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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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For more information, please see full course syllabus of Algebra 1
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Lecture Comments (8)

0 answers

Post by Rose A on February 22 at 09:07:56 AM

My advice, dont bother with the practice questions... Do that on a seperate website or book...

0 answers

Post by Farhat Muruwat on March 24, 2014

Whoever typing the practice questions are placing the equations VERY confusing.

Here is an example:

Q. 5m + 6n = − 299m − n = 6

When in fact it should be on SEPARATE LINES.

Q. 5m + 6n = − 29
  9m − n = 6

3 answers

Last reply by: Professor Eric Smith
Mon Jan 5, 2015 12:10 PM

Post by misrak taye on March 16, 2014

you are right 20 + -6 = 14

0 answers

Post by misrak taye on March 16, 2014

-6+-20=-14 you are joking

0 answers

Post by Delores Sapp on February 10, 2014

concepts presented well -- a  grammar hint. number of solutions rather than amount of solutions

Solving a System Using Substitution

  • To solve a system using substitution
    • Use one equation to isolate a variable.
    • Substitute this variable into the second equation
    • Solve the new equation created
    • Use back-substitution in one of the original equations
    • Solve this equation for the other variable
    • Check you solution by substituting both values into the system to see if it makes all equations true.
  • If the system has only one solution, the method will find the x and y values that work.
  • If the system has no solutions, then the substitution method will create a false statement.
  • If the system has an infinite amount of solutions, then the substitution method will create a true statement. All points on the line are considered a solution to the system.

Solving a System Using Substitution

j = 6 - 11k
3j - 7k = 92
  • 3(6 − 11k) − 7k = 92
  • 18 − 33k − 7k = 92
  • − 40k = 74
  • k = − [74/( − 40)]
k = − [37/20]
5m + 6n = - 29
9m - n = 6
  • − n = 6 − 9m
  • n = − 6 + 9m
  • 5m + 6( − 6 − 9m) = − 29
  • 5m − 36 − 54m = − 29
  • − 49m = 7
  • m = - [1/7]
  • 9 ( − [1/7] ) − n = 6
  • − [9/7] − n = 6
  • − n = 7[2/7]
n = − 7[2/7]
m = − [1/7]
8a + 2b = 10
4a - 6b = - 23
  • Let's start by solving for b in the first equation: 8a + 2b = 10
  • 2b = 10 − 8a
  • b = 5 − 4a
  • Plug b into the other equation 4a − 6(5 − 4a) = − 23
  • 4a − 30 + 24a = − 23
  • 28a = 7
  • a = [7/28] = [1/4]
  • 4g( [1/4] ) − 6b = − 23
  • 1 − 6b = − 23
  • − 6b = − 24
  • b = 4
a = [1/4]
x = 4y + 9
5x + 3y = 14
  • 5(4y + 9) + 3y = 14
  • 20y + 45 + 3y = 14
  • 23y + 45 = 14
  • 23y = − 31
  • y = − [31/23]
  • x = 4y + 9x = 4( − [31/23] ) + 9
  • x = − [124/23] + 9
  • x = − 115[9/23] + 9
x = - 106[9/23]
x = 5y + 21
7x - 3y = 16
  • 7(5y + 21) − 3y = 16
  • 35y + 147 − 3y = 16
  • 32y + 147 = 16
  • 32y = − 131
  • y = − [131/32]
  • x = 5( − [131/32] ) + 21
  • x = − 20[15/32] + 21
x = [17/32]
x = 4y - 7
5x - 3y = 15
  • 5(4y − 7) − 3y = 15
  • 20y − 35 − 3y = 15
  • 17y − 35 = 15
  • 17y = 50
  • y = [50/17] = 2[16/17]
  • x = 4( [50/17] ) − 7
  • x = [200/17] − 7
x = 4[13/17]
x - 3y = 6
2x - 5y = 10
  • x = 6 + 3y
  • 2(6 + 3y) − 5y = 10
  • 12 + 6y − 5y = 10
  • 12 + y = 10
  • y = − 2
  • x = 6 + 3yx = 6 + 3( − 2)
  • x = 6 − 6
x = 0
5x - 6y = 12
x - 2y = 8
  • x = 8 + 2y
  • 5(8 + 2y) − 6y = 12
  • 40 + 10y − 6y = 12
  • 40 + 4y = 12
  • 4y = − 28
  • y = − 7
  • x = 8 + 2yx = 8 + 2( − 7)
  • x = 8 − 14
x = − 6
24x - 4y = 60
3x - y = 12
  • − y = − 3x + 12
  • y = 3x − 12
  • 24x − 4(3x − 12) = 60
  • 24x − 12x + 48 = 60
  • 12x + 48 = 60
  • 12x = 12
  • x = 2
  • y = 3x − 12y = 3(2) − 12
  • y = 6 − 12
y = − 6
8x - 6y = 14
3x + 3y = 9
  • 3x + 3y = 9
  • 3x = 9 − 3y
  • x = [(9 − 3y)/3]
  • x = 3 − y
  • 8x − 6y = 148(3 − y) − 6y = 14
  • 24 − 8y − 6y = 14
  • 24 − 14y = 14
  • − 14y = − 10
  • y = [( − 10)/( − 14)] = [5/7]
  • x = 3 − yx = 3 − [5/7]
x = 2[2/7]

