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 take a look at systems of linear equations, pairs of linear equations
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and get into their solutions and how we can graph them.
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More specific things that we want to know is, what exactly is a system of equation and how can we find their solutions?
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One of the greatest ways that we can do to find their solutions is just looking at their graphs.
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Some special things that we want to know in terms of their types of solutions is how do we know when there is going to be no solution
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and how we will know when they will actually be an infinite amount of solutions.
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Watch out for those two as we get into the nuts and bolts of all this.
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What is a system of equations?
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A system is a pair or more of an equation coupled together.
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Sometimes you will see a { } put onto equations just to show that they are all being combined and coupled together.
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Here I have a pair, but I could have 3, 4, 5 and many more than just that.
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Now if all of the equations that are being coupled together are lines, then we will call these linear equations.
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You will notice in the examples that I have up here, both of these are lines in standard form.
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I have a system of linear equations.
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In order to be a solution of a system, it must satisfy all of the equations in that system all simultaneously, all at once.
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If it only satisfies the first equation then that is not good enough to be a solution of the system, it must satisfy all of them.
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Let us quickly look at how we can determine if a point is a part of the solution or not.
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We are going to do this by substituting it into an actual system.
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My system over here is 2x + 3y = 17 and x + 4y = 21.
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Is the point 7, 1 a solution or not?
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Let us go ahead and write down our system.
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Let us first put it into equation 1, we have (2 × 7) + (3 × 1) is it equal to 17, I do not know.
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Let us find out.
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2 × 7 would be 14 + 3 and sure enough I get 17.
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I know it satisfies the first equation just fine.
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Let us take it and substitute it into this second equation.
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I have 7x + (4 × y value × 1).
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4 × 1 = 4, 7 + 4, that one does not work out because 11 is not equal to 21.
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Even though it only satisfies one of the equations, I can say that it is not a solution.
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It only satisfies one equation.
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Let us try this with the point 1, 5 and let us see if that is a solution of the system.
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We will put in 2 or 3 if we want to know if this is equal to 17.
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I will put in 1 for x and we will put in 5 for y and let us simplify it.
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2 × 1 is 2 ,3 × 5 is 15 and 2 + 15 =17 and this one checks out.
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Now let us substitute it into the second equation.
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I will take 1 and put in for x, we will take 5 and put it in for y.
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I have 1 + 20 this that equal 21? It does 21 equals 21.
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I can say it is a solution and it satisfies both of the equations.
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There are few ways that you can go about solving a system of equations and we will see them in later lessons
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or the first ways that you can sense what the solution should be is to the graph the equations that are present.
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If these are linear equations that you are dealing with then you can use the techniques of graphing your lines to make this a much easier process.
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Let us work on finding our solutions graphically.
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To do this, all you have to do is graph the equations and then find out where those equations cross, when they do that is going to be our solution.
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There is one downside to this so be very careful.
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In order for this process to work out, you must make accurate graphs.
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If your lines are a little wavy or you do not make them very accurate then you might think you know where it cross, but actually identify a different point.
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Be careful and make accurate graphs.
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Since finding solutions using a graph is as nice and visual, we definitely will start with it
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but we will use some of those later techniques for solving because the accuracy does tend to be an issue with the graphs.
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Let us get into the graphing and finding a solution that way, I will try.
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Here I have a system of linear equations and it looks like both of them are almost written in standard form.
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The second one is not quite as in that form because it has a negative sign out front.
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I think we will be able to graph it just fine.
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I'm going to graph it by using its intercepts.
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We will make a little chart for our first equation and I'm going to identify where it crosses the y axis and the x axis by plugging in an appropriate value of 0.
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If I use 0 in my equation for the first one, I just need to solve and figure out what y is.
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Putting in a 0, we will limit that term and solving the rest out, divide it by 3 and I will get that y =2.
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I know that is going to be one point on my graph, 0, 2.
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We will go ahead and plug in 0 for y and we can see that it will eliminate our y terms and now we just have to solve 2x = 6.
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If we divide both sides of that by 2, we will get x =3.
