WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to work on solving a system of linear equations using the method of elimination.
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We focus on how to use this elimination method or how it differs from substitution
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and how we can recognize the many different types of solutions that could happen.
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Do we get one solution, and infinite amount of solutions, or possibly no solution or whatsoever.
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The goal with the elimination method is to combine the equations in such a way that we eliminate one of our variables.
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In that way, we will only have one that we will need to solve for.
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In order to do this, sometimes we will have to multiply one of the equations or even both of the equations by some sort of a constant.
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Now what we are looking to do when we multiply is we want a pair of coefficients that will end up canceling each other out.
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Do I have - 3x and 3x?
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When we combine those they would end up canceling out and that is what I want.
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Once I get a pair of coefficients that I know will cancel, we will go ahead and actually add the 2 equations together.
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In fact, some people call the elimination method the addition method because we end up adding the equations.
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We will get a new equation after doing that and we will only have one type of variable in it.
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We will end up solving that new equation.
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We will have half of our solution at this point so we will be able to take that and substitute it back into one of the originals.
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In fact, if you look at all the rest of stuffs from here on out, these are the same as the substitution method.
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We simply have half of our solution and we are back substituting so we can find the other value.
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Once we are all the way done through this entire process, it is not a bad idea to check the solution to make sure that does satisfy both of the equations.
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Let us walk through this method, so you get a better sense of how it works.
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Here I have 2x – 7y = 2 and 3x + y = -20.
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What my goal is to either eliminate these x values here or I need to eliminate my y values.
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If I was just going through and I decide that I was going to add them together as they are,
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you will see that this would not be a good idea because nothing happens.
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2x + 3x would give me 5x and – 7y + y would give me -6y.
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2 and -20 is -18, so I definitely created a new equation, but it still has x’s and it still has y’s.
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There is not much that I can do with it.
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What I want to happen is for some things to cancel out, so I only have one type of variable to solve for.
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Let us go ahead and a do a little work on this so that we can get something to cancel out.
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The first thing is we get to choose what we want to cancel out.
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Is it going to be the x's or y’s?
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I want to choose my y since one of them is already negative and the other one is positive.
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In order to make these guys came flat, let us take everything in the second equation and multiply it by 7.
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The first equation exactly the way it is, no changes, and everything in the second equation will get multiplied by 7.
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3 × 7 = 21x, I have 7y, -20 × 7 would be -140.
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I still have a system and I still have a pair but now I noticed what is going on here with the y.
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Since 1 is -7 and one is 7, when I add those together, we will end up canceling each other out.
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Let us do that, let us go ahead and get to the addition part of the elimination method.
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Adding up our x's, we will get 23x.
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When we add up the y, we will get 0y so that term is gone.
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Over on the other side - 138 and in this new equation the only thing I have in here is simply an x.
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I can solve this new equation for x and figure out what it needs to be which I can do by dividing both sides by 23.
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How many times 23 is going to 138?
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6 × 3 = 18, looks like 6 times so x = -6.
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That is half of our solution and now we need to work on back substitution.
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We will take this value here for our x and plug it back into one of the original equations.
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It does not matter which one you plug this into, just as long as you put it into one of them it should turn out okay.
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I end up solving this one for y, 2 × -6 = -12 and we are adding 12 to both sides would give me 14 and y divided by -7, I have -2.
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I have both halves of the solution.
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I have x = -6 and that y = -2.
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In the elimination method, we are working to eliminate one of our variables.
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You might be curious what would have happened if we would have chosen to get rid of those x's instead.
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We could have done it, but you would have to multiply your equations by something different.
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I’m not going all the way through this, but just to show you how you could have started.
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For example, if you multiply the first one by 3, it would have given you 6x - 21y = 6.
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If you multiply the second equation by -2, - 6x - 2y = 40.
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You would see that using those multiplications, the x's would have cancel out when you add them.
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In the elimination method, eliminate one of your variables.
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Onto the next important thing with the elimination method.
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We have many different things that could happen.
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We could have one solution, no solution or an infinite amount of solutions.
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Visually, we can see that the lines either cross, do not cross, or perhaps they are the same line.
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The way you recognize this when using the elimination method is that you might go through that method and everything will work out just fine.
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You actually be able to find your solution.
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That is when you know it has one solution, everything is good.
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If you go through the system and all your work looks good, but you create a false statement
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then you will know that there is actually no solution to the system.
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In fact, they are 2 parallel lines and they never cross.
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I think I mentioned this earlier, but go ahead and check your work if it looks like you are getting a false statement
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just to make sure that the false statement is created from them being parallel and not from you making a mistake.
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If you go through and you create a true statement, then this is an indication they have an infinite amount of solution.
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Again, check your work, but make sure that you know whether it has an infinite amount or not.
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These are exactly the same criteria that using the substitution method, so they are nice and easy to keep track of.
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False statement, no solution, true statement, infinite amount of solutions and everything works out normally, then you just have one solution.
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Let us get into some examples and see this in action.
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In this first example, we will look at solving 2x - 5y = 11 and 3x + y = 8.
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Think about our goal with the elimination method.
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We want to get rid of these x's or we want to get rid of the y.
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You get to choose which one you want rid of just a matter how you will end up manipulating these to ensure that they do cancel each other out.
