving y

= y

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

Many or No Solutions 

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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…

Equations with many solutions 3(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 The x’s on the left cancel each other out

- 3x

-3x 6=

6

The x’s on the right cancel each other out also

Equations with many solutions 6

=

6

What you are left with is a true statement 6 does equal 6 What does the solution mean? 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 This type of linear equation is called an identity

Equations with no solutions x+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

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

Solving Complex Equations 

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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..

Example #3 4(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…. Don’t forget…at the level you are required to be able to recognize and know how to work with the distributive property…. I must take care of the distributive property first before I do anything else 4(1 –x) + 3x = -2(x – 1)

Don’t forget that -2 times -1 = +2

4 – 4x + 3x = -2x + 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…..

Example #3 (continued) 4 – 4x + 3x = -2x + 2 4 – x = -2x + 2 -4x + 3x = -1x. We write -1x as just -x

Ok…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…

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

4 – x = -2x + 2 +2x 4+x =

+2x +2

The 2x’s cancel out leaving 2

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…

The 4’s on the left cancel out leaving just x

4+x =

+2

-4

-4 x =

+2–4=-2

-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!

Comments 

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

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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…

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

Your Turn 1. 2. 3. 4. 5.

4x + 27 = 3x -2m = 16m – 9 12c – 4 = 12c 12p – 7 = - 3p + 8 -7 + 4m = 6m - 5

Your Turn 1. 2. 3. 4. 5.

24 – 6r = (4 – r) -4(x – 3) = -x -2(6 – 10n) = 10(2n – 6) ¼ (60 + 16s) = 15 + 4s ¾ (24 – 8b) = 2(5b + 1)

Your Turn Solutions 1. 2. 3. 4. 5.

-27 ½ No solution 1 -1

1.

2. 3. 4.

5.

Identity; all real numbers 4 No solution Identity; all real numbers 1

Summary 

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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…

Credit 

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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…