8th grade (Eureka Math/EngageNY)
- Intro to equations with variables on both sides
- Equations with variables on both sides: 20-7x=6x-6
- Equations with variables on both sides
- Equation with variables on both sides: fractions
- Equations with variables on both sides: decimals & fractions
- Equation with the variable in the denominator
- Equations with parentheses
- Equations with parentheses
- Equations with parentheses: decimals & fractions
- Number of solutions to equations
- Worked example: number of solutions to equations
- Number of solutions to equations
- Creating an equation with no solutions
- Creating an equation with infinitely many solutions
- Number of solutions to equations challenge
Worked example: Learn to solve the equation 2x + 3 = 5x - 2. Created by Sal Khan.
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- At2:26, When Sal said that you could subtract 5x from both sides, and you subtract 5x from 2x, wouldn't the number result as a negative?(22 votes)
- At that point in the video, his equation is 2x + 3 = 5x - 2. He actually subtracts 2x, but when you subtract 5x from both sides, you have:
2x - 5x + 3 = 5x - 5x - 2
On the left side, you have 2x - 5x, which is -3x, and then you have that + 3, so you have -3x + 3 for the left hand side. For the right hand side, you have 5x - 5x - 2. The 5x - 5x part cancels out, leaving just a -2 for the right hand side.
-3x + 3 = -2
The next thing we can do is subtract 3 from both sides. We have the "+ 3" on the left side that can be eliminated there. Then the -2 on the right side becomes a -5. Subtracting a positive number from a negative numbers makes that number more negative.
-3x = -5
Now, I don't like having negative numbers in my equations, and since both sides are negative, I'll multiply both sides by -1 to convert these to positive numbers. That's the whole point in multiplying by -1, that we'll get positive numbers instead of negative numbers, and the equation is still valid because we'll be doing it to both sides. We get
3x = 5
Now, we can just divide both sides by 3. We get
X = 5/3
Thus, we've arrived at the same answer. I got slightly screwed up as I was trying to type this up, and that was because I had forgotten about the + 3 on the left side pretty early. I think it helps not to forget about that as you are trying to group the terms together. It's pretty easy to make that mistake when you group terms like that. That's what happened to me.
I hope this helps!(56 votes)
- I don't understand why he put 2 thirds if there was only one third at8:45.(16 votes)
- it doesn't look right because he didn't draw it completely to scale (the right size), there was actually 1 and 2 thirds there(9 votes)
- i can solve this, and i appreciate the demo. but my assignments don't give me problems like this, despite the fact that both of these are "solving equations with variables on both sides". why is that?(17 votes)
- I'm a little confused on how you go back and check if you got the correct answer since you're left off with x=1 2/3. How would you go through and check your solution?(15 votes)
- plugin 5/3 for x then solve and see if the statement is true for example if you get 4=9 then 5/3 is wrong but if you get something like 4=4 or 0=0 then you are correct.(8 votes)
- What practice set goes with this video? It would be great if there was a link to it, like there is from the practice sets to the appropriate video. (sorry if anyone already suggested this or if there is an easy way to find out that I just didn't see) -Protomas(13 votes)
- Equations with variables on both sides. If you are watching this in the 8th grade (US)- Solving equations section it should be under this video on the list of videos and practice tasks.(6 votes)
- hi i have a question my math teacher said that you can also turn fractions into decimals is that true? and will that be on future tests? if so should i choose the fraction answer or decimal answer?(12 votes)
- The decimal for is neater but Khan tends to ask for the fraction form of an answer. To turn a decimal into a fraction put it over 10 100 1000 depending on how many places it's asking for then simplify.(6 votes)
- I still don't understand how to solve the unknown , why 2x + 3 = 5x - 2, can anyone explain it in detail? Thank you.(6 votes)
- Let's go through it step-by-step:
1.) Write down the equation: 2x + 3 = 5x - 2
2.) Isolate the variable "x" on the right hand side:
[2x ( - 2x )] + 3 = [5x ( - 2x )]
3.) We get: 3 = 3x - 2, now er can add 2 to both sides:
[3 ( + 2 )] = 3x [- 2 ( + 2 )]
4.) We now get: 5 = 3x, now divide both sides by 3:
[5 ( / 3 )] = [3x ( / 3 )]
5.) We have our answer: x = 5/3, or x ≈ 1.7.
