7.2 Whole numbers: subtraction (Page 2/4)

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$\begin{matrix} 534 \\ \underset{̲}{- 203} \end{matrix}$

331

$\begin{matrix} 857 \\ \underset{̲}{- 43} \end{matrix}$

814

$\begin{matrix} 95,628 \\ \underset{̲}{- 34,510} \end{matrix}$

61,118

$\begin{matrix} 11,005 \\ \underset{̲}{- 1,005} \end{matrix}$

10,000

Find the difference between 88,526 and 26,412.

62,114

In each of these problems, each bottom digit is less than the corresponding top digit. This may not always be the case. We will examine the case where the bottom digit is greater than the corresponding top digit in the next section.

Subtraction involving borrowing

Minuend and subtrahend

It often happens in the subtraction of two whole numbers that a digit in the minuend (top number) will be less than the digit in the same position in the subtrahend (bottom number). This happens when we subtract 27 from 84.

$\begin{matrix} 84 \\ \underset{̲}{- 27} \end{matrix}$

We do not have a name for $4 - 7$ . We need to rename 84 in order to continue. We'll do so as follows:

Vertical subtraction. 84 - 27 is equal to 8 tens + 4 ones, over 2 tens + 7 ones.

Vertical subtraction. 7 tens + 1 ten + 4 ones, over 2 tens + 7 ones.

Vertical subtraction. 7 tens + 10 ones + 4 ones, over 2 tens + 7 ones.

Our new name for 84 is 7 tens + 14 ones.

Vertical subtraction. 7 tens + 14 ones, over 2 tens + 7 ones = 5 tens + 7 ones.
$\begin{matrix} = 57 \end{matrix}$

Notice that we converted 8 tens to 7 tens + 1 ten, and then we converted the 1 ten to 10 ones. We then had 14 ones and were able to perform the subtraction.

Borrowing

The process of borrowing (converting) is illustrated in the problems of Sample Set C.

Sample set c

84 - 27 = 57. The 8 in 84 is crossed out, with a 7 above it. There is a 14 above the ones column.

Borrow 1 ten from the 8 tens. This leaves 7 tens.
Convert the 1 ten to 10 ones.
Add 10 ones to 4 ones to get 14 ones.

672 - 91 = 581. The 6 in 672 is crossed out, with a 5 above it. The 7 in 672 is crossed out, with 17 above it.

Borrow 1 hundred from the 6 hundreds. This leaves 5 hundreds.
Convert the 1 hundred to 10 tens.
Add 10 tens to 7 tens to get 17 tens.

Practice set c

Perform the following subtractions. Show the expanded form for the first three problems.

$\begin{matrix} 53 \\ \underset{̲}{- 35} \end{matrix}$

The solution is 18. The subtraction can be broken into the quantity 5 tens + 3 ones, minus the quantity 3 tens + 5 ones. 5 tens + 3 ones can be broken down to 4 tens + 1 ten + 3 ones, or 4 tens + 13 ones. The difference is 1 ten + 8 ones, or 18.

$\begin{matrix} 76 \\ \underset{̲}{- 28} \end{matrix}$

The solution is 48. The subtraction problem can be expanded to the quantity 7 tens + 6 ones, minus the quantity 2 tens + 8 ones. 7 tens + 6 ones can be expanded to be 6 tens + 1 ten + 6 ones, or 6 tens + 16 ones. The sum becomes 4 tens + 8 ones, or 48.

$\begin{matrix} 872 \\ \underset{̲}{- 565} \end{matrix}$

The solution is 307. The subtraction problem can be expanded to be the quantity, 8 hundreds + 7 tens + 2 ones, minus the quantity, 5 hundreds + 6 tens + 5 ones. 8 hundreds + 7 tens + 2 ones can be expanded to 8 hundreds + 6 tens + 1 ten + 2 ones, or 8 hundreds + 6 tens + 12 ones. The difference is 3 hundreds + 0 tens + 7 ones, or 307.

