Solve the equation by splitting method....#maths#education#solution#thinking#equation#learn

Added: 2026-05-12

[00:00:00]This is a must- test final exam question and it has stumped quite a few top elementary students. Let's look at the problem. The question asks us to calculate 1/3 + 16 + 1/10th + 115th + 121st plus another 128th. If we try to find a common denominator directly, it's especially troublesome. You might spend ages on it and still not get the right answer. So for problems like this, let's first look for any patterns. In fraction calculations, there's a method called term decomposition.

[00:00:31]What does term decomposition mean? Let me give you an example like this kind of fraction.

[00:00:37]Ah, if I write it as 2 * 1/3, what form can it be decomposed into? It can be written as 1/2 minus 1/3. So, you see term decomposition splits it into two fractions, right? Oh, and look, when can we use term decomposition?

[00:00:53]When the reciprocals of two numbers are multiplied, you can decompose it into the difference of the reciprocals of those two numbers and these two numbers are consecutive. Right? So the difference between them is just one. Oh, that's when you can use term decomposition. Now let's see if these fractions can be decomposed this way.

[00:01:12]First I need to look at these fractions.

[00:01:16]Their denominators are all written as the product of two numbers. Right? For example, what can I write this as? 1* 1/3. The first one doesn't quite work since those two numbers aren't consecutive. Oh, and then for example, this 16th, 2 * 1/3, that works, right?

[00:01:32]But 2 * 1/5 does not seem to work. And 3 * 1/5 or 3 * 17th, none of these seem to follow any pattern. But we know even though there are many fractions, I can scale all of them up or down at the same time. Let's take a look. Say I shrink this whole sequence of fractions by a factor of two. That is multiplying all of them by 1/2. Isn't that what it means? So for example, let's put in some parentheses first.

[00:02:01]After multiplying all of them by 1/2, what do they become? It becomes 16th, then 112th, then 120th, and further down 130th, 142nd, and then 156th. But after you multiply by 1/2, it's not the same as the original expression anymore. So after multiplying by 1/2 in order to make it equal to the original what else do you need to do? You have to multiply by two again. Now let's see if I can list out the terms inside the parentheses. For example, the first one I can write it as 2 * 1/3. That seems to work. The next one can be written as 3 * 1/4. What about the next one? It's 4 * 1/5. Okay. And the next one is 5 * 16.

[00:02:46]Then the next one is 6 * 17th. Then 7 * 1/8.

[00:02:53]Close the parenthesis and multiply by two. Again, let's list out each term one by one and take a look. Starting from the first, it can be written as 1/2 - 1/3. The next one can be written as 1/3 - 1/4, then 1/4 - 1/5, and so on.

[00:03:10]Continuing in this way, some students might say, "If you list it out like this, doesn't it get more complicated?" Right? It seems like there are so many fractions now, but if you look closely, the minus 1/3 and plus 1/3 cancel each other out, don't they? Minus 1/4 and plus 1/4 cancel out as well. And as you continue, everything cancels out one after another. In the end, what's left? I'm left with 1/2 minus 1/8. And then outside the parentheses, there's a multiplication by two.

[00:03:47]1/2 minus 1/8 is the same as 48 - 1/8.

[00:03:52]Right? That gives 3/8. 3/8 * 2. And the final result is 3/4s. All right, that's the end of this problem. I suggest everyone give it a like and save it to watch a few more times. Otherwise, if you want to use such a useful method again, you won't be able to find