Addition and subtraction — carrying, borrowing, word problems: Worked examples

This page is the "leave-no-case-behind" companion to the parent note on addition and subtraction. There we built the engine: place value, carrying, borrowing. Here we drive it over every kind of road — every way a column can behave, every trap a word problem can hide. First we map the terrain, then we walk each cell of the map with a fully worked example.

If any symbol below feels unfamiliar, it was defined in the parent note (a carry is the small $1$ that climbs to the next column when a column sum reaches $10$ or more; a borrow is the $1$ — worth $10$ — you take from the next column when the top digit is too small to subtract).

The scenario matrix

Think of every addition/subtraction problem as a machine that can hit one of a small number of behaviours in its columns. Here is the complete list. If we work one example of each row, nothing can surprise the reader.

#	Case class	What makes it special	Example that hits it
A	Clean, no carry/borrow	every column stays $0$ – $9$	Ex 1
B	Addition with a chain of carries	carry ripples several columns	Ex 2
C	Carry that grows a new column	the sum is longer than both inputs	Ex 2 (top digit)
D	Subtraction, single borrow	one column short, neighbour lends	Ex 3
E	Subtraction, borrow across zeros	a $0$ (or many) must borrow-then-lend	Ex 4
F	Degenerate: subtract equal numbers / subtract zero / add zero	edge values	Ex 5
G	Result crosses below zero (top < bottom overall) → a negative	limiting/sign case	Ex 6
H	Word problem, mixed add & subtract	translation + ordering	Ex 7
I	Exam twist: unknown digit / missing addend	reasoning backwards	Ex 8

Every numeric answer on this page is machine-checked in the verify block.

Case A — clean columns, nothing overflows

Case B & C — a chain of carries that also grows a new column

Worked example

$B, C$ · $968 + 875$ Forecast: both numbers are 3 digits and near $1000$ , so the sum should be about $1800$ — that means a 4th digit appears. Predict carries in the ones AND tens columns.

Before computing, study the figure below: it shows the same sum written in four labelled columns (thousands, hundreds, tens, ones). Watch each pink "1" — that is a carry, and the pink arrow shows it hopping left into the next-higher column. The rightmost carry is born in the ones, the next in the tens, the last one has nowhere to land except a brand-new thousands column (shown in yellow). Read the figure top-to-bottom as top number, bottom number, then result; read the carries right-to-left as the ripple.

Ones: $8 + 5 = 13$ . Write $3$ , carry $1$ . Why this step? Using the base-ten split $13 = 10 \cdot 1 + 3$ (Division Algorithm): $3$ stays, the $10$ becomes one ten next door. In the figure this is the first pink "1" leaving the ones column.
Tens: $6 + 7 + carry 1 = 14$ . Write $4$ , carry $1$ . Why this step? Now three things pile up: both tens digits plus the incoming carry. $14 = 10 \cdot 1 + 4$ — repackage again. This is the middle pink arrow.
Hundreds: $9 + 8 + 1 = 18$ . Write $8$ , carry $1$ . Why this step? Same overflow logic; the carry keeps rippling. This is the leftmost pink arrow, pointing into empty space.
Thousands: nothing was written here originally, but a carry arrived → write $1$ . Why this step? This is Case C: the carry had no home, so it creates a new place — the yellow digit in the figure. That is why a 3-digit + 3-digit sum can be 4 digits.

Answer: $1843$ . Verify: $1843 - 875 = 968$ . ✓ Estimate: $\approx 970 + 880 = 1850$ , and $1843$ is right beside it. ✓

Case D — subtraction with a single borrow

Case E — borrowing across zeros (the chain)

Worked example

$E$ · $4003 - 1547$ Forecast: the ones digit $3$ needs to borrow, but the columns above it are $0, 0$ — empty! Predict a borrow chain where each $0$ turns into a $9$ .

