3 + (−2) = 1. Each yellow is +1 and each red is −1; two yellow–red pairs cancel, leaving 1 yellow. The sentence writes the moves in the order they were made.

It flips to red: 1 − 1 + (−1) = −1. A flip takes away a +1 and puts in a −1, so the value drops by 2, and flipping back adds 2. Many students expect a flip to change the value by 1.

4 − 1 = 3, and putting it back removes the “− 1” again. Taking a counter away writes a subtraction; putting one in writes an addition.

One yellow and one red go in: 2 + 0 = 2. A zero pair is worth 0, so it never changes the value. The tool writes it as “+ 0”.

5, then −1. The new step is cancelling 2 zero pairs: adding a negative brings in red counters, and they cancel some of the yellows.

3 + (−5) = −2. The line starts at 3, and putting in 5 reds is one amber jump of 5 to the left, ending at −2. The counters and the line always give the same value.

2. Each ring holds a yellow and a red, worth 0; there are 3 rings and 2 yellows left over. The value is the number of yellows take away the number of reds.

Still 2: moving counters around doesn’t change what they’re worth. A good strategy is to cross off one yellow and one red at a time, which is what 0 Cancel zero pairs does.

4 yellows and 1 red, 5 and 2, 6 and 3. Not with 6: yellows − reds = 3 and yellows + reds = 6 would need 4½ yellows. The value and the number of counters are both odd or both even.

6 works (3 zero pairs); 7 can’t. A value of 0 needs as many yellows as reds, so the number of counters must be even.

4 pairs cancel and 3 reds are left: the value −3 doesn’t change. With P yellows and N reds, the smaller of P and N is the number of pairs.

Still 2: the sentence reads 2 + 0. However many zero pairs go in, the value stays the same, which is why zero pairs can make taking away possible later.

4 reds; 1 yellow and 5 reds; 2 yellows and 6 reds, and so on. Adding any number of zero pairs to 4 reds keeps the value −4.

The pair vanishes but the sentence doesn’t change: it still says 0. Cancelling zero pairs tidies the mat; it isn’t a calculation.

4 + (−2) = 2. The 2 reds cancel 2 of the yellows. Adding a negative makes the value smaller.

0: every counter ends up in a zero pair. A number added to its opposite always gives 0.

−7. Adding reds to reds just makes more reds; there are no yellows, so nothing cancels. “Two negatives make a positive” is about multiplying, and even there it needs explaining.

−4. The 2 yellows cancel 2 of the 6 reds, leaving 4 reds.

4, −4, −4, 4. The answer takes the colour there is more of, and its size is the difference. Swapping the order of an addition doesn’t change the answer.

Both −3, with 3 reds left. Addition is commutative with negative numbers too.

No: 6 + (−2) = 4, 2 + (−6) = −4 and 4 + (−4) = 0. The answer is positive when the positive number is bigger in size, negative when the negative number is, and 0 when they match.

3. Two yellows go into Taken away and three stay on the mat. No zero pairs are needed when there are enough counters of the colour you take.

−2. Taking away 5 yellows needs 5, but there are only 3, so 2 zero pairs go in (the value is still 3). Take away 5 yellows and 2 reds are left.

5. There are no reds to take, so 2 zero pairs go in (still 3). Taking away the 2 reds leaves 5 yellows. Taking away a negative makes the value bigger.

7: put in 3 zero pairs, then take away the 3 reds. A student who takes away 3 yellows has done 4 − 3 = 1, and the caption says so.

1, 4 and 7. There are no reds in 6 yellows, so you need a zero pair for every red you take away. The answers are 7, 10 and 13.

3 and −3. (−2) − (−5) needs 3 zero pairs so there are 5 reds to take; (−5) − (−2) just takes 2 of the 5 reds. Swapping the order of a subtraction changes the sign of the answer.

0, 8 and −8. Taking 4 yellows from 4 yellows empties the mat. Taking 4 reds needs 4 zero pairs and leaves 8 yellows. Taking 4 yellows from 4 reds needs 4 zero pairs and leaves 8 reds.

