Linear Equation Solving A Review Common Core Algebra 1 Homework
Hello and welcome to another common core algebra one lesson by E math instruction. My name is Kirk weiler. And today we're going to be doing unit two lesson three linear equation solving a review. Before we get into this, just remember that you can find the worksheet that goes with this lesson. Along with a homework, by clicking on the video's description. As well, don't forget about the QR code at the top of the sheet that will allow you to take your smartphone or your tablet, scan it and come directly to this video. Let's get right into it. All right. Now, in theory, in common core math 8, you learned how to solve every linear equation under the sun. A linear equation is simply one where X or Y or T or whatever variable you have is simply to the first power. So there hasn't been anything squared or anything funny going on access to the first. All right? So what we're going to be doing in this particular lesson is solving a whole variety of different linear equations. We're going to be going through it rather quickly because you've had a lot of practice on these. And we're going to be starting with a lot of two step linear equations. In each one of these though, what I'm going to be doing is doing what we did in the last lesson where we used structure to solve equations by simply reversing the order of operations. So let's jump right into it. Okay? In letter ray, what we see is that we've taken the variable divided by three subtracted 7 and got negative two. In order to undo that, what we're first going to do is add 7 to both sides. We undo what's happened to X in the opposite order. Negative two plus 7 is positive 5. To undo the division by three, then we'll multiply both sides by three. And we'll get X equals 15 as our solution. Again, we could easily check to see if this is a solution by substituting it back into the equation. And seeing if it makes the equation true. All right? Letter B, we've taken X multiplied by four and added three. We've got negative 17. To undo that, we'll first subtract the three, undo what you did last. Negative 17 minus three be careful is negative 20. Then we'll undo multiplying by four by dividing by four. Giving the X equals negative 5. All right. Finally, letter C, we've taken X, we've multiplied it by 5, then we've added 12. We've got an 87. To undo adding 12, we'll subtract 12 from both sides. That'll give me 5 X is equal to 75. All right, to undo multiplication by 5, we'll divide both sides by 5. And we like 15 today. X is 15 again. All right, so a couple of 15s to begin this problem to begin this lesson. Okay? We're going to start to turn you loose pretty quickly, but pause the video now if you need to to review any of those simple equations solving techniques. All right, you ready to move on? Let's clear it out. Besides, two of those things were 15 anyway. How hard is that to remember? I think I could come up with an answer besides 15. Okay, let's keep going. In each case, there's only two things that we're going to be doing. And we always want to identify what comes first, what comes second. So how about this? Why don't you take a shot at D E and F? Okay? Pause the video, maybe take up to 5 minutes to complete all three, and then we'll revisit it. All right, let's go through them. In Leonard D, again, take a look. We've taken X added 7, then divided by three. So the first thing that we're going to do is we're going to rewrite the problem down, so that we have a little bit of room. But we're going to multiply both sides of the equation by three to get rid of that division by three. That gives me X plus 7 is equal to 6. Now we can get rid of the addition of 7 by subtracting 7 and getting X equals negative one. All right? Now letter E, a little bit trickier. Here we've taken X and we've subtracted one, and then we've multiplied by negative 6. You could certainly distribute the negative 6 through, but for me, since multiplying by negative 6 K last, I'm going to divide by negative 6 first. 18 divided by negative 6 be careful is negative three. Now the first thing I did was subtract one so the last thing I'm going to do is add one. Negative three plus one is negative two. And then we have an X equals negative two. Letter F pretty easy to interpret. We've got 8. We've multiplied X by 8, then we've added two. So we'll get X all by itself, isolating X by subtracting two first. Be careful, some people will tell me that negative two minus two is zero, but negative two and negative two gives me negative four. Divide both sides by 8 and be careful. Okay? Be easy to think that negative four divided by 8 is negative two, but really we just want to reduce this fraction by dividing numerator and denominator by four, and we see that X is negative one half. Be careful. All right, be careful with fractions. Four out of three people don't understand it. Okay, so I'm going to clear out the text. Copy down what you need to. And here we go. Math humor. If there's an oxymoron in this world, math humor is definitely it. All right, let's do another one. Oh, oh, fractions. All right, let's work G together. And then we'll have to do H and I on your own. All right. So what have we done here? We've taken X, multiplied it by three fourths, then subtracted 5, and gotten four. So we're going to add 5 to both sides. And we're going to get three fourths X equals 9. Now I know in theory what we really should do now is divide by three fourths. But that's just, to me, is a little weird. So what I'm going to do is I'm going to multiply by four thirds. This will give us a little more practice with fractions. The reason I do that is that I then