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We have a second point that we can go ahead and put on our graph, 3, 0.
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Now that we have two points, let us be accurate on graphing this out.
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If you want to make it even more accurate, one thing you can do is actually use more points to help you graph this out.
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We have one line on here, it looks pretty good.
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Let use another chart here and see if we can graph out the second one.
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Once we have both lines on here, we will go ahead and take a look at where they cross.
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This will be for line number two, we will get x and y. What happens when we put in some 0?
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-0 + y = 7 that is a nice one to solve.
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-0 is the same as 0, the only thing left is y = 7.
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Let us put on the point 0, 7 right up there.
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If we put in 0 for y, -x + 0 = 7, I will get that –x =7 or x = -7.
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There is another point I can put on there.
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Now that I have two points here, let us go ahead and connect the dots.
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We can see where our two lines cross.
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This is from our first line and the blue line is from our second equation.
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Right here, it looks like they definitely cross.
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That is the point -3, 1, 2, 3, 4.
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The reason why we went over a little bit of work on testing solutions is, if our graph was inaccurate and this was incorrect,
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we can take it and put it into our system just to double check that it actually works out.
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If you want you can take -3 and 4, plug it back in and let us see what happens.
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Starting with the first equation 2, 3y = 6 and let us plug in our x and y.
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Checking to see if this is the solution, to see if it satisfies our first equation.
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-6 + 12 does that equal 6? -6 + 12 does equal 6.
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It checks out for our first equation.
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Let us put it into the second one, our x value is -3 our y value is 4.
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Negative times negative would be 3.
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I’m sure enough 3 + 4 does equal 7 so it satisfies the second equation as well.
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I know that -3, 4 is definitely my solution.
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Let us try and find a solution to another system.
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Here I have -2x + y = -8 and y = -3x + 2.
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The first one I can go and ahead and make a chart for and maybe figure out to where its intercepts are.
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That seems like a good way to find out that one.
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What happens when x = 2, what value do we get for y?
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This term would drop away because of the 0 and I have only be left with y = -8.
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I know that is one point I will need.
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Let us go ahead and put a dot on there.
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x =0, y = -1, 2, 3, 4, 5, 6, 7, 8 way down here.
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Now we put in a 0 for y, -2x + 0 = -8, let us see how this one turns out.
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My 0 term will drop away and we continue solving for x by dividing both sides by -2, -8 ÷ -2 is 4.
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There is another point, 4, 0, 1, 2, 3, 4, 0.
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Now that we have two points, let us go ahead and connect them together and we will see our entire line.
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We got one of them down and now let us graph our second one.
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Notice how the second equation in our system is written in slope intercept form
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which means we will take a bit of a short cut by using just the y intercept and also using its slope.
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I will make things much easier.
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Our y intercept is 2, I know it goes through this point right here.
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Our slope is -3 which I can view as -3/1.
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Starting at our y intercept we will go down 3 and to the right 1.
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1, 2, 3 into the right 1, there is a point.
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I’m bringing out my ruler and we will connect these.
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Now we can check if each of them cross.
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Let us highlight which equations go with which line, there we go.
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Also, it looks like they do cross and I would say it crossed down here at 2, -4.
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Even though I made some pretty accurate graphs, it does look like it is a hair larger than 2, and sometimes this happens.
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Even when we do make some pretty accurate graphs, sometimes where they cross it looks like it is just a hair off.
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The frustrating part is the actual solution may be different from 2, -4.
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About the only way I'm sure when using this graphing method is to check it by putting it in to the system.
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Later on we will look at more accurate systems where we would not have to check quite as much.
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Let us go ahead and put in our values for x and y to see if it satisfies both equations.
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x is 2 and y is -4, -2 × 2 = -4 + -4 is a -8.
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It looks like the first one checks out, everything is nice and balanced so it satisfies the first equation.
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Let us go ahead and put it into the second one and see if that one also works.
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X is 2 and y is -4, let us see if this balances out.
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-4 is not equal to -6 + 2 I think it is, because when you add -6 and 2 you will get -4.
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It satisfies both equations.
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2, -4 is a solution.
