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Algebra 1 Solving Linear Equations in One Variable

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Section 2: Linear Equations: Lecture 2 | 28:52 min

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Lecture Comments (32)

	0 answers Post by Jessica Zhang on March 12, 2022 In example IV don't you add 2 and 4
	2 answers Last reply by: Steven Tue Mar 17, 2020 5:14 PM Post by Genevieve Carisse on October 21, 2017 Professor, what is the order to use? For example, 2x+6=4x-8. Instead of -2x first, why not do -6 or + 8 first? Thank you!
	1 answer Last reply by: Professor Eric Smith Tue Jun 6, 2017 11:22 AM Post by Carl Kellogg on June 6, 2017 Why did you pursue the path of teaching mathematics?
	0 answers Post by Karen Johnson on March 17, 2017 Example 4, you distributed the 2 to the 1, but not to 4 - 2x?
	2 answers Last reply by: AJH Yang Mon Sep 21, 2020 12:34 AM Post by Karen Johnson on March 17, 2017 Eric Smith, In example 2, you changed the -1 to - 5/5? how did you do this?
	1 answer Last reply by: Professor Eric Smith Fri Aug 26, 2016 7:07 PM Post by Raymond Hayden on February 23, 2016 Some of your questions have incorrect answers. For example: 40+x=-22 is x=-62, you have x=-64 as the answer. There are other errs as well, I just can't call them to mind right now. This is confusing and should be corrected. Thank you.
	1 answer Last reply by: Professor Eric Smith Tue Sep 22, 2015 4:16 PM Post by Oscar Prado on September 22, 2015 Hi there on your second example how did you have up with 20 times 3/4 is 15?
	1 answer Last reply by: Professor Eric Smith Sat Mar 28, 2015 5:29 PM Post by antonio cooper on March 26, 2015 Sir, I would like to inform you that on your practice questions (#1)40+X=-22. Your answer is X=-64, however unless I am mistaken -40+-22 is -62. Thank you for your lessons.
	1 answer Last reply by: Professor Eric Smith Thu Nov 13, 2014 10:31 AM Post by TAHA sakor on November 13, 2014 Why do we have to multiplication the equation twenty
	1 answer Last reply by: Professor Eric Smith Thu Nov 13, 2014 10:29 AM Post by TAHA sakor on November 13, 2014 what is consistent
	1 answer Last reply by: Professor Eric Smith Sun Jul 6, 2014 2:27 PM Post by ajaa trevino on July 2, 2014 In Example 4 why did you multiply everything by 2?
	2 answers Last reply by: Professor Eric Smith Mon Jun 23, 2014 1:25 PM Post by dzung tran on June 16, 2014 What if someone thought, on Example 1,instead of adding a negative 2x,they divided a negative 2x?The problem would look like x+6=2x-8. If I continued to solve it, it would turn to 6=x-8, and then to 14=x. I'm not saying that I did this, but I followed the rules and whatever I did to one side, I did to the other. Why is it different when I do a different operation?
	1 answer Last reply by: Professor Eric Smith Thu Jan 9, 2014 3:59 PM Post by Mohamed Elnaklawi on December 30, 2013 In example one, how did you simplify x+3, and get 6?
	1 answer Last reply by: Professor Eric Smith Thu Jan 9, 2014 4:14 PM Post by Araksya Fernandes on December 7, 2013 in Example V i got -7 as well, but from 4x-5=9+6x I said 4x-6x=5+9 -2x=14 x= 14:(-2) x=-7 If i do it this way, am I assured that I will get the answer right?
	2 answers Last reply by: Professor Eric Smith Tue Sep 3, 2013 6:39 PM Post by steven schwartzle on August 31, 2013 Do you still use PEMDAS for all Linear Equations??

