Class 8 (Foundation)
Let's practice some two step equations, some of which require merging terms and using the distributive property. Created by Sal Khan and CK-12 Foundation.
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- Math needs to solve its own problems. I'm not a therapist.(121 votes)
- At2:19, Sal kept 0.6 as a decimal. Could you also make that into a fraction (6/10) and then multiply both sides by 10 (getting 6x = 120), then dividing both sides by 6, getting x = 20? It seems much easier that way.(21 votes)
- Gauri, the way you described is a completely correct way of doing it, and I agree: it is easier. Usually, if you can covert decimals to fractions because fractions are easier to work with, that's a good way to do it.
Hope that helps!(18 votes)
- At2:14, why is it necessary to do the division? Couldn't you just do "Well, I know 6 goes into 12 2 times, but this is a 0.6, which is 10 times less that 6, so just multiply that quotient by ten and boom: 20 is the answer."?(10 votes)
- Because even if this certain division problem was easy, he was demonstrating it so that you could understand and do harder ones more easily… I know- this reply is coming a little to far in the future. Outdated but- hope I helped.(9 votes)
- Why do us students have to learn all subjects, but teachers can't teach all subjects?(13 votes)
- They specialize in one topic. Thats why each teacher is so good at one subject cause they excelled in that area and are able to teach us well.(3 votes)
- Thank You for these tips for two step equations. Is there other ways to do this?(8 votes)
- In two steps equations one need to solve one part of the equation first to solve the whole equation . This is going to be a good way to do one way equation.(8 votes)
- What is 13=-4x+9(5 votes)
- I will demonstrate the steps towards solving the equation:
13 = -4x + 9
13 - 9 = -4x
x = (13 - 9) / -4
x = -1
13 = -4(-1) + 9, passes
| x = -1 |
And that's our answer!(10 votes)
- At5:14why couldn't you divide both sides by 3/8? Thanks(6 votes)
- i don't get why he changed 5x-(3x+2)=1 to 5x-3x-2=1(8 votes)
Let's do a few more examples of solving equations. And I think you're going to see that these equations require a few more steps than the ones we did in the last video. But the fun thing about these is that there's more than one way to do it. But as long as you do legitimate steps, as long as anything you do to the left-hand side, you also do to the right-hand side, you should move in the correct direction, or you shouldn't get the wrong answer. So let's do a couple of these. So the first one says-- I'll rewrite it-- 1.3 times x minus 0.7 times x is equal to 12. Well, here the first thing that my instinct is to do, is to merge these two terms. Because I have 1.3 of something minus 0.7 of that same something. This is the same variable. If I have 1.3 apples minus 0.7 apples, well, why don't I subtract 0.7 from 1.3? And I will get 1.3 minus 0.7 x's, or apples, or whatever you want to call them. So is equal to 12. You could imagine that I did the reverse distributive property out here. I factored out an x. But the way my head thinks about it is, I have 1.3 of something minus 0.7 of something, that's going to be equal to 1.3 minus 0.7 of those somethings, that x. And of course 1.3 minus 0.7 is 0.6 times x of my somethings is equal to 12. And now, this looks just like one of the problems we did in the last video. We have a coefficient times x is equal to some other number. Well, let's divide both sides of this equation by that coefficient. Divide both sides by 0.6. So the left-hand side will just become an x. X is equal to-- and what is 12 divided by 0.6? 0.6 goes into 12-- let's add some decimal points here-- that's the same thing as 6 going into 120. 6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus 0.7, times 20. Let's verify that that is equal to 12. So I'll take the calculator out, just so you don't have to trust my math. So we have 1.3 times 20 is equal to 26. So this piece right here is 26. And then 0.7 times 20. I don't need a calculator for that. That is 14. 26 minus 14 is 12. So it checks out. We got the right answer for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2 is equal to 1. This looks very complicated. And whenever something looks daunting, just do steps that look like they're simplifying the equation. And over time, as long as you do legitimate steps, you should be able to make some progress. So the first thing I want to do, is I want to distribute this negative 1 over here. So this is the same thing as 5x minus 3x, minus 2. Right? I just did the distributive property on the 3x and the 2. This is a negative 1 times 3x plus 2. So it's negative 1 times 3x, plus negative 1 times 2. Or negative 3x minus 2. And that is going to be equal to 1. Now, I have 5 of something minus 3 of that same something. So that's going to be equal to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And then I have the minus 2 is equal to 1. And now, I like to get into the form where I have 2x, or I have something times x is equal to something. So I want to get rid of this negative 2 on the left-hand side. The best way I know how to do that is to add 2 to both sides. So add a 2 on the left-hand side. If I do it to the left-hand side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out, and you're going to get 2x is equal to 1 plus 2, is equal to 3. And now you can divide both sides by 2, and you get x is equal to 3/2. And I'll leave it for you to verify that this is indeed the correct answer. Let me draw a little line here so that our work doesn't get messy, although that might have made it even messier. So here we have to solve for s. And look, we have a fraction and 2 s terms. How do we do that? Well, just do it the same way. We have 1 times s minus-- you can view this as 3/8 times s is equal to 5/6. You could view this as 1 times s, minus 3/8 times s is equal to 5/6. You could factor out an s, if you like. Maybe I'll do it this way. I'll factor it onto the left-hand side. This is the same thing as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8 minus 3/8 is 5/8, times s. You could switch the order of multiplication. 