Hope this helped!(11 votes)
- So, I was doing the practice... How do I know which number I have to add to the variables?(6 votes)
- Uhhhh can you elaborate?
Idk if this will help but treat each number you have to add as a separate term and combine them if there are multiple ones. Ex. 4x + 5 = 10 - x
The +5 and 10 are like terms, so to eliminate them you would subtract 5 and get: 4x = 5 - x, then just add the x (because x and 4x are like terms) to get: 5x = 5, then divide by 5 to get x = 1. Sorry if this is kind of confusing :/(6 votes)
- Hi I am in 8th grade and in April I have the New York state exam it contains word problems with fractions and everything. Word problems are my biggest weakness where do I start(5 votes)
- The way I do it on exams is change the word problem into an equation, solve the equation and add the appropriate ending.(2 votes)
- Why do we subtract 2x?(5 votes)
- You need to get all the Xs on one side of the equation. This means you need to move either the 2x or the 5x to the other side. It doesn't matter which you elect to move. Just pick one and use the opposite oepration to move it. Sal chose to move 2x. Since 2x is positive, he must subtract 2x from both sides.
Once you have only one term containing X, any number on the same side as X must be moved to the other side, again by using opposite operations.
Hope this helps.(4 votes)
Let's try to solve a more involved equation. So, let's say that we have 2x plus 3, 2x plus 3 is equal to is equal to 5x minus 2. So this might look a little daunting at first. We have x's on both sides of the equation. We're adding and subtracting numbers. How do you solve it? And we'll do it in a couple of different ways. The the important thing to remember is, we just want to isolate an x. Once you've isolated an x, you have x equals something. Or x equals something. You're done, you've solved the equation. You can actually go back and check whether that works, So what we're going to do is just do a bunch of operations on both sides of the equation, to eventually isolate the x. But while we do those, I actually want to visualize what's occurring. Because I don't want you just say, oh what are the rules or the steps of solving equations. And I forgot whether this is a allowed or that isn't allowed. If you visualize what's happening, it'll actually be common sense what's allowed. So let's visualize it. So we have 2x right here on the left-hand side. So that's literally, that's x plus x. And then you have plus 3. Plus 3, I'll do it like this. So that's equal to plus 1, plus 1, plus 1. That's the same thing as 3. I could've drawn 3 circles here as well. Let do the same color. Plus 3. And then that is equal to 5 x's. Do that in blue. That is equal to 5 x's. So, 1, 2, 3, 5, 6. And I want to make it clear. You would never actually have to do it this way when you're solving the problem. You would just have to do the algebraic steps. But I'm doing this for you so you can actually visualize what this equation is saying. the left-hand side is these two orange x's plus 3. The right-hand side is 5x minus 2. So minus 2, we could write as -- so let me do this in a different color, I'll do it in pink. So, minus 2, I'll do as minus 1 and minus 1. Now, we want to isolate the x's on the same side of the equation. So, how could we do that? Well, there's two ways of doing it. We could subtract these two x's from both sides of the equation. And that would be pretty reasonable. Because then you'd have 5 x's minus the 2 x's. You'd have a positive number of x's on the right-hand side. Or, you could actually subtract 5x from both sides. And that's what's neat about algebra. As long as you do legitimate operations, you will eventually get the right answer. So let's just start off subtracting 2x from both sides of the equation. And what I mean there, I mean we're going to remove 2 x's from the left-hand side. And if we were to move 2 x's on the left-hand side, we have to remove 2 x's the right-hand side. Just like that. So what does that give us? We're subtracting 2 x's. 2 x's from the left. And we're also going to subtract 2 x's from the right. Now, what does our left-hand side simplify to? We have 2x plus 3 minus 2x. The 2 x's cancel out. So you're just left with -- you're just left with the 3. And you see that over here. We took 2 of these x's away. We're just left with the plus 1, plus 1, plus 1. And then on the right-hand side, 5x minus 2x. We have it over here. We have 5 x's minus 2 x's. You only have 1, 2, 3, x's left over. 3 is equal to 3x. And then you have your minus 2 there. You have your minus 2. So, normally if you were to do the problem, you would just have to write what we have here on the left-hand side. So what can we do