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## 6th grade

### Course: 6th grade > Unit 7

Lesson 3: One-step addition & subtraction equations- One-step addition & subtraction equations
- One-step addition equation
- One-step addition & subtraction equations
- One-step addition & subtraction equations
- One-step addition & subtraction equations: fractions & decimals
- One-step addition & subtraction equations: fractions & decimals

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# One-step addition & subtraction equations: fractions & decimals

In this math lesson, we practice solving equations with variables and fractions. We learn to isolate the variable by adding or subtracting terms from both sides of the equation. This helps us find the value of the variable, making the equation true. Keep practicing and have fun!

## Want to join the conversation?

- This video is good and all, but what if the denominator of the fractions are not the same. Then what do I do then, please explain. Cause I have a problem that doesn't have the same denominator.(32 votes)
- You still move the fraction away from the side with the variable. Then to do the addition/subtraction on the other side, you will need to use the correct steps for adding & subtracting fractions:

1) find a common denominator

2) convert each fraction to the common denominator

3) then, add/subtract the fractions.

If you look at the other questions posted under this video, you will see some examples of problems with different denominators.

Hope this helps.(24 votes)

- distance maybe?(30 votes)
- check again your answer maybe not write though(3 votes)

- Sal literally just said, "Let's do another one, this is too much fun!" I think this man is crazy(16 votes)
- Does Sal ever get anything wrong?(7 votes)
- Yes - Sal is human. If you watch the videos in regular mode (not full screen), then you will see correction boxes that pop up to correct content in the video.(17 votes)

- I'm horrible with fractions sooooo problem because I do not understand them and that's my problem so can you try to explain it a little better please?(8 votes)
- Hello there!

Fractions are quite easy to solve if you keep yourself reminded that they need a common denominator.

Unlike with multiplication and division, they need the same number on the bottom. For example, 1/4 cannot add or subtract with 1/2, so they need a common denominator.

In the same problem, you could list out the multiples. 4 is a multiple of 2 as well as itself. That means you'd have to make a half 2/4, as they're the same fraction.

If you were adding the problem, you would do 1/4 +2/4 = 3/4.

In this type of addition, they would have something like 1/4 + a =3/4.

You would still have to do the same thing, and switch to an inverse operation.

Hope this helps ^^

(If you have any questions regarding this, I'm always happy to help)(3 votes)

- Ill give you one to solve. 35= x/7(5 votes)
- You have "x divided by 7" on the right side. To move the 7, you use the opposite operation to division. You need to multiply both sides by 7.

See if you can finish the problem.

Hope this helps.(10 votes)

- What if the denominators are different like 11/6=n+7/9?

Thank you! :3(9 votes) - Does anyone else think that this is easy?(8 votes)
- what if there are 2 different denominators?(5 votes)
- I'll keep the explanation brief while hopefully still adequately covering the basics.

So, I'm looking to simplify a problem. The problem is 7/12 + 3/5. These denominators unfortunately do not match, so I cannot add them immediately. I'll have to find a way to make them match. I can do this by multiplying both numbers of the fraction by a number that matches the two. An easy way to do this is to multiply the fractions by each other's denominators, which is guaranteed to work, but not always simplest form. Now, once I do this, I will have 35/60 + 36/60. Now I can add these, and get 71/60, or 1 11/60.

I hope this helps you!(3 votes)

- If you have X+5=13 you could just subtract 5 from 13 instead of adding or subtracting something to both sides.(3 votes)
- You are just skipping work steps. The property that lets you do what you are suggesting says that you must add/subtract the same value from both sides. You are just skipping the step of writing out that +5-5 = 0 on the left side, which is what makes the 5 disappear on that side.(7 votes)

## Video transcript

- [Voiceover] Let's give
ourselves some practice solving equations. So let's say we had
the equation 1/3 plus A is equal to 5/3. What is the A that makes
this equation true? If I had 1/3 plus this
A, what does A need to be in order for 1/3 plus
that to be equal to 5/3? So there's a bunch of
different ways of doing this, and this is one of the fun
things about equations is there's no exactly one right way to do it. But let's think about
what at least I think might be the simplest way. And before I work through
anything, you should always try to pause the video, and do it on your own. So what I like to think
about is can I have just my A on one side of the equation? And since it's already
on the left-hand side, let's see if I can keep
it on the left-hand side, but get rid of this 1/3 somehow. Well the easiest was I can
think of getting rid of this 1/3 is to subtract
1/3 from the left-hand side of the equation. Now I can't just do that from the left-hand side of the equation. If 1/3 plus A is equal to 5/3, and if I just subtract 1/3
from the left-hand side, then they're not going
to be equal anymore. Then this thing is going to be 1/3 less, which this thing isn't going to change. So then this thing on
the left would become less than 5/3. So in order to hold the
equality, whatever I do on the left-hand side I have to do on
the right-hand side as well. So I have to subtract 1/3 from both sides. And if I do that, then
on the left-hand side, 1/3 minus 1/3, that's the whole reason why I subtracted 1/3 was
to get rid of the 1/3, and I am left with A is
equal to 5/3 minus 1/3, 5/3 minus 1/3, minus 1/3, and what is
that going to be equal to? I have five of something,
in this case I have 5/3, and I'm gonna subtract 1/3. So I'm gonna be left with 4/3. So I could write A is equal to 4/3. And you could check to
make sure that works. 1/3 plus 4/3 is indeed equal to 5/3. Let's do another one of these. So let's say that we have
the equation K minus eight is equal to 11.8. So once again I wanna solve for K. I wanna have just a K
on the left-hand side. I don't want this subtracting
this eight right over here. So in order to get rid of
this eight, let's add eight on the left-hand side. And of course, if I do
it on the left-hand side, I have to do it on the
right-hand side as well. So we're gonna add eight to both sides. The left-hand side, you
are substracting eight and then you're adding eight. That's just going to cancel out, and you're just going to be left with K. And on the right-hand
side, 11.8 plus eight. Well, 11 plus eight is 19,
so it's going to be 19.8. And we're done, and once again,
what's neat about equations, you can always check to see
if you got the right answer. 19.8 minus eight is 11.8. Let's do another one,
this is too much fun. Alright, so let's say that
I had 5/13 is equal to T minus 6/13. Alright, this is interesting
'cause now I have my variable on the right-hand side. But let's just leave it there. Let's just see if we can
solve for T by getting rid of everything else on the right-hand side. And like we've done in the
past, if I'm subtracting 6/13, so why don't I just add it? Why don't I just add 6/13? I can't just do that
on the right-hand side. Then the two sides won't be equal anymore, so I gotta do it on the
left-hand side if I wanna hold the equality. So what happens? So what happens? On the left-hand side I have, let me give myself a
little bit more space, I have 5/13 plus 6/13, plus 6/13 are equal to, are equal to... Well, I was subtracting
6/13, now I add 6/13. Those are just going to add to zero. 6/13 minus 6/13 is just
zero, so you're left with T. So T is equal to this. If I have 5/13 and I add to that 6/13, well I'm gonna have 11/13. So this is going to be
11/13 is equal to T, or I could write that
the other way around. I could write T is equal to 11/13.