No, it’s still −4. Extra zero pairs are worth 0, so they never change the answer; they just leave more pairs to cancel. The caption points out that 4 would do.

Put in 3 zero pairs, then take away the 3 yellows: 0 − 3 = −3. Taking a number away from 0 gives its opposite.

Both end with 8 yellows. Taking away 3 reds leaves behind the 3 yellows that came in with their zero pairs, which is exactly what putting in 3 yellows does.

Counters: 2 + 0 + (−6), so 2 − 6 = −4 is the same as 2 + (−6). Every subtraction can be written as adding the opposite.

3 + (−2) = 1 and 3 − 2 = 1. Adding a negative and subtracting a positive give the same answer.

Both give 3: 1 − (−2) = 3 leaves 3 yellows, and 1 + 2 = 3 leaves 5 yellows and 2 reds. Cancel the zero pairs and the two mats match.

2, 3 and 4, then 1 − (−4) = 5. Each time one more red is taken away, which leaves one more yellow. Subtracting a more negative number gives a bigger answer.

6 and −6. 3 × (−2) puts in 3 groups of 2 reds. The second line shows it as repeated addition: −2 + (−2) + (−2).

−6. It means take away 3 groups of 2. At 0 there is nothing to take, so 6 zero pairs go in; taking away 6 yellows leaves 6 reds.

6. Take away 3 groups of (−2) from 6 zero pairs: the 6 reds go, leaving 6 yellows. A negative times a negative is taking away negatives, which (like 3 − (−2)) makes the value bigger.

Both −8. 2 × (−4) puts in 2 groups of 4 reds; (−4) × 2 takes 4 groups of 2 yellows away from 8 zero pairs. Different stories, same answer: multiplication is commutative.

−6, −4, −2, 0 and 2: each answer is 2 more than the one before. The pattern says (−2) × (−1) has to be 2, and the counters agree.

12 and 10: one zero pair for each counter taken away. The answers are −12 and 10.

6. Put in at least 6 zero pairs, then take away 6 reds. A student who puts reds in has done 2 × (−3) = −6, and the caption shows the difference.

Both are 0: no groups at all, or 5 groups with nothing in them to take away. Nothing goes in and nothing comes out.

The line starts at 3. The dashed circle is the zero pairs going in: + 0, no move. Then a teal jump of 2 to the right ends at 5. Taking away a negative moves right.

Both are an amber jump of 5 to the left, from 2 to −3. Adding a negative and subtracting a positive are the same move.

3 becomes −(3) = −3: the dashed jump is a reflection through 0. Flipping every counter gives the opposite number.

3 − 1 + (−1) = 1. A flip takes away a +1 (one jump left) and puts in a −1 (another jump left), so the value falls by 2.

−2. Many students write 2, “the difference”. 2 zero pairs go in, 10 yellows are taken away, and 2 reds are left.

Any subtraction of a negative: 3 − (−4) = 7. Taking away reds removes negatives, so the value goes up.

−5. Taking away 2 yellows needs 2 zero pairs; then the yellows go and 5 reds are left. The student treated −3 as if it were 3.

−2. The 3 yellows cancel 3 of the 5 reds. Adding the sizes only works when the two numbers have the same sign.

0. Taking 2 reds from 2 reds empties the mat, with no zero pairs needed: any number take away itself is 0.

−2: 7 zero pairs and 2 reds left over. Deciding which colour there is more of is a quick check on any answer.

16, from (−5) + 10 up to 10 + (−5). Going through the first number from −5 to 10 in order means none can be missed (and the second number is always 5 minus the first).

Only when B = 0, with any A. Putting in B and taking away B leave the same mat only if B is no counters at all.

Whenever B is negative: 3 − (−2) = 5 but 3 + (−2) = 1. Subtracting a negative adds, and adding a negative subtracts.

16: 12 subtractions (take away 5 from 0, −1, …, −5, or take away (−5) from 0, 1, …, 5) and 4 multiplications: (−1) × 5, (−1) × (−5), (−5) × 1 and (−5) × (−1).

The best ones force zero pairs: subtracting a negative, taking away more than there is, or a negative number of groups.