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Solving Linear Equations in One Variable

If only one value will make an equation true it is conditional. If any number will make the equation true it is an identity. If no number will make the equation true, it is a contradiction.
To solve a linear equation we work to isolate the variable on one side of the equation. To help out we use the Addition and Multiplication property of equality.
The addition and multiplication property of equality says we may add the same number to both sides of an equation, and we may multiply the same number on both sides of an equation (as long as it is not zero.)
To help out with the solving process you can multiply all terms by a common denominator. This clears out fractions.
It is a good idea to check the solution in the original problem to ensure that it is correct.

Solving Linear Equations in One Variable

40 + x = - 22

40 + x - 40 = - 22 - 40

x = - 64

10 = 54 - y

10 - 54 = 54 - y - 54
- 44 = - y

44 = y

r − [1/5] = [2/5]

r − [1/5] + [1/5] = [2/5] + [1/5]

r = [3/5]

[5/8] = [2/8] − t

[5/8] − [2/8] = [2/8] − t − [2/8]
[3/8] = − t

− [3/8] = t

Thirty - five is six more than a number. Find the number.
x = unknown number

35 = 6 + x
35 - 6 = 6 + x - 6

29 = x

Twelve less than a number is sixty six. Find the number.
n = unknown number

n - 12 = 66
n - 12 + 12 = 66 + 12

n = 78

Eighteen is thirty - two less than a number. Find the number.
n = unknown number

18 = n - 32
18 + 32 = n - 32 + 32

50 = n

The sum of 6, 8, and a number is equal to 30. What is the number?
x = unknown number

6 + 8 + x = 30
14 + x = 30
14 + x - 14 = 30 - 14

x = 16

The difference of 55 and a number is equal to 23. What is the number?
x = unknown number

55 - x = 23
55 - x - 55 = 23 - 55
- x = - 32

x = 32

The sum of 21.8, 60.1 and a number is equal to 111.5. What is the number?
x = unknown number

21.8 + 60.1 + x = 111.5
81.9 + x = 111.5
81.9 + x - 81.9 = 111.5 - 81.9

x = 29.6

[y/12] = 3

12 ×( [y/12] ) = 3 ×12

y = 36

[g/9] = 17

9 ×( [g/9] ) = 17 ×9

g = 153

22i = 132

[22i/22] = [132/22]
i = [132/22]

i = 6

( 3[1/3] )s = [2/4]

[10/3]s = [2/4]
( [3/10] ) ×[10/3]s = [2/4] ×( [3/10] )
s = [6/40] = [3/20]

s =[(3)/(20)]

( 4[2/7] )l = [3/5]

[30/7]l = [3/5]
( [7/30] ) ×[30/7]l = [3/5] ×( [7/30] )
l = [21/150] = [7/50]

l =[(7)/(50)]

− 12j = 480

[( − 12j)/( − 12)] = [480/12]

j = 40

Four and one fifths times a number is equal to two and three eighths. What is the number?

4[1/5]x = 2[3/8]
[21/5]x = [19/8]
( [5/21] ) ×[21/5]x = [19/8]v ×( [5/21] )

x =[(95)/(168)]

Three tenths of a number is equal to three and seven eighths. What is that number?

[3/10]d = 3[7/8]
[3/10]d = [31/8]
( [10/3] ) ×[3/10]d = [31/8] ×( [10/3] )
d = [310/24] = [155/12]

d =[(155)/(12)]

Fifty - two divided by a number is four.

[52/h] = 4
h ×[52/h] = 4 ×h
52 = 4h

13 = h

( 12[4/9] )w = [11/13]

[112/9]w = [11/13]
( [9/112] ) ×[112/9]w = [11/13] ×( [9/112] )

w =[(99)/(1456)]

3y - 11 = 37

3y - 11 + 11 = 37 + 11
3y = 48

y = 16

89 = 17 - 6x

89 - 17 = 17 - 6x - 17
72 = - 6x
12 = - x

- 12 = x

[(a − 7)/4] = 11

4( [(a − 7)/4] ) = 11(4)
a - 7 = 44
a - 7 + 7 = 44 + 7

a = 51

16 = [(45 − t)/3]

