# Common Core Algebra I.Unit #5.Lesson #1.Solutions to Linear Systems and Solving by Graphing

Jun 02, 2021Hello and welcome to another

## common

### core

#### algebra

1## lesson

from emathinstruction. My name is Kirk Weiler and today we are going to do### unit

5,## lesson

1,#### solutions

to### systems

and## solving

using graphs before we begin. Let me remind you that you can find a worksheet to go with this video as well as a homework assignment by clicking on the video description or by visiting our website at WWE Math Quiet Instruction, don't forget the QR codes at the top of each worksheet that the QR code allows you. to scan with your cell phone or tablet and bring you directly to this video.Okay, let's jump into it. We've seen before what a solution to a system of equations is, but I'd like to remind you what it means. Since that is the entire

### unit

we are turning to now, this entire unit is dedicated to### systems

of equations and systems of inequalities, so a point X Y is a solution to a system if it makes all of the equations in the system true , a set of#### solutions

. to a system is the collection of all the pairs that are solutions of the system see point 1 very well, then it is very, very simple something will be a solution to a system if it makes all the equations true and the complete set of solutions to a system will be all the pairs that make them true good let's look at exercise 1 determine if point 2 point 5 is a solution for each of the systems given show the work that takes each one to your answer for each good I think that You already know enough about how to do this to give it a try, so pause the video right now and see if you can figure out if 2 point 5 is a solution for these two different systems.More Interesting Facts About,

## common core algebra i unit 5 lesson 1 solutions to linear systems and solving by graphing...

Let's do it now. Remember you have to do both. of them are true so let's start with y equals 4x minus 3 I'm going to put 5 for y I'm going to put 2 for okay now I'm going to check this one I'm going to put 2 + 4 B and minus for If I give you a system of equations consisting simply of two or more equations and ask you whether or not a given XY matches a solution to that system, you should be able to tell me whether it makes all the equations true at the same time. Alright, I'm going to clarify this, so pause the video now if you have to.

Here goes, let's see what we have next. Now let's consider the system of equations shown below y equals 2x plus 5 and y equals. 2 minus Well, I'm going to graph them without having to use the

### graphing

calculator. I'm just going to use the slope and the y-intercept, so for this first line I know my slope is 2/1 and my y-intercept is 5, so 1 2 3 4 5 now, unfortunately here, I'll probably have to go instead of going right 1 and up 2, I'm going to have to go down left 1, but we're used to that right, so let me graph that, actually, the guy I use uses my prefab and it says tag then and it's equals 2x plus 5 here we go, let me read into the other one, this one is harder, in fact, I might want to rewrite it using the commutative property as negative x plus 2, which will allow me to see the slope. is negative 1 divided by 1 and the y-intercept is 2, so now I can go down one to the right, one to the right, etc. also on the left 1 up 1 and then I'll put our line in red, okay and good, it should Probably have arrows on that other one too, but that's not really the point of the problem and it's equal to 2 minus X.The letter B asks in at what point the two lines intersect. Let's go to the green intersection right there, at minus 1 point 3 and Part C asks us, whoa! Applying the letter C asks us to demonstrate that this point is the solution to the system. It's ok no problem. Well, show that this point is the solution to the system just as we did in the first exercise. I want to put three for y and minus 1 for for for first and most fundamental ways to do it. One of the first and most fundamental ways to solve the system of equations is by

### graphing

both equations.Now the logic of why graphing the two equations and finding their intersection points works is absolutely beautiful. You know, you may have learned this. You should have learned it in 8th grade math, but you may never have gone through the exact logic, so I'm going to solve this problem and pause the video if I have to. There it goes and let's review the logic of

## solving

a system by graphing well first. let's make sure we remember something about graphing equations correctly, any pair of okay, so remember that, right? Every graph is as it is the graph is a picture of every XY pair that makes an equation true and any XY pair that isn't true doesn't end up on that graph.Now we put it together with what we know about systems of equations and you can understand why graphing two equations and finding their points of intersection solves the system, so let's take a look at exercise three. Exercise three says that you can now put the definition of a graph of an equation together with the definition of a system. Fill in the blank with one of the words shown, so this is our word bank problem, we have four words with true intersection solutions and they are both good, so each of those words goes in one of these blanks just once, okay, what I'd like you to say.