5/8 times s is equal to 5/6. And you might be able to go straight from that. If I have 1 of something minus 3/8 of that something, I have 8/8 of that something minus 3/8 of that something, I'm going to have 5/8 of that something. And now, to solve for s, I can multiply both sides by the inverse of this coefficient. So I multiply 8 over 5 times 5/8s. If I do it to the left-hand side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so that those cancel out and those cancel out. And you are left with s is equal to-- right, this is just a 1-- is equal to-- well, the 5's we can divide. Divide the numerator and the denominator by 5. Divide the numerator by 2 and the denominator by 2. You're left with-- sorry, divide the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus 7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just 5 over 12 times q minus 7 is equal to 2/3. And what I want to do with this video, is to show you that I can do it two different ways. But as long as I do legitimate operations, I should get the same answer. So the first way I'm going to do it, is I'm going to multiply both sides of this equation by the inverse of 5/12. So I'm going to multiply both sides by 12 over 5. Because I wanted to get rid of this 5/12 on the left-hand side. It makes everything look a little bit messy. And I multiply it by 12 over 5, because these are going to cancel out. The 5 and the 5 cancel out, the 12 and the 12 cancel out. So the left-hand side of my equation becomes q minus 7 is equal to the right-hand side, 2/3 times 5/12. If you divide the 12 by 3, you get a 4. You divide the 3 by 3, you get a 1. So 2 times 4 is 8 over 5. And now we can add 7 to both sides of this equation. So let's add-- I want to do that in a different color-- add 7 to both sides of this equation. These two 7's cancel out. That was the whole point of adding the 7. And you are left with q is equal to 8/5 plus 7. Or we could write 8/5 plus 7 can be written as 35/5. And so this is going to be equal to 8-- well, the denominator is 5. 8 plus 35 is 43. So my answer, going this way, is q is equal to 43/5. And I said I would do it two ways. Let's do it another way. So let me write the same problem down. So I have 5/12-- actually, let me just do it a completely different way. Let me write it the way they wrote it. 5 times q minus 7, over 12 is equal to 2/3. Let me just get rid of the 12 first. Let me multiply both sides of this equation by 12. I just don't like that 12 sitting there, so I'm going to multiply both sides by 12. So these are going to cancel out, and you're going to be left with 5 times q minus 7 is equal to 2/3 times 12. That's the same thing as 24 over 3. So this is, let me write this. 2 over 3 times 12 over 1 is equal to-- if you divide that by 3, you get a 4, divide that by 3, you get a 1-- is equal to 8. So you get 5 times q minus 7 is equal to 8. And then instead of dividing both sides by 5, which would get us pretty close to what we were doing over here, let me distribute this 5, I just want to show you, you can do it multiple legitimate ways. So 5 times q is 5q. 5 times negative 7 is minus, or negative 35, is equal to 8. 5q minus 35 is equal to 8. Now, if I want to get rid of that minus 35, or that negative 35, the best way to do it is to add 35 to both sides. I did that so that these cancel out, and I'm left with 5q is equal to 8 plus 35, which is 43. Now I can multiply both sides of this equation by 1/5, which is the same thing as dividing both sides by 5. And these cancel out. You get q is equal to 43 over 5. So there's a bunch of ways you can do these problems. But as long as you do legitimate steps, you will get the right answer. And I'll leave it to you to verify that this truly is the right answer for q. This is the q that will satisfy this equation. Let's do one word problem here. Jade is stranded downtown with only $10 to get home. Taxis cost $0.75 per mile, but there's an additional $2.35 hire charge. Write a formula and use it to calculate how many miles she can travel with her money. All right. So the total cost of a cab ride is going to be equal to just the initial hire charge, which is $2.35, plus the $0.75 per mile, times the number of miles. We're letting m is equal to the miles she travels. Miles traveled. So this is the equation. We know that she only has $10 to get home. So her cost has to be $10. So we have to say, the cost has to be $10. So 10 is equal to 2.35 plus 0.75m. So how do we solve for m, or the number of miles Jade can travel? Well, we can get rid of the 2.35 on this right-hand side by subtracting that amount from both sides of this equation. So let's do that. So let's subtract minus 2.35 from both sides. These will cancel out. That was the point. The left-hand side-- what is 10 minus 2.35? Now, these will cancel out. Now what is 10 minus 2.35? 10 minus 2 is 8. 10 minus 2.3 is 7.7. So it's going to be 7.65. If you want to believe me, let's do it. 10 minus 2.35. 7.65. And that is going to be equal to 0.75m. Let me write that in that same color. It's nice to see where different things came from. 0.75m. I have, like, five shades of this purple here. so this is that, that is that, and then these two guys canceled out. Now to solve for m, I can just divide both sides by 0.75. So if I divide that side by 0.75, I have to do it to the left-hand side as well. 0.75. That cancels out, so on the right-hand side, I'm left with just an m. And on the left-hand side-- I'll have to get my calculator out for this one-- I have 7.65 divided by 0.75, which is equal to 10.2. m is 10.2, so Jade can travel 10.2 miles.