3(16) = ( [(45 − t)/3] )3
48 = 45 - t
48 - 45 = 45 - t - 45
3 = - t

- 3 = t

9 = [(17 + b)/5]

5(9) = ( [(17 + b)/5] )5
45 = 17 + b
45 - 17 = 17 + b - 17

28 = b

Three consecutive numbers have a sum of fifty - four. Find the numbers.

x, x + 1 ,x + 2 represent the three consecutive numbers
x + (x + 1) + (x + 2) = 54
3x + 3 = 54
3x + 3 - 3 = 54 - 3
3x = 51

x = 17

Five consecutive numbers have a sum of one hundred. Find the numbers.

x, x + 1, x + 2, x + 3, x + 4 represent the five consecutive numbers.
x + (x + 1) + (x + 2) + (x + 3) + (x + 4)
5x + 10 = 100
5x + 10 - 10 = 100 - 10
5x = 90

x = 18

[(3m + 10)/4] + 6 = 18

[(3m + 10)/4] + 6 − 6 = 18 − 6
[(3m + 10)/4] = 12
4( [(3m + 10)/4] ) = 12(4)
3m + 10 = 48
3m + 10 - 10 = 48 - 10
3m = 38

m = [38/3]

− 8 = [(5 − c)/2] + 10

− 8 − 10 = [(5 − c)/2] + 10 − 10
− 18 = [(5 − c)/2]
2( − 18) = ( [(5 − c)/2] )2
- 36 = 5 - c
- 36 - 5 = 5 - c - 5
- 41 = - c

41 = c

[(6y − 3)/10] − 12 = 36

[(6y − 3)/10] − 12 + 12 = 36 + 12
[(6y − 3)/10] = 48
10( [(6y − 3)/10] ) = 48(10)
6y - 3 = 480
6y - 3 + 3 = 480 + 3
6y = 483

y = 80.5

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*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Solving Linear Equations in One Variable

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Objectives

Solving Linear Equations in One Variable

Solving Linear Equations in One Variable Cont.

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Intro 0:00
Objectives 0:08
Solving Linear Equations in One Variable 0:34

Conditional Cases
Identity Cases
Contradiction Cases

Solving Linear Equations in One Variable Cont. 2:00

Addition Property of Equality
Multiplication Property of Equality
Steps to Solve Linear Equations

Example 1 4:22
Example 2 8:21
Example 3 12:32
Example 4 14:19
Example 5 17:25
Example 6 22:17

Algebra 1 Online Course

Section 1: Properties of Real Numbers
	Basic Types of Numbers	30:41
	Operations on Numbers	19:26
	Order of Operations	12:06
	Properties of Real Numbers	18:52
Section 2: Linear Equations
	The Vocabulary of Linear Equations	12:22
	Solving Linear Equations in One Variable	28:52
	Solving Formulas	12:02
	Applications of Linear Equations	28:41
	Applications of Linear Equations, Motion & Mixtures	24:26
Section 3: Graphing
	Rectangular Coordinate System	22:55
	Slope & Graphing	27:58
	Linear Equations in Two Variables	20:36
Section 4: Functions
	Introduction to Functions	21:24
	Graphing Functions	16:12
Section 5: Systems of Linear Equations
	Systems of Linear Equations	25:54
	Solving a System Using Substitution	20:01
	Solving a System Using Elimination	19:40
	Applications of Systems of Equations	24:34
Section 6: Inequalities
	Solving Linear Inequalities in One Variable	17:13
	Compound Inequalities	16:13
	Solving Equations with Absolute Values	14:12
	Inequalities with Absolute Values	17:07
	Graphing Inequalities in Two Variables	15:33
	Systems of Inequalities	21:13
Section 7: Polynomials
	Integer Exponents	44:51
	Adding & Subtracting Polynomials	18:33
	Multiplying Polynomials	25:07
	Dividing Polynomials	44:56
Section 8: Factoring Polynomials
	Greatest Common Factor & Factor by Grouping	28:27
	Factoring Trinomials	21:44
	Factoring Trinomials Using the AC Method	30:09
	Special Factoring Techniques	30:14
Section 9: Quadratic Equations
	Solving Quadratic Equations by Factoring	23:38
	Solving Quadratic Equations	29:27
	Equations in Quadratic Form	16:47
	Quadratic Formulas & Applications	29:04
	Graphs of Quadratics	26:53
	Polynomial Inequalities	21:42
Section 10: Rational Equations
	Multiply & Divide Rational Expressions	26:41
	Adding & Subtracting Rational Expressions	20:24
	Complex Fractions	18:23
	Solving Rational Equations	16:24
	Rational Inequalities	18:54
	Applications of Rational Expressions	20:20
	Variation & Proportion	27:04
Section 11: Radical Equations
	Rational Exponents	14:32
	Simplify Rational Exponents	15:12
	Adding & Subtracting Radicals	17:22
	Multiply & Divide Radicals	19:24
	Solving Radical Equations	15:05
	Complex Numbers	29:16