$\begin{matrix} 441 \\ \underset{̲}{- 356} \end{matrix}$

$\begin{matrix} 775 \\ \underset{̲}{- 66} \end{matrix}$

709

$\begin{matrix} 5,663 \\ \underset{̲}{- 2,559} \end{matrix}$

3,104

Borrowing more than once

Sometimes it is necessary to borrow more than once . This is shown in the problems in [link] .

Sample set d

Perform the Subtractions. Borrowing more than once if necessary

641 - 358 = 283. the 4 in 641 is crossed out, with a 3 marked above it. Above the 1 in 641 is 11. The 6 in 641 is then crossed out, with a 5 marked above it. The 3 above the 4 is crossed out, with a 13 marked above it.

Borrow 1 ten from the 4 tens. This leaves 3 tens.
Convert the 1 ten to 10 ones.
Add 10 ones to 1 one to get 11 ones. We can now perform $11 - 8$ .
Borrow 1 hundred from the 6 hundreds. This leaves 5 hundreds.
Convert the 1 hundred to 10 tens.
Add 10 tens to 3 tens to get 13 tens.
Now $13 - 5 = 8$ .
$5 - 3 = 2$ .

534 - 85 = 449. The 3 in 534 is crossed out, with a 2 above it. Above the 4 is a 14. The 5 in 534 is then crossed out, with a 4 marked above it. The 2 above the 3 in 534 is crossed out, with a 12 above it.

Borrow 1 ten from the 3 tens. This leaves 2 tens.
Convert the 1 ten to 10 ones.
Add 10 ones to 4 ones to get 14 ones. We can now perform $14 - 5$ .
Borrow 1 hundred from the 5 hundreds. This leaves 4 hundreds.
Convert the 1 hundred to 10 tens.
Add 10 tens to 2 tens to get 12 tens. We can now perform $12 - 8 = 4$ .
Finally, $4 - 0 = 4$ .

$\begin{matrix} 71529 \\ \underset{̲}{- 6952} \end{matrix}$

After borrowing, we have

71529 - 6952 = 64577. Above the 5 is a 4, and above the 2 is a 12. Above the 1 is a 0, and above the 7 is a 6. The 0 and the 4 are crossed out, with a 14 written above the 4, and a 10 written above the 0.

Practice set d

Perform the following subtractions.

$\begin{matrix} 526 \\ \underset{̲}{- 358} \end{matrix}$

168

$\begin{matrix} 63,419 \\ \underset{̲}{- 7,779} \end{matrix}$

55,640

$\begin{matrix} 4,312 \\ \underset{̲}{- 3,123} \end{matrix}$

1,189

Borrowing from zero

It often happens in a subtraction problem that we have to borrow from one or more zeros. This occurs in problems such as

$\begin{matrix} 503 \\ \underset{̲}{- 37} \end{matrix}$

and
$\begin{matrix} 5000 \\ \underset{̲}{- 37} \end{matrix}$

We'll examine each case.

Borrowing from a single zero.

Consider the problem $\begin{matrix} 503 \\ \underset{̲}{- 37} \end{matrix}$

Since we do not have a name for $3 - 7$ , we must borrow from 0.

Vertical subtraction. 503 - 37 is equal to 5 hundreds + 0 tens + 3 ones, minus 3 tens + 7 ones.

Since there are no tens to borrow, we must borrow 1 hundred. One hundred = 10 tens.

Vertical subtraction. 4 hundreds + 10 tens + 3 ones, minus 3 tens + 7 ones.

We can now borrow 1 ten from 10 tens (leaving 9 tens). One ten = 10 ones and 10 ones + 3 ones = 13 ones.

Vertical subtraction. 4 hundreds + 9 tens + 13 ones, minus 3 tens + 7 ones = 4 hundreds + 6 tens + 6 ones, equal to 466.

Now we can suggest the following method for borrowing from a single zero.

Borrowing from a single zero

To borrow from a single zero,

Decrease the digit to the immediate left of zero by one.
Draw a line through the zero and make it a 10.
Proceed to subtract as usual.

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Source: OpenStax, Contemporary math applications. OpenStax CNX. Dec 15, 2014 Download for free at http://legacy.cnx.org/content/col11559/1.6

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