*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.


Solving a System Using Substitution

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 0:00
  • Objectives 0:09
  • Solving a System Using Substitution 0:32
    • Substitution Method
    • Substitution Example
    • One Solution, No Solution, or Infinite Solutions
  • Example 1 9:45
  • Example 2 12:48
  • Example 3 15:01
  • Example 4 17:30

Transcription: Solving a System Using Substitution

Welcome to www.educator.com.0000

In this lesson, we are going to work on solving a system of equations using the substitution method.0002

What we want to focus on is how we use the substitution method in order to find solutions and I will walk into that process.0013

We will also keep in mind that since sometimes there is no solution or there is an infinite amount of solutions,0021

how we can recognize those cases when using the substitution method.0026

We are called the reason why we need more methods for solving a system is that when we use the graphing method,0035

you have to be very accurate in order for that method to work and if your lines are a little bit wavy0042

or you do not make them just right then you may not find a solution of that system.0048

Since, accuracy is such a problem that is why we are going to focus more on those algebraic methods.0054

How can we manipulate the system in such a way so that it gives us a solution.0060

Two of these methods of our algebraic methods are substitution and elimination.0067

In this lesson, we focus on substitution and then look at elimination in the next one.0073

Let us see how the substitution method works.0080

In the substitution method, what we first end up doing is taking one of our equations and solving it for one of the variables.0086

Once we have taken one equation and we have solved it for variable, we will take that and substitute it into the other equation.0095

What this will do is it will give us an entirely new equation with only one type of variable in it0105

and will end up solving for that one remaining variable that is in there.0110

We will have half of our solution but since we are looking for a system and we need an x and y coordinate,0117

then we will end up using back substitution to find the rest of our solution.0123

We will take what we have found and end up substituting back into one of the original equations.0129

This will help us find the other variable.0135

As long as everything works out good, that it should be our solution, but it is not a bad idea to just take that and check it in the original system.0138

If you do make a mistake, sometimes you would not catch it, so playing it back into the original system is always a good idea.0147

Let us walk through this in substitution method using the following example.0157

I have - x + 3y = 10 and 2x + 8y = -6.0162

I want to start solving for at least one of the variables in one of the equations.0170

I have lots of different choices that I can do.0175

I can solve for x over here.0178

I can solve for y in the first one or you can even pick on the second equation a little bit and solve for its x or y.0180

It does not matter which one you solve for, just pick a variable and go ahead and solve for it.0187

I only choose x in the first equation, it looks like it will be one of the easier ones to get all by itself.0194

What I will do is solve for x and I'm moving everything to the other side.0200

I'm subtracting 3y on both sides.0209

It is almost all by itself but it looks like I have to multiply it by -1 to continue on from there.0213

I’m multiplying through by -1 and I will end up with x = 3 and y -10.0220

This entire thing right here is equal to x.0228

For the substitution method happens is I will take in the x in the other equation and replace it with what it is equal to, 3y – 10.0235

It will leave lots of room for this so you can see what it is going to turn that second equation into.0250

Instead of writing x, I’m going to write 3y -10.0258

I have taken that second equation, substituted in those values and notice how this new equation we created it only has y.0269

Since this new equation only has y’s in it, we will be able to solve for y and be able to figure out what that is.0279

We still have a little bit of work to do, of course we will have to simplify and combine terms but we will definitely be able to solve for that y.0287