Transcription: Solving Linear Equations in One Variable

Welcome back to www.educator.com.0000

In this lesson we will look at more of the nuts and bolts of solving linear equations.0002

Specifically some of the things that we will actually do is we will look at types of solutions.0010

What does it mean to actually have a solution versus when it is consistent or inconsistent?0015

The things that will make this handy is how you can also deal with fractions when they are in the solving process.0023

It is something that my students always ask me.0031

What a solution means is a value that when you substitute it in for the variable, it makes the equation true.0039

We are looking for that magic value that would make the whole thing true.0046

In some cases, only one value would end up making the whole thing true.0051

If you only have one value that will do that or sometimes only two or three, we call this conditional.0057

The condition is that this variable must be that value.0064

There are other cases where we actually end up where any value will do.0069

It does not really matter what the value is as soon as you substitute it in for the variable it will make it a true statement.0075

If any value will do we say that the equation is an identity.0083

Lastly, there are very few types of equations that no matter what you try and substitute in for that variable, nothing seems to work.0089

No matter what that value is if nothing works then we say that that equation is a contradiction.0100

I have hidden a couple of those into the examples later on so watch for the one that is an identity and one that is a contradiction0106

and how we actually pick that out of just one that actually have a solution.0116

How do you exactly solve a linear equation?0123

You will use a few properties to help you out in the solving process.0126

One of them is your addition property of equality.0130

It is a handy property but what it says is that you can add the same amount to both sides of the equation.0133

In fact, many people like to look at this as almost like a balance scale.0140

Whatever you add to one side of your equation then you better be sure to add it to the other side as well to make sure that it all balances out.0146

If you are going to add + 4 over here, make sure you also add + 4 to the other side.0155

There is also the multiplication property of equality.0163

For this one you are allowed to multiply both sides by the same value and it will keep the equation the same.0168

You do have to be a little bit more careful with that property because it works as long as you do not multiply both sides by 0.0175

But you can use any other number you want.0184

You can multiply both sides by 3, multiply both sides by (x) as long as (x) is not 0, it will keep that equation the same.0186

One thing that I like to do when solving equation is to think of this process,0197

it often help with fractions as well as to get the variable that we are looking for all alone.0203

The very first thing that I like to do is scan it over and see if it does contain any fractions0209

and immediately clear them out if I can using some sort of common denominator.0213

After I have done that, I like to simplify each side of the equation as much as possible before shifting things to one side or the next.0217

We will simplify each side as much as possible first.0225

After we simplify both sides, we will try and isolate the variable on one side of the equation.0230

It may be difficult to do especially if there is more than one copy of that variable but if we can get them together usually we can isolate it just fine.0236

Lastly, it is always a good idea to check the solution just to make sure it actually works in the original problem.0244

That is always a good thing to do in case we make a mistake in one of these earlier steps.0251