next? Remember, we want to isolate the x's. Well, we have all of our x's on the right-hand side right here. If we could get rid of this negative 2, off of the right-hand side, then the x's will be alone. They'll be isolated. So how can we get rid of this negative 2, if we visualize it over here. This negative 1, this negative 1. Well, we could add 2 to both sides of this equation. Think about what happens there. So, if we add 2, so I'm going to do it like this. Plus 1, plus 1. So you could literally see. We're adding 2. And then we're going to add 2 to the left-hand side. 1 plus, 1 plus. What happens? Let me do it over here as well. So we're going to add 2. We're going to add 2. So what happens to the left-hand side? 3 plus 2 is going to be equal to 5. And that is going to be equal to 3x minus 2 plus 2. These guys cancel out. And you're just left with 3x. And we see it over here. We have the left-hand side is 1 plus 1 plus 1 plus 1 plus 1. We have 5 1's, or 5. And the right-hand side, we have the 3 x's, right over there. And then we have the negative 1, negative 1. Plus 1, plus 1, negative 1, these cancel out. They get us to 0. They cancel out. So we're just left with 5 is equal to 3x. So we have 1, 2, 3, 4, 5 is equal to 3x. Let me clear everything that we've removed, so it looks a little bit cleaner. These are all of the things that we've removed. Let me clear that out. And then let me clear that out, like that. Edit. Clear. So now we are just left with 1, 2, 3, 4, 5. Actually, let me move this over. So I could just move this over right over here. We now have 1, 2, 3, 4, 5. These are the two that we added here, is equal to 3x. These guys canceled out. That's why we have nothing there. Now, to solve this, we just divide both sides of this equation by 3. And this is going to be a little hard to visualize over here. But if we divide over here both sides by 3, what do we get? We divide the left by 3. We divide the right by 3. The whole reason why we divided by 3 is because the x was being multiplied by 3. 3 is the coefficient on the x. Fancy word, it literally just means the number multiplying the variable. The number we're solving, the variable we're solving for. So these 3's cancel out. The right-hand side of the equation is just x. The left-hand side is 5/3. So 5/3, we could say is is equal to 5/3. And this is different than everything we've seen so far. I now have the x on the right-hand side, the value on the left-hand side. That's completely fine. This is the exact same thing as saying 5/3 is is equal to x is the same thing as saying x's equal to 5/3. Completely equivalent. Completely equivalent. We sometimes get more used to this one, but this is completely the same thing. Now, if we wanted to write this as a mixed number, if we want to write this as a mixed number, 3 goes into 5 one time with remainder 2. So it's going to be 1 2/3. So it's going to be 1 2/3. So we could also write that x is equal to 1 2/3. And I'll leave it up for you to actually substitute back into this original equation. And see that it works out. Now, to visualize it over here, you know, how did he get 1 2/3, let's think about it. Instead of doing 1, I'm going to do circles. I am going to do circles. Actually, even better, I'm going to do squares. So I'm going to have 5 squares on the left-hand side. I'll do it in this same yellow color right here. So I have 1, 2, 3, 4, 5. And that is going to be equal to the 3 x's. x plus x plus x. Now, we're dividing both sides of the equation by 3. We're dividing both sides of the equation by 3. Actually, that's where we did it up here, we divided both sides by 3. So how do you do that right-hand side's pretty straightforward. You want to divide these 3 x's into 3 groups. That's 1, 2, 3 groups. 1, 2, 3. Now, how do you divide 5 into 3? And they have to be thought through even groups. And the answer tells us. Each group is going to be 1 2/3. So, 1 2/3. So it's going to be 2/3 of this, the next one. And then we're going to have 1 2/3. So this is 1/3. We're going to need another. Another 1, so this is 1 1/3. We're going to need 1 more 1/3, so this is going to be right here. And then we're left with 2/3 and 1. So we've broken it up into 3 groups. This right here. Let me make it clear. Let me make it clear, this right here is 1 2/3. 1 2/3. And then this right here, this 1/3. That's another 1/3, so that's 2/3, and then that's 1 right there. So that's 1 2/3. And then finally this is 2/3 and this is 1, so this is 1 2/3. So when you divide both sides by 3 you get 1 2/3. Each section, each bucket, is 1 2/3 on the left-hand side. On the left-hand side, or 5/3. And on the right-hand side we just have an x. So it still works. A little bit harder to visualize with fractions.