have four divided by four. And three divided by three, which leaves me with just one X here, remember we can do 9 divided by three and get three, and then do three times four and get 12. All right? So see if you can do that on H letter I doesn't really require it, but there is a fraction involved. So try to do H and I pause the video now, see what you get. All right, let's go through it. All right, letter H is the same deal. We take negative 5 halves, we multiply it by 6 by X sorry, and then we add 6. So we start to undo that by subtracting the 6 on both sides. We get negative 5 halves X, be careful. This negative 5. I'm going to get rid of multiplying by negative 5 halves by multiplying by negative two fifths. All right. Canceling all around. Here, watch this. We'll do 5 into 5 and get one. And then we have negative one times negative two, gives me a positive two. All right. Finally, simple enough, right? Subtract three from both sides. 6 X negative one minus three, negative four, be careful. Divide both sides by 6. Reduce this fraction by dividing numerator and denominator by two, almost looks like a division sign. And we'll get negative two thirds. All right. So there we go. Some simple two step equations probably the hardest ones are those ones where you've got a fraction multiplying X but if you multiply by two reciprocal, take the fraction and flip it, multiply by that, it will eliminate it. All right. So pause the video now if you need to, and then we'll erase the text. Here we go. Scrub in. And it's gone. Okay. So now, let's get into some that are a little more complicated. This is very, very similar to what we did yesterday. Very similar to what we did yesterday. Right. We want to look at what's happened to X and just reverse everything, right? So what was the first thing? Let's say we subtract three. So we did this in the last lesson. We subtracted three, then we multiplied by 5, then we divide by 8. And then we added two. So let's reverse all of that, right? I love doing these types of problems. Sometimes when I have them in my pre calculus class, they involve things that are more complicated, let's say than just the four operations, but still it's ultimately the same. You just have to reverse what's been done. So the first thing I'm going to reverse is this addition by two, so I'll subtract two from both sides, simple enough. That will sim looks like a giant division bar. Sorry about that. To make that more complicated than it had to be. 7 minus two is 5. So I took care of that. Now we'll undo division by 8. By multiplying both sides by 8, all right, now give me 5 times X minus three. Is equal to 40. Okay, great. Now we can undo multiplication by 5 by dividing both sides by 5. Giving me X minus three is equal to 8. And then we can undo subtraction of three. By adding three to both sides. And we get X equals 11. Man, that took a lot of work. And yet, what it really took was just looking at what had been done to X and reversing it so that we got X all by itself. Okay? So pause the video now if you need to, and then I'm going to clear out the text. All right, here it goes. And it's gone. Okay. Now, what we haven't tackled yet are situations where there are variables on both sides of the equation. Now, ultimately, when we solve these equations, what we're going to be doing is we're going to be applying property equality to both sides of the expression. All right, in order to solve it. So on the left hand side, so let's just take a look at this. We'll walk through. It says by using the distributive property right equivalent expressions for both sides of the equation show the work below. All right, so we're not solving the equation here. I'm just going to rewriting a technically. All right. All we're really doing is rewriting both sides of the equation using the distributive property and some other things. So for instance, I can rewrite that as 5 X -15 plus two X here I can distribute the four and I can get four X plus 12. I'm going to do a little bit more work and I'm going to combine the like terms here, 5 X and two X 7 X all right? And now I'm sitting here. All right, so I'm going to, I'm going to rewrite this up here. Okay. Now, unlike other equations, here we have an X on both sides. So what we're going to do is we're going to use the property of equality. That says that we can subtract the same thing from both sides of the equation and still have something that's true. So 7 X minus four X will give us three X -15, and then of course four X minus four X is just zero. So we end up at 12. Now we can add 15 to both sides. And get three X is equal to 27. And then divide both sides by three. And that'll get me X equals 9. And there's our solution. Now, last but not least, we can certainly check to make sure that this is a solution. How do we do that? Why don't we do that by using what we learned a couple of lessons ago, which is that something is a solution to an equation if it makes the equation true. Well, let me box off a little space here and then I can put my head down here, high but let's check it. Let's put the 9 back into the original equation. When it all possible, you always want to check in what you started with. So in other words, I'm taking this guy and everywhere there's an X I'll put in a 9. So let's do it. Here, order of operations tells me to do the subtraction first. That almost looks like a one. Over here, I'll do this addition. 5 times 6 is 30. Two times 9 is 18. Four times 12 is 48. 