We will definitely be able to find that out by the four steps when we check it by substituting it back into the equation.0256

Let us get into the examples and see how the solving process actually works.0264

I want solve (2 × x) +3 = 4x – 8, our goal is to figure out what the value for x is that would make this entire thing true.0270

In order to help this process out, I can see that it does not have any fractions, I’m going to try and simplify both sides and make it as compact as possible.0281

This means I'm going to use my distributive property on the left so they do not have my x inside parentheses 2x + 6.0291

Looking at both sides of the equation now, there is not much I can simplify if I was looking at just left or if I was looking at just the right.0310

What I want to do is to try and isolate my variable, in other words try and get it all alone.0320

Since I have the next on the left and on the right, I have to work on getting these together first.0326

In order to get them together, I will use my addition property of equality to add the same thing to both sides.0333

What I will add is -2x, we will put that on both sides and see the results.0340

2x and -2x, those two would cancel each other out and get rid of each other and just be left with 6.0351

On the right side, I have 4x - 2x so only 2x left – 8.0358

That is good, I worked out that way because now I only have one x to deal with and I can work on isolating that and getting it all by itself.0368

How am I going to do that? I better move this into the other side by adding 8 to both sides.0377

6 + 8 that will give me 14 and -8 + 8 will cancel each other out and be gone.0384

I have 14 = 2x, we just divide both sides by 2, I have 7 = x.0400

It looks like we have found a solution something that will make our equation true.0415

My question is does it actually work or not?0420

On to that last step, the one we actually see if this is the solution and we do that by substituting it back into the original problem.0424

I’m going to write the original problem, I’m going to write it all except for those x’s.0431

I’m going to take that value and put it in wherever I saw an x, let us see.0443

I had an x + 3 put 7 there and 4 × x now 4 × 7.0451

I'm doing here is simplify both sides of our equation now, let us see if it balances out.0458

2 × 7 + 3 is 10, 2 ×10 that would be 20, I’m not sure if these are equal, if we continue we find out.0465

4 × 7 is 28, 28 - 8 is 20, sure enough it looks like the value of 7 does make this equation true.0479

I know that 7 = x is my solution.0494

Let us try another this one, 3/4x – 1 = 7/5.0498

This one contains fractions so remember how I suggested taking care of them and you do not have to worry about quite as much.0507

We are going to try and find a common denominator that we can just multiply it by and get rid of our fractions.0515

Looking at our denominator, I have 4 and 5.0522

A common denominator in this case would be 20.0529

I’m going to multiply both sides by that 20, 20 and 20.0541

Let us write down the rest of our equation as well, 3/4x -1 and 7/5.0555

Sometimes it is no fun to deal with numbers like 20 but watch what it does when I start distributing it on the left side here,0567

and we will multiply the 20 and 7/5 together.0573

20 × 3/4 would be a 15x and 20 × 1 being – 20.0578

We do not have to deal with fractions on the outside of the equation anymore.0591

Perfect, I like it.0593

20 × 7/5 if we want to cancel out 5 from there, when you are looking at 4 × 7 = 28.0595

I just have to solve 15x – 20=28, let us work on getting an x all by itself by isolating it.0608

I will add 20 to both sides giving us 15x equal to 48 and then we will divide both sides by 15.0622

This will give us that x is equal to 48/15 but if we want we can even reduce that a little bit further.0650

I think 3 goes into the top and into the bottom.0657

This will use 16/5, not bad.0665

You have x equals 16/5, now get in front of that last step.0672

Does it equal to 16/5? Does it not? Let us find out by playing it back into the original equation.0676

I’m just testing it out and see if it holds true.0682

The original equation is ¾ and I have that x, that is where we will put our number -1 and we will see if it is equals 7/5.0686

Let us see our number was 16/5, check and see what happens.0700

I can take out a 4 from the 16 so this would be 12/5 – 1, it will give me a common denominator of 5.0707

I would have 12/5 – 5/5 and how do we know? That is equal to 7/5.0734

This one definitely checks out, I do know that 16/5 is my solution, it looks good.0743

