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8/14/2019 Solving Eq W-Variables on Both Sides
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Algebra 1
Ch 3.4 Solving Equationswith Variables on Both Sides
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Objective
Students will solve equations withvariables on both sides of the equation
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General Rule
It really doesnt matter which side youcollect the variable to.however, thegeneral rule for solving equations withvariables on both sides of the equals sign isto collect the variables to the side withthe largest variable The reason we use the general rule is tominimize the amount of work with negativenumbersstudents are not always proficientwith working with negative numbers.
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Example 17x + 19 = - 2x + 55
Lets analyze this equation first
Deciding where to start.I see that on the left side is 7x and on theright side is 2x.
In this instance I will use the general rule and collect the xs to the leftside of the equationI do that by working on the right side of theequation and by adding 2x to both sides of the equationwhich lookslike this:
7x + 19 = - 2x + 55+2x +2x
9x + 19 = + 55
The 2xs on theright side cancelout leaving +55
7x + 2x = 9x
After collecting the variables to the right side I am left with a 2-stepequation. Add/Subtract first and then multiply or divide
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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 9
The 9scancel outleaving x x = 4
36 9 = 4
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Example #290 9y = 6y
Lets analyze this equation
I see that there is a 9y on the left and a +6y on the rightIm going tomake the decision to collect the variables to the right side of the equation.
To undo the 9y, add 9y to both sides of theequationit looks like this:
90 9y = 6y
+ 9y +9y
90 = 15y
The 9yson the left
cancelout
leaving90
6y + 9y = 15y
You are left with a 1 step equation
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Example #2 (continued)90 = 15y
To undo the multiplication of 15ydivide both sides by 15
90 = 15y
15 15
The 15s cancelout leaving y
6 = y
90 15 = 6
The solution that will make the equation true is y = 6
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Many or No Solutions
When working with linear equations you willcome across situations in which there aremany or no solutions to the equationLater on in the course when we plot systemsof equations you will find that the many or nosolution equations mean somethingmore
about that later in the course.Lets look at some examples
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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
equationwe must take care of that before we do anything else3(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 xs to either side of the equationI choose the left
3x + 6 = 3x + 6
- 3x -3x
6 = 6
The xs on theleft cancel each
other out
The xs on theright cancel
each other out
also
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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 andsubstitute it for x you will alwaysget a true statement
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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 equationIt doesntmatter which side you collect the variables to.I will collect themto 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 . Towrite this correctly you would use the symbol as follows:
2 4
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Equations with no solutions2 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
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Solving Complex Equations
Okyou have all the information you need tosolve any equation
The key is to analyze the equation first and makesome decisionsIts ok if you dont get it right the first timegoback and problem solve to see where you madeyour errorI will walk you through a more complex equationso that you can see my thought process..
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Example #34(1 x) + 3x = -2(x 1)
OkI notice right off that I have the distributive property on both the leftand right side of the equation.
I must take care of the distributive property first before I do anything else
Dont forgetat the level you are required to be able to recognizeand know how to work with the distributive property .
4(1 x) + 3x = -2(x 1)
4 4x + 3x = -2x + 2
Dont forget that
-2 times -1 = +2
Oknow I see that I have multiple xs on the left side and -2x onthe rightIm going to make the decision to combine the xs on theleft before I collect all the xs to the same side My new equation
will look like this..
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Example #3 (continued)4 4x + 3x = -2x + 2
4 x = -2x + 2
-4x + 3x = -1x.We write -1x as
just -xOknow I have xs on both sides of the equation
I need to get the xs on the same sideI am going to use thegeneral rule and collect the xs to the left side.thats because
x is bigger than -2x and I dont want to have to work with
negative numbersto undo the -2x add 2x to both sides4 x = -2x + 2
+2x +2x
4 + x = + 2
The 2xs cancelout leaving 2
-x + 2x = +1x.This is written
as just +x
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Example #3 (continued)4 + x = + 2
Oknow all I have to do is isolate the x on the left and I have thesolution.to undo the +4, subtract 4 from both sides of theequation.like this
4 + x = + 2
-4 -4The 4s on the leftcancel out leaving
just xx = - 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!
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Comments
On the next couple of slides are some practiceproblemsThe answers are on the last slide
Do the practice and then check your answersIf you do not get the same answer you must question what you didgo back andproblem solve to find the error
If you cannot find the error bring your work tome and I will help
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Your Turn
1. 4x + 27 = 3x2. -2m = 16m 93. 12c 4 = 12c4. 12p 7 = - 3p + 85.
-7 + 4m = 6m - 5
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Your Turn
1. 24 6r = (4 r)2. -4(x 3) = -x3. -2(6 10n) = 10(2n 6)4. (60 + 16s) = 15 + 4s5.
(24 8b) = 2(5b + 1)
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Your Turn Solutions
1. -272. 3. No solution4. 15. -1
1. Identity; all realnumbers
2. 43. No solution4. Identity; all real
numbers5. 1
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Summary
A key tool in making learning effective is beingable to summarize what you learned in a lesson inyour own wordsIn this lesson we talked about Solving Equationswith Variables on Both Sides. Therefore, inyour own words summarize this lessonbe sureto include key concepts that the lesson covered aswell as any points that are still not clear to youI will give you credit for doing this lessonpleasesee the next slide
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CreditI will add 25 points as an assignment grade for you working onthis lessonTo receive the full 25 points you must do the following:
Have your name, date and period as well a lesson number as aheading.Do each of the your turn problems showing all workHave 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 headingWithout showing work for the your turn problemsWithout a summary in your own words
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