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I think the better ones to go after are probably these y, since one is already positive and one is already negative.
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The way we are going to do this is when I take the second equation and we are going to multiply everything there by 5.
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Let us see the result this will have.
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We have not touched the first equation and everything in the second equation by 5 would be 15x + 5y = 40.
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I can see that when we add together our y values, they will cancel each other out.
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2x + 15x that is 17x, -5y + 5y they would cancel each other and get 0y.
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11 + 40 =51, so in this new equation I can see that we only have x to worry about and we can continue solving for x.
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We will do that by dividing both sides by 17, x =3.
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Now that I have half of my solution and I know what x is, let us take it and end up substituting it back into one of our original equations.
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(2 × 3) - 5y = 11.
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Our goal here is to get that y all by itself.
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We will multiply the 2 and 3 together and get 6, then we will subtract 6 from both sides, - 5y is equal to 5.
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Now, dividing both sides by -5, we will have y = 8, -1.
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We have both halves of our solutions and I can say that our solution is 3, -1.
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Let us try another one.
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For this next one we will try solving 3x + 3y = 0 and 4x + 2y = 3.
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In this one, it looks like all of my terms on x and y are all positive.
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It is not clear which one will be the easier one to get rid off, we just have to pick one and go with it.
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Let us go ahead and give these x's a try over here and see if we can eliminate them.
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Some things that you can do to help you eliminate some of these variables, is first just try and get them to be the exact same number.
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What some of my students realizes is that if you just take this coefficient and multiply it up here,
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then take the other coefficient and multiply it into your second equation.
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That is usually enough to make them exactly the same .
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I’m going to do that here.
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I’m going to take the entire second equation and we will multiply it by 3.
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We will take everything in the first equation and we will multiply that by 4.
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Let us see, 4 × 3 = 12x + 12y, 0 × 4 = 0, so that will be my new first equation.
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Everything in the second one by 3, 12x + 6y = 9 will be the second one.
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We are off to a good start and I can see at least that the x's are now exactly the same.
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We want them to cancel out, let us go ahead and take one of our equations and multiply it entirely through by a negative number.
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Let us do it to the second one, everything through in the second one by -1.
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It will give us -12x - 6y = -9, the first equation is exactly the same so we would not touch that one.
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By multiplying it just that way, we are doing a little bit of prep work, now I can be assured that my x's will definitely cancel out.
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Let us add these 2 equations together and end up solving for y.
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12x - 12y that will give us 0x, that is gone.
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12y - 6y will be 6y and 0 + -9 = -9.
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y = -9/6 which we might as well just reduce that call it to say – 3/2 and there is half of our solution.
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Let us go ahead and take this, plug it back into the original and see if we can find our other half.
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I’m just going to choose this one right here, 4x + 2 =- 3/2 equals 3.
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In this equation, the only thing I need to solve for is that x, let us go ahead and do that.
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2 × -3/2 = - 3, adding 3 to both sides here I will be left with 4x = 6 and dividing both sides by 4, I will have that x = 3/2.
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My final solution here, x = 3/2 and y = -3/2.
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As long as you do proper prep work, you should be able to get your final solution just fine.
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Remember that it has 2 parts, the x value and y value.
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Let us see if we can solve this one using elimination.
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I have my x's that are both positive and my y’s are both negative, but they are almost the same.
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Let us start off by multiplying the second one by 5 and see if it gets a little bit closer to canceling.
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5x - 5y, 5 × 12 = 60 we multiply the entire second equation by 5.
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It still looks like nothing will cancel out, so I will also multiply the second equation by -1.
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-5x + 5y = - 60 looking pretty good here.
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I will add these together and I can eliminate my x's but you will notice that it looks like the y will also eliminate.
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If the x's and y’s are gone, the only thing on the left side is just 0.
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What is on the other side? 3 - 60 would be a -57 and that is a bit of a problem because the 0 does not equal -57.
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Let us say that this is a false statement.
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This is one of these situations and it is a good idea to go back through your work, check it and make sure it is all correct.
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All of our steps do look good here, we multiplied it by 5, we multiplied by - 1.
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We do not actually have a whole lot of spots where we could make mistakes.
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What this false statement is telling us here is that there is no solution to the system.
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These 2 lines are actually parallel and they are not crossing whatsoever.
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Watch out for those special cases.
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Let us try one last example, x – 4y =2 and 4x – 16y = 8.
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I want to try and maybe eliminate, let us do the x's.
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We will do this by multiplying the first equation by -4.
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-4x, -4 × -4 =16y equals -8.
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In the second equation I do not need to manipulate that one, I will leave it exactly the same.
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This is a lot like our previous example.
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The x's are going to cancel out when they are added together, but then again, so are the y values.
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Both of them are going to cancel out.
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I have 0 on one side of my equal sign.
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This one actually is not quite so bad because if I look at my -8 and 8, they will cancel each other out as well.
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I will get 0 = 0, which happens to be a true statement.
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Now, one statement is true but what is your x and y?
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In indicates that the lines are exactly the same and there are on top of one another.
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We have an infinite number of solutions.
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Using one of these algebraic methods is a great way to figure out what the solution is.
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Keep in the back of your mind what the graphs of these look like
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so we can interpret what it means to have one solution, no solution, or an infinite amount of solutions.
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