30 plus 18 is 48. So that is a true equation, so yes it is a solution. All right. Great. So you're supposed to be able to solve linear equations. And we'll work on them quite a bit this year, even though that you've learned them a lot, a lot a lot in 8th grade common core math. But we'll do more of them in here, obviously. For now, pause the video if you need to write any of this down. All right, here it goes. Okay. Moving on. So, wow. Now we've got some complicated ones. Let's give these a shot. What I'd like you to do is really test yourself here. Okay, we're going to draw a line right down the middle of the paper. What I'd like you to do is pause the video at this point. Take as much time as you need to and see if you can solve these equations. Then we'll come back and we'll go through them, okay? All right. Let's get some practice. So many places to begin. I think we'll start by doing distribution. All right? It can't hurt. When I often tell my students, is do anything you want to do, just make sure it's not wrong. I know that might seem weird, but there are oftentimes many different paths that you can get or that you can take in order to solve a math problem. Many, many different paths. But one thing you can't do is do things that are wrong. All right? But there's many things that are right. And you can always distribute. You can always distribute that as multiplication over addition and subtraction. So let's take a look at what we have here. We'll have 7 X and then we'll have 7 times two, which is 14. Now here it's a little bit tricky because there's this subtraction, so distribute the subtraction as well. Negative three times X would be negative three X and then negative three times positive three would be negative 9. This side of the equation is a little bit easier 5 times X is 5 X 5 times negative three is negative 15. And then we have plus X now what we're going to do is we're going to just keep writing equivalent expressions, okay? I'm going to do this a little bit quicker, 7 X minus three X is four X watch out. We have negative 14 and negative 9. So that's negative 23. All right? On this side, we have 5 X plus X, which is 6 X, and then -15. Now, a lot of times people like getting it so that the X is on the left hand side of the equation. So I'm going to do that. I'm going to subtract 6 X from both sides. Now four X -6 X that's a little bit tricky. It's negative two X -23 is equal to negative 15. I can then get everything without an X on the other side by adding 23 to both sides. That'll give me negative two X is equal to positive 8. And then I can divide both sides by negative two and get X equals negative four. All right. We really should go back and check our final answers, but for the sake of time, I'll let you do that on your own. So X equals negative four. All right, let's take a look at B under B is a little bit tricky because visually speaking a lot of students will look at that. They'll say, well, 9 -6. Three. And 9 -6 is three. Unfortunately, we don't have 9 -6 anywhere. All right? That's actually 9 minus 6 times this. Because remember, that multiplication absolutely positively has to come before the subtraction. Now to make our lives easier again, it's best to take this negative 6 and distribute it all at once. So negative 6 times X is negative 6 X and negative 6 times positive one is negative 6. Here, not so much of an issue. Okay? Positive two times X is positive two X positive two times negative four is negative 8. And then we have plus 27. All right, let's rearrange this a bit. I'm going to rewrite this as negative 6 X plus 9 plus negative 6, just so that I'm just flip flopping these two. I commutative property and all that. Okay, here not so much of a problem. In fact, I could combine these at all at once, negative 8 plus 27 is positive 19. Think about that for a minute. All right? Over here now, I can combine these two, positive 9, negative 6 is positive three. All right. Let me just show you by contrast. Some people don't mind having X end up on the right hand side. And to avoid negative sometimes it's nice, right? Because now I'll have two X plus 6 X, which will give me 8 X. Of course, avoiding negatives only will last so long because when I subtract 19 from both sides, three -19 is negative, 16. Then dividing both sides by 8. Gives me X equals negative two. All right. And we can check that as well. So it's very, very important that you review these equations solving techniques and that you understand how to solve linear equations. Common core math really almost demands that we are fluent with the previous course material before we go on to the next course. And I know that that can be tough, especially when you had a long summer break between 8th grade and 9th grade or between whatever two levels of math your transitioning between. So make sure that you get up to speed on these, make sure that you're comfortable with them. Pause the video now if you need to copy anything down. All right, here we go. It's gone. Let's finish this lesson up. So really, today's lesson should have all been review. We reviewed essential two step techniques to solving linear equations when the variable is only on one side of the equation. And we also we also solved equations where the variables show up on both sides and you have to use the distributive property to help write equivalent expressions. But ultimately speaking, you use things like the associative property, the commutative properties, the distributive property, to rewrite both sides of the equation to a point where you can actually isolate the X variable, or whatever variable you're working with. Okay. Thank you for joining me for another common core algebra one lesson by E math instruction. My name is Kirk weiler and until next time. Keep thinking and keep solving problems.
linear equation solving a review common core algebra 1 homework