Study the figure before the steps: the four columns are labelled, and above each original top digit a yellow number shows what that digit becomes after the borrow passes through it ( $4 \to 3$ , $0 \to 9$ , $0 \to 9$ , $3 \to 13$ ). The blue arrows across the top show the borrow request travelling right-to-left, one column at a time. The pink bottom row is the final answer. The key lesson the picture teaches: a $0$ can't lend, so it first borrows (becoming $10$ ), immediately lends $1$ onward, and is left showing $9$ .

Ones: $3 - 7$ is short. Borrow from the tens — but the tens is $0$ , so the request passes up the line. After the chain settles, ones becomes $13 - 7 = 6$ . Why this step? A $0$ has nothing to lend, so it must first borrow from its neighbour. This is the yellow $13$ over the ones in the figure.
Tens: it borrowed to help the ones, so it briefly held $10$ , immediately lent $1$ down, and is left as $10 - 1 = 9$ . Bottom digit is $4$ : $9 - 4 = 5$ . Why this step? This is the "borrow to lend" move — every middle $0$ becomes $9$ (the yellow $9$ over the tens).
Hundreds: same again — the $0$ became $9$ after passing the chain; bottom is $5$ : $9 - 5 = 4$ . Why this step? The single original borrow is still being paid for, one column at a time — the second yellow $9$ .
Thousands: the chain finally lands: $4$ lent $1$ → becomes $3$ (yellow $3$ ); bottom is $1$ : $3 - 1 = 2$ . Why this step? The $- 10$ that started way back in the ones is settled here as a $- 1$ .

Answer: $2456$ . Verify: $2456 + 1547 = 4003$ . ✓ Estimate: $4000 - 1500 = 2500$ , close to $2456$ . ✓

Case F — degenerate inputs (zero and equal numbers)

Case G — when the answer goes below zero

Worked example

$G$ · $312 - 500$ Forecast: the whole top number ( $312$ ) is smaller than the whole bottom ( $500$ ). No amount of borrowing can conjure a $5$ in the hundreds from a $3$ . Predict a negative result.

Recognise the sign first. Since $312 < 500$ , the true answer is negative. Rewrite as $- (500 - 312)$ . Why this step? Column borrowing assumes the top is at least the bottom overall. When it isn't, you swap and negate — the size is $500 - 312$ , the sign is minus. See Negative Numbers and Integers.
Now subtract the larger from the smaller-labelled bottom, i.e. $500 - 312$ .
- Ones: $0 - 2$ → borrow chain (Case E territory): $10 - 2 = 8$ .
- Tens: $0$ became $9$ after lending: $9 - 1 = 8$ .
- Hundreds: $5$ lent $1$ → $4$ ; $4 - 3 = 1$ . So $500 - 312 = 188$ . Why this step? We compute the positive gap, then reattach the minus.
Attach the sign: $312 - 500 = - 188$ . Why this step? The gap is $188$ and $312$ is the smaller, so the difference points negative.

Answer: $- 188$ . Verify: add the difference back to the subtrahend: $- 188 + 500 = 312$ . ✓ The check identity $(a - b) + b = a$ still holds with negatives.

Case H — a mixed word problem

Worked example

$H$ · warehouse stock A warehouse starts with $2050$ crates. On Monday it ships out $1268$ . On Tuesday a delivery adds $940$ . How many crates now? Forecast: "ships out" lowers the count, "delivery adds" raises it. Predict a number below $2050$ after Monday, then back up on Tuesday — roughly $2050 - 1270 + 940 \approx 1720$ .

