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### Course: Operations and Algebraic Thinking 229+ > Unit 4

Lesson 3: Equivalent systems of equations- Why can we subtract one equation from the other in a system of equations?
- Worked example: equivalent systems of equations
- Worked example: non-equivalent systems of equations
- Reasoning with systems of equations
- Reasoning with systems of equations
- Equivalent systems of equations review

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# Worked example: non-equivalent systems of equations

Sal analyzes a couple of systems of equations and determines whether they have the same solution as a third given system.

## Want to join the conversation?

- At1:51, how did you get y = 2x - 1?(7 votes)
- Sal started with the equation: 14x - 7y = 7

He then subtracted 14x from both sides, creating: -7y = -14x + 7

Then, he divide the equation by -7. I'll write it out so you can see the steps:

-7y/(-7) = -14x/(-7) + 7/(-7)

y = 2x - 1

Hope this helps.(16 votes)

- wouldn't hansel's 2nd equation be equivalent to the teacher's 2nd equation divided by -7?(4 votes)
- Yes, Hansol's 2nd equation is equivalent to the teacher's 2nd equation. However, Hansol's first equation is not equivalent to the teachers. So, the system is not equivalent.(7 votes)

- How do we graph non linear equations and solve.(5 votes)
- - Scarlett and Hansol's teacher gave them a system of linear equations to solve. They each took a few steps that lead to the systems shown in the table below. So this is the teacher system. This is what Scarlett got after taking some steps. This is what Hansol got. Which of them obtained a system that is equivalent to the teacher's system? And just to remind ourselves, an equivalent system is a system that has, or at least for our purposes, is a system that has the same solution, or the same solution set. So if there is a certain xy that satisfies this system in order for Scarlett's system to be equivalent it needs to have the same solution. So let's look at this. So Scarlett, let's see, let's see if we can match these up. So her second equation here, so this is interesting, her second equation 14x - 7y = 2 over here the teacher has an equation 14x - 7y = 7. So this is interesting because the ratio between x and y is the same, but then your constant term, the constant term is going to be different. And I would make the claim that this alone tells you that Scarlett's system is not equivalent to the teacher. And you're saying, well, how can I say that? Well, these two equations if you were to write them into slope intercept form, you would see because the ratio between x and y, the x and y terms is the same. You're going to have the same slope, but you're going to have different Y intercepts. In fact, we can actually solve for that. So this equation right over here we can write it as if we, let's see, if we subtract 14x from both sides you get -7y - 14, woops, -7y =, is equal to -14x + 7 and we could divide both sides by -7. You get y = 2x -1 so that's this... All I did is algebraically manipulate this. This is this line and I could even try to graph it so let's do that. So I'll draw a quick coordinate. This is just going to be very rough. Quick coordinate axis right over there, and then this line, this line would look something like this. So its y intercept is -1 and it has a slope of 2. So let me draw a line with a slope of, a line with a slope of 2 might look something like that. So that's this line right over here or this one right over there, and let's see. This one over here is going to be, if we do the same algebra, we're going to have - 7y = -14x + 2, or y = , I'm just dividing everything by -7, 2x - 2/7 so this is going to look something like this. Its y intercept is -2/7 so it's like right over there. So this line is going to look something like, I'm going to draw my best, my best attempt at drawing it, it's going to look something... Actually that's not quite right. It's going to look something like... I'll actually just start it right over here. It's going to look something like, something like this. It's going to have the same slope, and obviously it goes in this direction as well. Actually, let me just draw that. So it's going to have the same slope, but at different y intercepts. That doesn't look right, but you get the idea. These two lines are parallel. So these two lines are parallel so any coordinate that satisfies this one is not going to satisfy this one. They have no points in common. They are parallel. That's the definition of parallel. Since this and this have no points in common, there's no way that some solution set that satisfies this would satisfy this 'cause any xy that satisfies this can't satisfy this or vice versa. They're parallel. There are no points. These two things will never intersect. So Scarlett does not have an equivalent system. Now what about Hansol? Well, we see Hansol has the same thing going on here. 5x - y, 5x - y, but then the constant term is different, -6, positive 3. So this and this also represent parallel lines. Any xy pair that satisfies this, there's no way that it's going to satisfy this. These two lines don't intersect. They are parallel. So Hansol's system is not equivalent either(1 vote)

- "
*the ratio*" at

between x and y is the same,

but then your constant term,

the constant term is going to be different.

And I would make the claim

that this alone tells you that Scarlett's**system**

is not equivalent to the teacher.0:54

I suspect this is not valid reasoning. I guess the correct statement is:

"*this alone tells you that Scarlett's*"**2nd equation**

is not equivalent to teacher's**2nd equation**.

