- Same thing to both sides of equations
- Representing a relationship with an equation
- Dividing both sides of an equation
- One-step equations intuition
- Identify equations from visual models (tape diagrams)
- Identify equations from visual models (hanger diagrams)
- Solve equations from visual models
Dividing both sides of an equation
Let's get a conceptual understanding of why one needs to divide both sides of an equation to solve for a variable. Created by Sal Khan.
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- could you take 2x from both sides to find what x is?(27 votes)
- in an equation such as 1/2x=3+17 you could turn it into x=6+34 if that's what you mean(21 votes)
- why is a balance there(0 votes)
- to show how an equation works with the example of an old fashioned scale.
If you keep the scale balanced, the equation is correct, but if one side is heavier than the other side, you would not have a true equation.(77 votes)
- why didn't you do it the old fashion way 3x=9 and solve by doing in inverse opperation(6 votes)
- because he want us to learn the concept.(14 votes)
- At2:16he says that 1/3 of this total mass is equal to 1/3 of that total mass. If the unknown mass was 5x and not 3x would you have to multiply by 1/5? Or if it was 8x would you have to multiply by 1/8?(6 votes)
- Yes, or instead you could divide by 5 or 8. In fact the division sign came from fractions...(6 votes)
- I don't understand the one third to one third part ,Can someone please help me?(5 votes)
- how to solve quardratic formula(5 votes)
- If you have an equation that looks like, "x^2+2x+1", you can either factor it out or use the quadratic formula to solve. If you want to use the quadratic formula, a = 1 (because there is 1 x^2), b =2 (since there are 2 x's), and c = 1 (since it is 1). All you have to do is plug those into the quadratic formula and simplify.(1 vote)
- how do you multiply an equation it don't make since: B(2 votes)
- Basically you just multiply the left part (everything before the equals sign) by something, and then do the same to the right side, and put the results together again with an equals sign.
For example, multiply x = 15 by 3. First the left side:
x * 3 = 3x.
Then the left side:
15 * 3 = 45.
Put them together again:
3x = 45.
(of course this example makes little sense if you're trying to find x; it's just to show you how it works)
There is a video with a practical example here: https://www.khanacademy.org/math/algebra/one-variable-linear-equations/alg1-one-step-mult-div-equations/v/solving-one-step-equations-2(6 votes)
- How do I go about this problem
- 4k/7 - 46 = 3k/11
- Get all the k's on one side. Subtracting 3k/11 and adding 46 to both sides will do the trick.
4k/7 - 3k/11 = 46
- Next, use the least common multiple (LCM) to get the fractions to have like denominators.
Finding the LCM:
7: 7, 14, 21, 28,... ...70, 77
11: 11, 22, 33,... ..., 66,77
- Multiply both the numerator and denominator of each fraction by 77 (the LCM) divided by the previous denominator.
44k/77 - 21k/77 = 46
- Multiply both sides by the denominator (77).
44k - 21k = 3542
23k = 3542
- Divide by 23 on both sides.
k = 154 <-- answer!
I hope this helps, and have a great day!(4 votes)
- why is the variable always X?(2 votes)
- It's not always x. But x is the most common one to use in simple problems.(4 votes)
- In the video he named different ways to solve this problems, does anybody have any other ways to solve it? or there is only one way to solve it(3 votes)
- there is usually more than one way to solve a problem.(1 vote)
So we have our scale again. And we've got some masses on the left hand side and some masses on the right hand side. And we see that our scale is balanced. We have the same total mass on the left hand side that we have on the right hand side. Instead of labeling the mystery masses as question mark, I've labeled them all x. And since they all have an x on it, we know that each of these have the same mass. But what I'm curious about is, what is that mass? What is the mass of each of these mystery masses, I guess we could say? And so I'll let think about that for a second. How would you figure out what this x value actually is? How many kilograms is the mass of each of these things? What could you do to either one or both sides of this scale? I'll give you a few seconds to think about that. So you might be tempted to say, well if I could end up with just one mystery mass on the left hand side, and if I keep my scale balanced, then that thing's going to be equal to whatever I have on the right hand side. And that part would actually be a true statement. But then to get only one of these mystery masses on the left hand side, you might say, well why don't I just remove two of them? You might just say, well why don't I just remove-- let me do it a good color for removing-- why don't I just remove that one and that one? And then I'll just be left with that right over there. But if you just removed these two, then the left hand side is going to become lighter or it's going to have a lower mass than the right hand side. So it's going to move up and the right hand side is going to move down. And then you might say, OK, I understand. Whatever I have to do to the left hand side, I have to do to the right hand side in order to keep my scale balanced. So you might say, well why don't I remove two of these mystery masses from the right hand side? But that's a problem too because you don't know what this mystery mass is. You could try to remove two from this, but how many of these blocks represent a mystery mass? We actually don't know. But you might then say, well let's see, I've got three of these things here. If I essentially multiply what I have here by 1/3 or if I only leave a 1/3 of the stuff here, and if I only leave a 1/3 of the stuff here, then the scale should be balanced. If this has the total mass as this, then 1/3 of this total mass is going to be the same thing as 1/3 of that total mass. So let's just keep only 1/3 of this here. So that's the equivalent to multiplying by 1/3. So if we're only going to keep 1/3 there, we're going to be left with only one of the masses. And if we only keep 1/3 here, let's see, we have one, two, three, four, five, six, seven, eight, nine masses. If we multiply this by 1/3, or if we only keep 1/3 of it there, 1/3 times 9 is 3. So we're going to remove these . And so we have 1/3 of what we originally had on the right hand side and 1/3 of what we originally had on the left hand side. And they will be balanced because we took 1/3 of the same total masses. And so what you're left with is just one of these mystery masses, this x thing right over here, whatever x might be. And you have three kilograms on the right hand side. And so you can make the conclusion, and the whole time you kept this thing balanced, that x is equal to 3.