Translate (Word Problem Translation): "ships out" → subtract; "adds" → add. $2050 - 1268 + 940.$ Why this step? Keywords fix the operations before any arithmetic.
Do the operations left-to-right, so the subtraction comes first: $2050 - 1268$ . Why this step? Addition and subtraction share the same precedence, so by the left-to-right convention we evaluate $2050 - 1268$ before adding $940$ . (This is a rule about reading order, not associativity — subtraction is not associative, so we may not regroup freely.) Now the borrow chain, walked one column at a time (top digits of $2050$ are $2, 0, 5, 0$ ):
- Ones: $0 - 8$ is short → borrow. The tens is $5$ , which can lend, so ones becomes $10 - 8 = 2$ .
- Tens: the $5$ lent $1$ , so it is now $4$ ; then $4 - 6$ is short → borrow again. The hundreds is $0$ , so this borrow must chain further. After borrowing, the tens digit becomes $14 - 6 = 8$ .
- Hundreds: it was $0$ and had to lend to the tens, so it first borrows from the thousands (becoming $10$ ), immediately lends $1$ down (leaving $9$ )... but it had already been reduced by that lend, so it shows $9 - ...$ — cleaner to say: the $0$ becomes $9$ after the borrow-then-lend, and the bottom digit here is $2$ , giving $9 - 2 = 7$ .
- Thousands: the $2$ lent $1$ down → becomes $1$ ; bottom is $1$ : $1 - 1 = 0$ (no leading zero written). So $2050 - 1268 = 782$ .
Add the delivery: $782 + 940$ .
- Ones: $2 + 0 = 2$ .
- Tens: $8 + 4 = 12$ → write $2$ , carry $1$ .
- Hundreds: $7 + 9 + 1 = 17$ → write $7$ , carry $1$ → new thousands digit $1$ . So $782 + 940 = 1722$ . Why this step? Standard carrying (Cases B/C) finishes the count.

Answer: $1722$ crates. Verify: run it backwards — $1722 - 940 + 1268 = 2050$ . ✓ Matches the starting stock, so the whole story is consistent (units: crates throughout).

Case I — exam twist: a missing digit

Worked example

$I$ · find the unknown digit $□$ In the correct addition below, one digit is hidden. Find it. $4 □ 6 + 158 624$ Forecast: you can't add left-to-right here — you must reason from the answer backwards, one column at a time.

Ones column: $6 + 8 = 14$ → digit $4$ (matches the answer's $4$ ), carry $1$ . Why this step? Confirms the ones and, crucially, hands a carry of $1$ up to the tens — we need that.
Tens column: $□ + 5 + carry 1 = 12$ (the answer shows $2$ in the tens, and there must be a carry into the hundreds because $4 + 1 = 5 \neq = 6$ ). Why this step? The hundreds of the answer is $6$ but $4 + 1 = 5$ , so a carry of $1$ must have come from the tens — meaning the tens sum was $\geq 10$ , i.e. it equals $12$ (to leave a $2$ ).
Solve: $□ + 5 + 1 = 12 \Rightarrow □ = 6$ . Why this step? Simple backward algebra once we know the tens must total $12$ .

Answer: $□ = 6$ , so the top number is $466$ . Verify: $466 + 158 = 624$ . ✓ Matches the given result exactly.

Recall Quick self-test across the matrix

Which case is $3009 - 12$ ? ::: Case E — borrow chain across the two zeros. Why can $487 + 765$ have four digits? ::: Case C — a carry out of the hundreds creates a new thousands place. What is $600 - 600$ and which cell? ::: $0$ , Case F (subtract equal numbers). If the top number is smaller overall, what sign is the difference? ::: Negative — Case G, swap and negate. First thing to do in a mixed word problem? ::: Translate keywords into $+$ and $-$ before computing (Case H).

Connections

Parent: Addition and subtraction — the engine these cases run on
Place Value and Base-Ten System — why columns cap at $9$ and carries are always $1$ .
Division Algorithm — $a + b = 10 q + r$ is the carry split (Cases B, C).
Negative Numbers and Integers — the sign logic of Case G.
Estimation and Rounding — the "is my answer reasonable?" sanity checks used throughout.
Word Problem Translation — keyword→operation mapping for Cases H and I.