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When you are going through finding different solutions for a system, several different things could happen.
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The first situation that could happen is a lot like the examples we just covered.
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You are going to go through the graphing process and you are going to find one spot where the two actually cross.
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You will notice this kind of case that you like to be in.
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However, when graphing these lines, you might also find that the lines are completely parallel.
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It does pose a little bit of a problem because it means that the lines would not cross whatsoever.
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With parallel lines, we will not have a solution to our system.
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Another thing that could possibly happen is you go to graph the lines and they turn out to be exactly the same line.
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In that situation, we will not only get solutions but we will get lots of solutions because if they are the same line, they will cross an infinite amount of spots.
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Basically everywhere on a line will be a solution.
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Watch for these to show up when you are making these graphs, you will either find one spot where they cross then you have a solution.
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You will see that the lines are parallel and they do not cross so you have no solution.
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You will find that they are exactly the same line and then you will have an infinite number of points as your solution.
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In this next example, I want to highlight that the two special cases that could happen.
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One thing I want to point out is that when you are looking at the system,
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it is sometimes not obvious whether it has no solution or an infinite amount of solutions.
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You got to get down into the graphing process or the solving process before you realize that.
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My first equation is x +2, y= 4 and the other one is 3x + 6, y =18.
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I’m going to graph these out using my x and y intercepts.
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What happens when x =0, what happens when y= 0?
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For the first one, I put in 0 for x and we will go ahead and we start solving for y.
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You can see my 0 term is going to drop away, which is good.
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I will divide both sides by 2 and I will get that y = 2.
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There is one point that I will mark on my graph 0, 2.
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I will put in 0 for y and let us see what happens with that one.
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The 0 term is going to drop out and the only thing I'm left with is x = 4.
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I will put that point in there, 4, 1, 2, 3, 4, 0.
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I will go ahead and graph it out.
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Now that we have one line, let us go ahead and graph the other.
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We will get its intercepts by putting in these 0 for x and 0 for y.
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(3 × 0) + 6 time to solve for y.
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Our 0 term will go away and I will just be left with 6y =18.
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6 goes into 18 three times so I have 0, 3 as one of my end points.
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Putting in 0 for y, let us see what happens there.
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3x + 6 × 0 = 18.
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It looks like that term will drop away and I will just have 3x =18.
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3 goes into 18 six times, x = 6.
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1, 2, 3, 4, 5, 6 I have that one.
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Let us go ahead and graph it out now.
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What we can see from this one is that it looks like the two lines are parallel and they are not crossing whatsoever.
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If it is a little off and it looks like one of your lines could potentially cross but maybe off the graph,
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one thing you can do is you can rewrite the system into slope intercept form so you can better check the slopes.
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I'm pretty sure that these are parallel, I’m going to say that there is no solution to the system.
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They do not cross whatsoever.
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Let us try one more and see if this one has a solution.
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This is y = 4x - 4 and 8x -2y =8.
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The first one is written in slope intercept form so I will graph it by identifying the y intercept and its slope.
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This starts at -4 and starting at the - 4 I will go up 4/1and just like that we can go ahead and graph it out.
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There is our first line.
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The second one is written more in standard form so I will go ahead and use its intercepts to track that one down.
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What happens when x =0 and what happens when y = 0.
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When x =0, that will drop away that term right there and I'm looking at -2y = 8.
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We will be dividing both sides by -2 and I will end up with -4.
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There is one point I can put on there, notice it is on the same spot.
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Putting in 0 for y, I can see the term that will go away.
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I am left with 8x = 8.
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Dividing both sides by 8, I am simply be left with x = 1.
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We will put that point on there and both points ended up on the other line.
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When I go to draw this out, you will see that one line actually ends up right on top of the other one.
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They are essentially the same line.
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In this case, we end up with an infinite amount of solutions because they still cross, but they cross in lots of different spots.
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I could say they cross everywhere, there are infinite number of solutions.
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One of the best ways that you can figure out a solution to a system of equations
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is simply graph both of the equations that are present in the system and see where they cross.
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We will look at some more accurate methods in the next lesson.
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