Let us give it a shot.0293

I’m going to distribute through by my 2 and put 6y – 20 + 8y = -6.0295

Let us go ahead and combine our y terms, so 6 + 8 is a 14y and then let us go ahead and add 20 to both sides.0307

14y – 6 + 20 is 14 divide both sides by 14 and get y = 1.0319

It looks like I have a solution but remember that this is only half of our solution.0330

We still need to figure out what x is.0333

What this point, now that we know what y is I can substitute it back into one of my equations and figure that out.0336

Let us do that, -x + 3 and I will put in what y is equal to 1.0346

I will end up solving this equation for x, -x + 3 = 10, so I'm going to subtract 3 from both sides.0355

I will go ahead and multiply both sides by -1 and get x = -7.0372

I have both halves of my solution.0383

The x is written first, so -7 and y is written second.0387

My solution is -7, 1.0393

If I'm looking at that and I'm curious if it is the solution or not, it is not a bad idea to just take it and plug it back into the original.0396

Let us see how that works out.0404

My value for x is -7 + 3 and my value for y is 1.0408

Does that really equal 10? Let us find out.0413

Also a- -7 would be 7 + 3 = 10, sure enough, 7 + 3 is 10.0416

That works out in the first equation.0427

Right now, it is still the same way with the second one.0430

(2 × -7) + (8 × 1) is that =equal to -6, let us find out.0432

-14 + 8 sure enough -6 is equal to -6, it checks out in both the equations.0440

I know that -7, 1 is my solution.0449

There is a lot of substitution, of course, in the substitution method, but is not as bad as you think.0454

You do have some choices on what you will solve for and where it goes.0460

One thing that we saw when we are finding solutions graphically is that several things could happen.0472

We might have one solution, we might have no solution or we actually might have an infinite amount of solutions.0477

Now when we are looking at all of these graphically, it was pretty simple on figuring out which case we were at.0483

We either looked at both our lines, actually cross and then we said they have a solution.0489

We could see if there was no solution if the lines were parallel and we can see that it was an infinite amount solution if they were on the same line.0495

Now that we are doing things more algebraically, we would not be able to visually see what case we are in.0503

How is it we figure out if it has no solution or an infinite amount of solutions?0509

It all depends on what values you will get while going through that solving process.0514

If you go through the substitution method and you can find an x and y that work, then you will know that it has a solution.0520

It has one solution, nothing strange happens.0526

If however you go through that substitution method and end up creating a false statement, it means that there is actually no solution to the system.0530

Now if you do come across a false statement, what I always tell everyone is check your work first to make sure that you have made no mistakes.0539

If all your work looks great and you still create a false statement then you can be sure that it has no solution.0548

If you go through and you create a true statement then you can not see what your solution is.0555

That is a good indication that the two lines are exactly the same and you have an infinite amount of solutions.0560

Again, check your work with this one, but if all your work looks good, then you know that you have an infinite amount.0566

Let us look at some examples like these ones and see how we can recognize what I mean by a false statement and what I mean by a true statement.0576

We want to solve the following system 5x - 4y = 9 and 3 - 2y = -x.0589

The first thing I need to do is solve for either x or y.0597

It does not matter which one we choose, it looks like it might be a good idea to solve for x down here since it is almost all by itself anyway.0601

We will just go ahead and multiply everything through by -1, -3 + 2y = x.0609

Now that we have what x is equal to, we will substitute it into the first equation for that x.0619

It will leave lots of space and we will just go ahead and put it in there, -3 + 2y.0627

We can see in this new equation it only has y, so I need to work on getting them together and isolating them.0636

-15 it will be from distributing through +10y -4y = 9.0645

It have some like terms I can put together 10y and -4 that will be 6y =9 and adding 15 to both sides, + 15 = 24.0656

Now dividing both sides by 6, I will get that y = 4.0680

We have half of our solution and we always figure out what the other half is by putting it back in one of the original equations.0685

I often have students ask me should you put it back in the original, after all could not I just stick that y back and up here.0692

What I tell them is yes you could put it back into this equation for y, but be very careful because that equation, we already have manipulated in some sort of way.0700