We will have something like this one, more fractions here and that is good.0755

4 + 5x = (5 x 3) + x and see what we can do this one.0758

Let us simplify both sides of the equation first and then go from there.0765

I'm going to distribute my 5 into the parentheses.0769

4 + 5x = 15 + 5x, it looks pretty good.0775

I want to get my x’s together and hopefully isolate it.0786

We will subtract 5x from both sides, 4 = 15.0791

Look at what happened there, when I subtract 5x from both sides, I lost all of my x’s completely.0805

Even worse than that than I thought I have left over this 4 -15 that is not true.0812

That does not make any sense, 4 does not equal 15.0819

What I was left here is known as a false statement.0823

What this is telling me is that since I did not make any mistakes that you do not know the value for x is going to work.0828

This is an example of one of those situations where we have a contradiction.0837

No matter what value for x you are trying use, it simply not going to work in this equation.0849

In fact, no solution exists for it.0854

Let us look at another one and see if we have any better luck.0860

In this one is -x + 3 = (1 + 4) - 2x/2.0863

We have a fraction, I'm going to try and get rid of those fractions first by multiplying everything through by a 2.0869

I will multiply it on the left side and I will multiply it on the right side, both by 2.0877

Just to keep things nice and balanced.0886

-x + 3, 1 + 4 - 2x / 2.0890

We multiplied both sides by 2, we can go ahead and distribute it.0901

Let us see what the result will be.0907

-2x + 6 equals, I have 2 + and when it distributes on this second part here, it is going to get rid of that 2 in the bottom, 4 - 2x.0911

Looking good, let us continue trying to combine our terms specifically those x’s and the numbers.0933

I will add 2x to both sides and you will get 6 = 2 + 4.0943

You will notice in this one that we lost our x’s but this one is a little different.0958

Instead of being left with other nonsense, this one we are left with 6 = 6 which is actually something that is true.0963

Before you get too worried, of course check all the steps and make sure they are all correct, which they are.0977

What this is trying to indicate here is that we have an identity.0982

It does not matter the value of x, you can use any value and the equation is still going to ring true.0985

Let us mark this one as an identity.0992

In short, that is a great way that we can identify both of those situations.0999

If it has an actual solution, it is a conditional equation.1003

You usually go through the solving process and you will be able to figure out what that value is and able to test it and see if it actually works.1007

If it is a contradiction, so nothing works, then you will go through the solving process and usually your variable drops away entirely,1014

and what you are left with is a false statement, it does not make sense.1022

That is even after all of your steps are completely correct.1026

If you have an identity like this one then you will go through that solving process and you will be left with a true statement1029

which indicates that you could use any value want for x.1036

Let us try some more examples.1042

In this next one I have 1/3x – 5/12 = ¾ +1/2 x.1046

There is a lot of fractions in here, let us see if we can take care of them all by getting some sort of common denominator.1053

What would 3/12, 4, 2 go into?1062

I think our common denominator would have to be a 12.1067

We are going to multiply both sides by 12.1072

Let us write everything we got here, 1/3x – 5/12 and I have ¾ + 1/2x.1084

Then we are going to distribute through on both sides and see what we get.1100

1/3 of 12 would be 4x, 12 × 5/12 would give us 5, then I will have 12 × 3/4.1108

That goes in there 9 times, then (12 × ½) + 6x.1124

Notice that common denominator, we clear out all those fractions then you do not have to worry about them.1134

What we have left here is 4x - 5 = 9 + 6x.1139

If you want to continue trying to simplify this by getting the x’s together and getting them isolated.1144

Our x over here and x over here let us get them together by subtracting 4x from both sides, -5 = 9.1153

