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Intro to equations with variables on both sides

Learn how to solve the equation 2x + 3 = 5x - 2 with the variable on both sides. We start by visualizing the equation, then isolate the variable by performing the same operations on both sides. Finally, we solve the equation to find the value of the variable. Created by Sal Khan.

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  • piceratops tree style avatar for user Isaac Min
    Hi, I had a math test few days ago and didn't get one question. Can anyone tell me the answer and steps to solve it?
    The question is: 3(1-2x) = 4-6x Thank You.
    (285 votes)
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    • hopper cool style avatar for user Chuck Towle
      Isaac,
      3(1-2x) = 4-6x First distribute the 3
      3*1 - 3*2x = 4-6x And multiply the terms
      3-6x = 4-6x Then add 6x to both sides.
      3-6x+6x = 4-6x+6x The -6x+6x becomes zero
      3-0 = 4-0 And subtract the 0 from each sides
      3=4
      3 does not equal 4, so this is a false statement.
      This means the problems has no solution.
      No matter what number you put in for x, the original equation can never be true.
      So the answer is "No solution"

      I hope that helps make it click for you.
      (493 votes)
  • leaf red style avatar for user Anise Akiko Ambuehl
    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?
    (23 votes)
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  • female robot amelia style avatar for user Zari Ball
    Can someone explain this a lot easier I'm just sitting and my brain is fried?
    (14 votes)
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    • stelly blue style avatar for user Kim Seidel
      Your goal is to create: x = a number.

      1) When you have the variable on both sides, start by moving all the Xs to one side of the equation. In the example from the video, this means you need to move either the 2x or the 5x to the other side. It doesn't matter which you pick to move. Just pick one and use the opposite operation to move it. Sal chose to move 2x. Since 2x is positive, he must subtract 2x from both sides.

      2) 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.
      -- Start by moving the constant term (the one without the variable. This is where Sal adds 2 to both sides of the equation.
      -- Then, if there is any coefficient other than 1 in front of the variable, then you would use division to move the coefficient to the other side. This is where Sal divides the equation by 3.

      Hope this helps.
      (28 votes)
  • blobby green style avatar for user Protomas Ludwig
    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
    (18 votes)
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  • blobby green style avatar for user mirandanimroe09
    For anyone who is confused, this is how to solve 2x+3 = 5x-2 easier

    2x+3 = 5x-2
    First off we need to isolate the similar terms on both sides.
    We can do this just by transposing it to the other side, but remember to change the sign whenever your transposing a number to the other side.
    We can change it in this format:
    (Terms with variables) = (Numbers only)
    so 2x - 5x (the positive 5x changed into a negative 5x since we are transposing it)
    2x - 5x = -2 -3 (same here)
    so 2x - 5x = -3x
    -2 - 3 = -5
    so we get -3x = -5
    divide by -3 to both sides to cancel out the negative 3
    x= -5/-3 which is also equal to 5/3 :))
    hope this helpss!
    (22 votes)
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  • blobby green style avatar for user patricia kankam
    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
    (8 votes)
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  • blobby blue style avatar for user ZHONG HEYAN
    I still don't understand how to solve the unknown , why 2x + 3 = 5x - 2, can anyone explain it in detail? Thank you.
    (8 votes)
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    • male robot hal style avatar for user RN
      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!
      (14 votes)
  • aqualine seed style avatar for user Noemi.Casas10
    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?
    (13 votes)
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  • aqualine ultimate style avatar for user evismiyamoto
    So, I was doing the practice... How do I know which number I have to add to the variables?
    (6 votes)
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    • starky sapling style avatar for user yun *
      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 :/
      (4 votes)
  • duskpin seedling style avatar for user Asher
    Why do we subtract 2x?
    (5 votes)
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    • stelly blue style avatar for user Kim Seidel
      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.
      (7 votes)

Video transcript

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.