What you have to do is pause and see if you can fill in each of those blank spaces with one of those words. Well, take a little time once you're done, read the four points as if they were a full paragraph because what you should do is create an argument a reason a recent argument for and graphing and finding intersections works well pause the video Now, let's review it number one to solve a system of systems of graphical equations that graphically you find the intersection of the two graphs this works because any point of intersection is found less, but with graphs because the points of intersection are found in both graphs, they must make both equations true and because the points of intersection are the majority, they make both equations true, they are solutions to the system of equations, it's not that cool and that's really the argument for why to graph two equations and finding their point or points of intersection solves the system correctly the point of intersection is found on both graphs correctly because it is found on both graphs makes the graphs or makes both equations true and that is the definition of solution of a system you know all the points that make all the equations true at the same time, okay, so I'm going to erase those words, write them down, pause the video, okay, let's move on, so let's do a practical application of this.

Know how to graph lines, so graphing a bunch of lines and figuring out where they intersect won't be time well spent. Let's take a look at one last applied problem. Janelle and Swather are taking fifty true or false test questions. in her history class okay Janelle started after the sweater and she had already finished a Trank, no I misread that Janelle started after the sweater she had already finished twelve questions so you know Swath is working she has 12 questions done and Janelle begins. Janelle answers questions at a rate of two per second while sweating that she answers them at a rate of four questions for every Sorry, five questions for every four minutes.

I can't seem to read this issue today, so we have Janelle at 2 per minute. it's a little more complicated five questions every four minutes Janelle eventually catches up to Swetha how many minutes does it take her and what question are they on when Janelle when Janelle reaches the letter well a asks us to create two

## linear

models for Janelle and Sweat those questions answered since Janelle started now remember Janelle starts second well it might be helpful to plot some points on graph paper shows the work you use so actually let's do it a little remember this is the number of minutes since Janelle started so just let's track the sweater.The sweater had finished 12 questions. This axis goes through two, so I'm going to put the sweater in blue. So what will be in blue? And she had finished twelve. What did she do two a minute? Every minute that goes by she answers another two, now that means we only go up one space for every minute because these go up two at a time, so this is sweating this graph and actually creating a

## linear

model for Swetha is going to be very , very easy, in fact, we could say force wathah, that the number of questions that she answered, I'll just call it Y is equal to two times X plus twelve, right, because her slope is two, two questions per minute and she starts with There are already twelve questions answered, so now let's go to Janel.Now Janel has a more complicated situation, but she's starting from scratch, so oh, I'm asking this question wrong. Janel answers 2 per minute, sweat she answers five questions every 4 minutes, so I messed it up. let me go back and delete this real quick. I don't want to start the video at all, okay, and I'm actually sorry that this is all wrong again, so cover my sweat of apologies that I answer you five questions every time. four minutes now she's starting at twelve so every four minutes she'll answer another five so we go through four minutes and then two four five one two three four one three five I know this is a little complicated one two three four two four five one two three four one three five one two three four two four five is complicated because the y axis goes, the y axis unfortunately goes here in twos, okay, that's a little bit better, now let's get our equation, now think about this, she's answering questions in five questions every four minutes, that's your slope, okay, five questions every four minutes, that's your slope 5/4 X plus 12 right now Janelle Janelle has the easy Janelle has the easy because Janelle is going to answer questions at 2 per minute but it starts at 0, okay, it starts fine without answering any of the questions, this is the 1, now it will look like it has a slope of 1, but it doesn't really write because what we're really doing goes over 1 and up 2 and we just move it and keep doing it and eventually we'll connect them with a line too, so here's Janelle's line on the right and Janelle's equation is and equals MA, which is just 2x , right, we could say 2x. plus 0 or we could just say 2x and you'll get the 2x, so we graphed the two equations right, so we took care of that and solved the problem.

What was the problem? How many minutes do they last? It takes him to catch up and what question are they on when Janelle catches up, well this is where they catch up and what time is that. Oh, it takes you 16 minutes to catch up and they're keeping an eye on question 32, okay? That's pretty good, aren't I sorry I messed it up in the first place? You know, it's very, very easy to overlook reading problems. You know the reason I took a breath was because I knew that in order for Janelle to catch up she had to be answering questions.

Faster than swaha, Janelle was answering two questions a minute. Shweta is actually answering them at less than two questions a minute, so she's almost only asking almost one question a minute, I mean, five questions every four minutes, you know, just a little bit. more than one question a minute to get Janelle up to speed like she was originally doing. Janelle wasn't going to catch up, so the question didn't really make much sense, although we did okay. This time I am going to delete the text, please copy everything you need; okay, here we go, let's finish, so what we did today was review what it means to be a solution for a system and that is simply an XY point or any pair of values that makes both equations true at the same time.

We also saw how a not easy but direct way to solve a system is to graph the two equations and find their points of intersection. visit this more in later lessons for now. I would like to thank youfor joining me for another

## common

### core

#### algebra

lesson from emathinstruction. My name is Kirk Weiler and until next time keep thinking and problem solving.If you have any copyright issue, please Contact