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## Operations and Algebraic Thinking 218-221

### Course: Operations and Algebraic Thinking 218-221 > Unit 5

Lesson 2: One-step equations intuition- 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

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# 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.

## Want to join the conversation?

- could you take 2x from both sides to find what x is?(29 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(5 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)

- 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.(2 votes)

- At2:27, wouldn't 3 of those small, yellow blocks equal 1 of those big, blue blocks?(3 votes)
- Dividing both sides of the equation is an attempt to make the equation simpler, not more complex!

Suppose you have an equation like: 3x=27. You can make it more simple by dividing both sides by 3: x=9.

On the other hand, if you have an equation like: 4x=7, you don't need to divide both sides by 2 to make 4x simpler! Just ISOLATE "x" from 4, and thus the solution is: x=7/4.

But the main thing to realize is that, it's an EQUATION! Both sides are equal! So you must divide or add or subtract or multiply BOTH sides together! If you don't, then the value/solution of the equation will be changed giving you a WRONG output! :)(3 votes) - Isn't this the same as lesson 3 in this section, ie "The why of algebra?"(3 votes)

## Video transcript

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.