Am I right?(5 votes) - So if scarlett's system of equation is NOT equivalent to the teacher's system of equation, How would Sal know that Hansol's system of equation is NOT equivalent to the teacher's system of equation if he didn't graph them like what he did in Scarlett's?(4 votes)
- Hansol had the same problem that Scarlett did, being that the lines were parallel. The way you can tell is that both Scarlett's and Hansol's right hand side was different from the teachers right hand side, while the left hand sides were the same. The mistake that they both made is that they forgot the golden rule of algebra, which is whatever you do to one side of an equation, you MUST do to the other side.(2 votes)

- What's the difference between substitution and elimination for the systems of equations?(3 votes)
- A substitute teacher fills in for the regular teacher, so substitution gets rid of a variable by filling in a value from one equation into the other equation. Elimination means cancelling out a variable, so this gets rid of a variable by additive inverses to cancel a variable.(3 votes)

- At1:05-1:11, sal said "I would make a claim that
**this alone**tells you that Scarlett's system of equations is NOT equivalent to the teacher?" What does Sal mean when he said "*this alone*"?(2 votes)- To have an equivalent system, you must have equivalent equation. Sal points out on the 2nd equations that the left sides match, but the right sides do not match. So, the equations can't be equivalent. The phrase "this alone" is referring to the fact that the 2 sides of the equations do not match. It is basically all the info you need to determine that the system is not equivalent to the original.

Hope this helps.(3 votes)

- I have seen a system like this in the practice exercise for the Teacher:

14x-7y=-28

-8x+4y=15

If I try to apply the elimination method I finally get 0 = -1.75. The question was if the equations of the other students are equivalent and one of them actually had the same weird result so they were equivalent. But I presume the exercise is wrong cause that equivalence doesn't make any sense.(1 vote)- Good observation. Let us rearrange our equations and see what happens:

𝑦 = 2𝑥 + 4

𝑦 = 2𝑥 + (15/4)

Notice that these are both linear equations meaning they graph lines. Also note that they both have a slope of 2. These means that both lines are parallel. However, both lines have different 𝑦-intercepts. Combining these two facts, we know that the two lines*never*intersect. This means for this system of equations, there are*no solutions*which explains the nonsensical results you were getting. Comment if you have any questions.(4 votes)

- Is there a way other than manipulating equations from which we can find out whether the equations are equivalent?(2 votes)
- you could substitute multiple points that you know is on one of the equations and see if it is valid for the other one also (which would be the same as creating tables with common x's).(2 votes)

- at1:50, why is it ok to divide x term from y term? contradicts all I have learned so far lol(2 votes)

## Video transcript

- Scarlett and Hansol's teacher gave them a system of linear equations to solve. They each took a few steps that lead to the systems shown in the table below. So this is the teacher system. This is what Scarlett got
after taking some steps. This is what Hansol got. Which of them obtained a system that is equivalent to
the teacher's system? And just to remind ourselves, an equivalent system is a system that has, or at least for our purposes, is a system that has the same solution, or the same solution set. So if there is a certain xy that satisfies this system in order for Scarlett's
system to be equivalent it needs to have the same solution. So let's look at this. So Scarlett, let's see, let's see if we can match these up. So her second equation here, so this is interesting, her second equation 14x - 7y = 2 over here the teacher has an equation 14x - 7y = 7. So this is interesting because the ratio between x and y is the same, but then your constant term, the constant term is
going to be different. And I would make the claim that this alone tells you
that Scarlett's system is not equivalent to the teacher. And you're saying, well,
how can I say that? Well, these two equations if you were to write them into slope intercept form, you would see because the ratio between x and y, the x
and y terms is the same. You're going to have the same slope, but you're going to have
different Y intercepts. In fact, we can actually solve for that. So this equation right over here we can write it as if we, let's see, if we subtract 14x from both sides you get -7y - 14, woops, -7y =, is equal to -14x + 7 and we could divide both sides by -7. You get y = 2x -1 so that's this... All I did is algebraically
manipulate this. This is this line and I could even try to graph it so let's do that. So I'll draw a quick coordinate. This is just going to be very rough. Quick coordinate axis right over there, and then this line, this line would look something like this. So its y intercept is -1
and it has a slope of 2. So let me draw a line with a slope of, a line with a slope of 2 might look something like that. So that's this line right over here or this one right over there, and let's see. This one over here is going to be, if we do the same algebra, we're going to have - 7y = -14x + 2, or y = , I'm just
dividing everything by -7, 2x - 2/7 so this is going to look
something like this. Its y intercept is -2/7 so
it's like right over there. So this line is going
to look something like, I'm going to draw my best, my best attempt at drawing it, it's going to look something... Actually that's not quite right. It's going to look something like... I'll actually just start
it right over here. It's going to look something like, something like this. It's going to have the same slope, and obviously it goes in
this direction as well. Actually, let me just draw that. So it's going to have the same slope, but at different y intercepts. That doesn't look right,
but you get the idea. These two lines are parallel. So these two lines are parallel so any coordinate that satisfies this one is not going to satisfy this one. They have no points in common. They are parallel. That's the definition of parallel. Since this and this have
no points in common, there's no way that some solution set that satisfies this would satisfy this 'cause any xy that satisfies this can't satisfy this or vice versa. They're parallel. There are no points. These two things will never intersect. So Scarlett does not have
an equivalent system. Now what about Hansol? Well, we see Hansol has the same thing going on here. 5x - y, 5x - y, but then the constant term is different, -6, positive 3. So this and this also
represent parallel lines. Any xy pair that satisfies this, there's no way that it's
going to satisfy this. These two lines don't intersect. They are parallel. So Hansol's system is
not equivalent either.