Worked example: Subtracting 3-digit numbers (regrouping twice)
About this video
The video subtracts 286 from 913, a problem that needs regrouping twice. First 10 is moved from the tens to the ones (3 becomes 13), which leaves 0 tens, so then 100 is moved from the hundreds to the tens. The answer, found both in expanded form and the standard way, is 627.
You'll be able to
- spot when both the ones and the tens places need regrouping.
- regroup twice in one subtraction problem.
- check that regrouping keeps the value of the number the same.
Key words
- regrouping
- Taking value from one place and giving it to another so you can subtract; also called borrowing.
- expanded form
- Writing a number as the sum of its place values, like 913 = 900 + 10 + 3.
- tens place
- The second digit from the right; 10 tens is the same as 100.
Transcript
So let's subtract 286 from 913. But first I'm going to do it in a slightly different way. I've taken each of these numbers, and I've expanded them out. This 9 in the hundreds place represents 900. This 1 in the tens place represents 10. This 3 in the ones place represents 3.
Likewise, 286 is the same thing as 200 plus 80 plus 6. So let's try to subtract going place by place. So if we start in the ones place, we have a problem immediately. 3 is less than 6. How do we subtract a larger number from a smaller number? We also have a problem in the tens place, 80 is larger than 10.
How do we subtract a larger number from a smaller number? And you might guess the answer here is regrouping, sometimes called borrowing. We're going to take value from one place and give it to another. So let's say this scenario right over here, where we have this 3, and we want to take some value from one of the other places.
Well, I could take 10 from the tens place, so then this is going to become 0. And if I give that 10 to the ones place, so 10 plus 3 is 13. Notice I haven't changed the value. 900 plus 0 plus 13 is still 913. Now, this solved the problem for the ones place. I can now subtract 6 from 13.
But it made the problem in the tens place even worse. I now have to subtract 80 from 0. What do I do? Well, luckily, I can go to the hundreds place. I could take 100 from 900, so then I'm left with 800. And I could you give it to the tens place. So if I give it to the tens place, then this is going to be 100.
Notice this still adds up to 913. 800 plus 100 plus 13 is 913. Why is this valuable? Well, now in every column, I'm subtracting a smaller number from a larger. You might say, wait, isn't there a positive sign here? But we have this negative out here. So we're subtracting 6 from 13.
We're subtracting 80 from 100, subtracting 200 from 800. So let's do it. 13 minus 6 is 7. 100 minus 80 is 20. 800 minus 200 is 600. So we're left with 600 plus 20 plus 7, which is 627. Now let's do the exact same thing here, but we're not going to expand out the numbers.
So 6 is greater than 3, what do we do? Well, we can regroup from the tens place. We can take 10 from here so we're left with 0 tens and give that 1 ten to the ones place. So you give 10 to the 3, it becomes 13. But now we have a problem in the tens place. How do we subtract 8 from 0?
Well, we could take 100 from the hundreds place, so 900 becomes 800, and give that 100 to the tens place. So you give the 100 to the tens place, 100 plus 0 tens is 100. 100 is the same thing as 10 tens. And so now we are ready to subtract. 13 minus 6 is 7, 10 minus 8 is 2.
Remember, this is really 10 tens minus 8 tens to get 2 tens. 100 minus 80 to get 20. And then finally, we have 800 minus 200 to get 600-- 627.
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