# 2.04 Subtraction with regrouping

Lesson

## Ideas

When we  subtract large numbers using the vertical algorithm,  it's important we write our numbers in the correct order. Do you remember how to do this?

### Examples

#### Example 1

Find the value of 4976-3782.

Worked Solution
Create a strategy

Use the standard algorithm method, with the first number at the top.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & & &4 &9 &7 &6 \\ &- & &3 &7 &8 &2 \\ \hline & \\ \hline \end{array}

Subtract the ones column first, we have 6-2=4:

\begin{array}{c} & & &4 &9 &7 &6 \\ &- & &3 &7 &8 &2 \\ \hline & & & & & &4\\ \hline \end{array}

In the tens column we can see that 7 is less than 8, so we need to trade 1 hundred from the hundreds place.

So we get 17-8=9 in the tens column and 9 hundreds becomes 8 hundreds in the first row.\begin{array}{c} & & &4 &8 &\text{}^1 7 &6 \\ &- & &3 &7 &8 &2 \\ \hline & & & & & 9 &4\\ \hline \end{array}

In the hundreds column, we have 8-7=1:\begin{array}{c} & & &4 &8 &\text{}^1 7 &6 \\ &- & &3 &7 &8 &2 \\ \hline & & & & 1& 9 &4\\ \hline \end{array}

And for the thousands place 4-3=1:\begin{array}{c} & & &4 &8 &\text{}^1 7 &6 \\ &- & &3 &7 &8 &2 \\ \hline & & & 1& 1& 9 &4\\ \hline \end{array}

So 4976-3782 = 1194.

Idea summary

We always start from the ones place, when we work in a vertical algorithm. If we don't have enough to subtract, we can regroup from the next place to the left.

## Subtraction of large numbers with regrouping

Let's see how we can solve subtraction where we may not have enough in one place to subtract. In this case, we need to regroup from the next place on the left. Let's see how to do this.

### Examples

#### Example 2

Find the value of 56\,196-44\,827.

Worked Solution
Create a strategy

Use the standard algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & & &5 &6 &1 &9 &6 \\ &- & &4 &4 &8 &2 &7 \\ \hline & \\ \hline \end{array}

In the ones column we can see that 6 is less than 7, so we need to trade 1 ten from the tens place.

So we get 16-7=9 in the ones column and 9 tens becomes 8 tens in the first row.\begin{array}{c} & && 5&6 &1 &8 &\text{}^1 6 \\ &- && 4&4 &8 &2 &7 \\ \hline & && & & & &9\\ \hline \end{array}

In the tens column, we have 8-2=6:\begin{array}{c} & && 5&6 &1 &8 &\text{}^1 6 \\ &- && 4&4 &8 &2 &7 \\ \hline & && & & & 6&9\\ \hline \end{array}

In the hundreds column we can see that 1 is less than 8, so we need to trade 1 thousand from the thousands place.

So we get 11-8=3 in the hundreds column and 6 thousands becomes 5 thousands in the first row.\begin{array}{c} & && 5&5 &\text{}^1 1 &8 &\text{}^1 6 \\ &- && 4&4 &8 &2 &7 \\ \hline & && & & 3& 6&9\\ \hline \end{array}

In the thousands column, we have 5-4=1:\begin{array}{c} & && 5&5 &\text{}^1 1 &8 &\text{}^1 6 \\ &- && 4&4 &8 &2 &7 \\ \hline & && & 1& 3& 6&9\\ \hline \end{array}

And for the ten thousands place 5-4=1:\begin{array}{c} & && 5&5 &\text{}^1 1 &8 &\text{}^1 6 \\ &- && 4&4 &8 &2 &7 \\ \hline & && 1& 1& 3& 6&9\\ \hline \end{array}

So 56\,196 - 44\,827 = 11\,369.

Idea summary

When subtracting, we must put the first number, our total, at the top of our algorithm. Unlike addition, which we can solve in any order, subtraction can only be solved in the order it is written.