If you want to be sure that you are not making mistakes on top of mistakes, take that and plug it into of the original ones,0711

something that you have not mess around with too much.0718

Let us see 3 – 2 and I will put the 4 in there.0721

We take this and solve for x, 3 -8 = -x.0729

3 - 8 would be -5 and dividing both sides by negative, I have that x = 5.0737

I have both halves of my solution x = 5, y = 4.0747

My solution is 5, 4.0755

Again, if you want to make sure that is the solution, feel free to plug it back into both equations and check it to make sure it works out.0759

In this next example, we will see if we can solve this one using the substitution method.0769

It says 4x - 5y = -11 and x + 2y = 7.0773

Let us see what variable shall we solve for.0780

This x looks like it would be nice and easy to get all alone, let us do that one.0784

We will go ahead and say x = -2y + 7.0790

I took the 2y and subtracted it from both sides.0797

I will take all of this and substitute it into the first one.0801

It will give me 4 - 5y = -11.0805

I left a big open spot so I can drop what x is equal to in there, 2y + 7, it looks good.0812

Now I’m going to distribute through with my 4, -8y + 28.0820

And I'm going to continue trying to solve for y.0832

I will also see what I got here, -8 - 5 putting those together I will end up with -13, subtracting 28 from both sides, I will end up with -39.0835

It looks like I can finally do dividing both by it sides by -13, in doing that y = 3.0853

Let us go ahead and take that and substitute it back into one of our original equations.0863

Let us go ahead and put in here, x + (2 × y).0867

Multiplying together the 2 and the 3 would be 6 and I'm left with x + 6 = 7.0878

Subtracting 6 from both sides x = 1.0886

My solution for this one is x = 1 and y = 3.0892

In some systems, remember we could have that situation of having no solution or an infinite amount of solutions.0903

Sometimes when you are looking at an equation, it is tough to tell whether you have that situation or not.0909

Watch very carefully what happens with this one.0914

I’m going to start off like I what did with the other examples.0917

We just want to take these and solve it for one of our variables.0920

I’m going to solve the first one for x since that one looks like it can be nice and easy to get isolated.0924

That would give me x - 2y from both sides, + 4.0932

Now that I have x isolated, let us substitute it into the second equation.0940

3 + 6y = 13 and we will drop in -2y + 4.0947

Now I only have y in this new equation so we will continue solving it by getting y all by itself.0959

Let us go ahead and distribute -6y + 12, + 6y = 13.0965

There is something interesting happening here.0975

Over here I have -6y and over here I have 6y.0977

Since they are like terms, I can put them together by -6 and 6 would give me 0y, it cancel each other out.0982

That means the only thing left is 12 on the left and 13 on the right but that does not make any sense.0993

12 is not equal to 13 so I’m going to write not true.0999

If you check through back all of my steps, I did not make any errors.1008

I moved the -2y to the other side that turnout okay and I did my distribution fine.1012

I know my steps worked out perfectly good.1019

What this statement is telling me at the very end, since it is not true is that these two lines are actually parallel.1023

There is no solution and they do not cross.1030

Watch out for these false statements that might come along the way and they will help you recognize when a system does not have a solution.1039

One more system, this one is 2y = 4x and the other equation is 4x - 2y = 0.1051

Let us take the first one and solve it for y.1061

We can easily do it just by dividing both sides by 2, y = 2x1065

Now that it is solved for y, let us go ahead and substitute that into the second equation.1075

You will see that this is turning out a lot like the previous example.1091

If I combine my x’s, the other since they are like terms, they will cancel each other out.1097

This is slightly different though because after they do cancel out, all that I get is 0 = 0, which happens to be a true statement 0 does equals 0.1103

Let us check our work and make sure you did everything okay.1118

I divide both sides by 2 and that worked out good and I combined my 2’s together and cancelled.1122

All my work looks great.1127

The fact that I have a true statement and all my steps were good means that what is going on here is that the lines themselves are exactly the same line.1129

We might even see that at the very beginning if we end up rewriting them.1139

When I moved the 2y to the other side on the second equation and then end up rearranging them, they are exactly the same equation.1146

Visually these would be right on top of one another.1160

This tells us we have an infinite number of solutions and everywhere on that line is a solution.1164

These more algebraic methods are definitely more accurate when it comes to finding solutions.1179

You want to be very familiar and being able to go through them.1184

Be on the watch out for these two special cases, if you get a false statement then you know you have no solution.1188

If you get a true statement then know you have an infinite amount of solutions.1195

Thank you for watching www.educator.com.1199