6x – 4x that would be 2x, that looks good.1170

I only have one x to deal with.1177

Let us go ahead and subtract 9 from both sides, -9 -9 -14 = 2x.1180

There is only one last thing to do, divide both sides by 2.1194

It looks like our solution is -7 or we are not too sure until we check it.1200

Let us substitute it back into the equation and see what happens.1211

I have 1/3 - 5/12 = ¾ + ½ and into all of those blank spots where we are used to have the x, let us go ahead and put that -7 in there.1215

Let us see if these things are true or not, also -7 × 1/3, that is -7/3 – 5/12, is that equal to ¾ - 87/2?1237

I do not know, we got to do a lot more simplifying before we can figure that out.1259

We will use our common denominator to help us out, the common denominator of 12.1263

Let us see if I need a common denominator over there that would be -28/12.1270

A common denominator of 12 on the other side would be 9/12 minus, top and bottom by 640, 2/12.1279

Let us see what we can do, -28 - 5 would be -33/12.1294

What is going on the other side? 9/12 – 42.1306

It looks like I have another -33/12.1310

They are exactly the same and what that means is that x does equal in -7.1314

You can see when you go through that process of checking it,1322

it can be a lot of work especially if you have lots of fractions and stuff that you are dealing with.1324

But it is a good idea just to make sure that the solution you come up with is the solution.1329

Let us do one more.1338

Here I have (3x -15)/4 + (x + 47)/7 = 11.1339

This one involves fractions, let us take care of them out of the gate by multiplying it by a common denominator, let us use 28.1347

A lot to keep track of in here, (3x -15)/4 + (x + 47)/7 = 28 × 11.1364

We will take that 28, we will actually distribute it to both parts on this left side.1379

We will take it to this giant fraction here and I will take it to this giant fraction over here.1384

What this is going to do is, it is going to allow us to cancel out those fractions in the bottom.1388

28/4 would give us 7, that is still going to be multiplied by that 3x -15 part.1394

When it distributes on the second piece, 28/7 reduces and that goes in there four times.1404

We still have a 4 multiplied by x + 47.1411

Over on the side we have 28 × 11 that is 308, some fairly big numbers.1417

Let us continue distributing and see what we can do.1426

Let us take this 3 × 7, we will also take the 15 × 7 and do the same thing with 4 on this side.1429

Hopefully free of those x’s and get them out of parentheses.1440

21x - (7 × 15) = 105 + 4x.1444

Now I have 4 × 47 = 188, again, some big numbers but just have to push on through.1461

I have more than one copy of x here let us get those guys together.1476

I also have some things that do not have an x, let us get those together as well.1480

21 + 4x would be 25x - 105 and 188, I think that would give us 83.1487

I only have a single x to deal with so I will simply work on getting that isolated.1508

Let us subtract 83 from both sides so that the 25x is the only thing on the left side.1516

308 – 83 will give us 225, almost done.1529

Let us go ahead and divide both sides by 25 now.1539

This will give us x = 9.1549

This one is a lengthy process to get all the way down to just x = 9.1554

Even when you get that far, go ahead and double check it just to make sure that it actually works out.1558

I have my (3 × x -15)/4 on one side and both in those blank spots, let us go ahead and put that 9.1571

Let us start to simplify and see what we will get.1589

3 × 9 = 27/4, I see a 9 + 47 = 56.1592

We are hoping that this will equal 11.1610

27 - 15 that is 12/4 that is 56/7, we are getting closer.1614

12/4 is 3, 56/7 let us see that goes in their 8 ×, 3 + 8 =11.1625

Sure enough this one checks out.1639

I know that x does equal 9.1642

When going through the solving process, you do have a lot to keep track of.1648

The most important thing is truly working getting those x’s all by themselves and isolate it so that they are the only thing on one side of the equation.1652

If you have to deal with fractions, use a common denominator to clear them all out.1660

And also definitely go back and check your solution to make sure it is a solution.1664

Thank you for watching www.educator.com1671

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Eric Smith

Solving Linear Equations in One Variable

Slide Duration:

Table of Contents

Section 1: Properties of Real Numbers

Basic Types of Numbers

30m 41s

Intro