Grade 4 · 4.NF.B.3C

Adding and subtracting mixed numbers with like denominators

At the school track meet, Ufuk runs 3\tfrac{2}{5} miles in the morning and 1\tfrac{4}{5} miles in the afternoon. How far is that in all, and how do you take one mixed number away from another when the fraction part you take is the bigger one?

You need

Mixed numbers and improper fractions: rewriting and comparingGrade 4 Adding and subtracting fractions with like denominatorsGrade 4

Where you are 4 in Mixed numbers and improper fractions

Fractions, ratio and percent3

Fractions15

Mixed numbers and improper fractions4

Mixed numbers and improper fractions: rewriting and comparing

Adding and subtracting mixed numbers with like denominators

Adding mixed numbers Subtracting mixed numbers

When you reach for it

Ufuk, Agata and Habiba train for the school track meet. They count how far they run in miles, and the distances are rarely whole miles. Agata jogs 2\tfrac{1}{5} miles to warm up and then runs 1\tfrac{2}{5} miles more. Habiba still has part of a long run left to go. How far in all? How far is left? How much farther did one runner go than another?

Each question joins two amounts or takes one away from another, and the amounts are Mixed numbers and improper fractions: a whole number and a fraction part together. A mixed number such as 2\tfrac{1}{5} means 2 + \tfrac{1}{5}. When every fraction part has the same denominator, the pieces are all the same size, so the work comes down to two things you already know: adding and subtracting whole numbers, and adding and subtracting fractions with a common denominator (Adding and subtracting fractions with like denominators). The one new move is trading. Sometimes the fraction parts add up to more than a whole, and sometimes a whole must be cut into pieces before you can take pieces away. Both trades rest on the two ways of naming an amount from Mixed numbers and improper fractions: rewriting and comparing.

The lesson takes three days, one part a day:

Part 1: adding mixed numbers, wholes with wholes and fraction parts with fraction parts.
Part 2: adding when the fraction parts make more than a whole.
Part 3: subtracting mixed numbers, trading a whole when needed, and word problems.

The procedure

Part 1: wholes with wholes, fraction parts with fraction parts

Agata jogs 2\tfrac{1}{5} miles, then runs 1\tfrac{2}{5} miles. How far did she go?

A mixed number is a sum. 2\tfrac{1}{5} is 2 + \tfrac{1}{5}, and 1\tfrac{2}{5} is 1 + \tfrac{2}{5}. So the whole question is one long sum of four numbers:

2\tfrac{1}{5} + 1\tfrac{2}{5} = 2 + \tfrac{1}{5} + 1 + \tfrac{2}{5}

In a sum you may change the order of the numbers. That is the commutative property of addition (properties of addition): \tfrac{1}{5} + 1 is the same as 1 + \tfrac{1}{5}. You may also change which numbers are grouped together first. That is the associative property. Together they let us move the whole numbers next to each other and the fifths next to each other:

\begin{aligned} 2\tfrac{1}{5} + 1\tfrac{2}{5} &= (2 + 1) + \left(\tfrac{1}{5} + \tfrac{2}{5}\right) \\ &= 3 + \tfrac{3}{5} \\ &= 3\tfrac{3}{5} \end{aligned}

The wholes add as whole numbers: 2 + 1 = 3 miles. The fraction parts are both fifths of a mile, so they add by counting fifths: 1 + 2 = 3 fifths. Agata went 3\tfrac{3}{5} miles.

Figure 1Start at 2\tfrac{1}{5}. One whole hop and two fifth hops land on 3\tfrac{3}{5}.

The number line shows the same thing. Start at 2\tfrac{1}{5}. Adding 1\tfrac{2}{5} is a hop of one whole and then two hops of \tfrac{1}{5}. The whole hop lands on 3\tfrac{1}{5}, and the two small hops land on 3\tfrac{3}{5}.

Why it works. Nothing in the sum was cut or changed. Every mile is still a mile and every fifth is still a fifth. We only counted the miles together and the fifths together, the way you count dimes with dimes and pennies with pennies. The denominator stays 5 for the reason it stayed the same in Adding and subtracting fractions with like denominators: the size of the piece did not change.

More than two numbers, or a whole number. The same properties let you add any number of mixed numbers at once: add all the wholes, add all the fraction parts. A whole number such as 4 is a mixed number with no fraction part, so 4 + 2\tfrac{3}{8} = 6\tfrac{3}{8}: the whole parts are 4 + 2 = 6, and the eighths are just the \tfrac{3}{8} that was already there.

Part 2: when the fraction parts make more than a whole

Ufuk runs 3\tfrac{2}{5} miles in the morning and 1\tfrac{4}{5} miles in the afternoon. Group the wholes and the fraction parts as before:

\begin{aligned} 3\tfrac{2}{5} + 1\tfrac{4}{5} &= (3 + 1) + \left(\tfrac{2}{5} + \tfrac{4}{5}\right) \\ &= 4 + \tfrac{6}{5} \end{aligned}

That is 4\tfrac{6}{5} miles. It is a true name for the distance, but it is not finished. The fraction part \tfrac{6}{5} is more than one whole: a mile has only 5 fifths, and here there are 6 (Fractions greater than one (improper fractions)).

Figure 2The fraction parts \tfrac{2}{5} and \tfrac{4}{5} are six fifths. Five of them fill one whole mile, and \tfrac{1}{5} is left over.

Five of the six fifths fill one whole mile. So split \tfrac{6}{5} into \tfrac{5}{5} and \tfrac{1}{5}, and give the full mile to the wholes. Moving the 1 over to the 4 is the associative property again:

\begin{aligned} 4 + \tfrac{6}{5} &= 4 + \left(\tfrac{5}{5} + \tfrac{1}{5}\right) \\ &= (4 + 1) + \tfrac{1}{5} \\ &= 5\tfrac{1}{5} \end{aligned}

So 3\tfrac{2}{5} + 1\tfrac{4}{5} = 5\tfrac{1}{5}. Ufuk ran 5\tfrac{1}{5} miles. This is the same trade you make when adding whole numbers: ten ones become one ten, and here five fifths become one whole. A finished mixed number has a fraction part less than 1, so it shows at once which two whole numbers the amount lies between. 5\tfrac{1}{5} is between 5 and 6.

Another route: improper fractions. You can also turn both mixed numbers into fifths first, as in Mixed numbers and improper fractions: rewriting and comparing. There, \tfrac{17}{5} was packed into 3\tfrac{2}{5}; now run it backward.

\begin{aligned} 3\tfrac{2}{5} &= \tfrac{3 \times 5 + 2}{5} = \tfrac{17}{5} \\ 1\tfrac{4}{5} &= \tfrac{1 \times 5 + 4}{5} = \tfrac{9}{5} \end{aligned}

Now both amounts count fifths, so add the counts: \tfrac{17}{5} + \tfrac{9}{5} = \tfrac{26}{5}. To read it as miles, pack the fifths five to a mile: 26 \div 5 = 5 \text{ R } 1, so \tfrac{26}{5} = 5\tfrac{1}{5}. The two routes agree, as they must, since both count the same fifths of a mile.

The wholes-and-parts route keeps the numbers small, which helps when the wholes are big. The improper route never needs a separate trade, which some people find calmer. Either one is correct.

Part 3: subtracting, and trading one whole

No trade needed. Habiba’s long run is 4\tfrac{5}{6} miles. She has run 1\tfrac{2}{6} miles so far. How far is left?

Taking away 1\tfrac{2}{6} means taking away 1 mile and \tfrac{2}{6} of a mile. Taking an amount away in two batches leaves the same as taking it all at once, so take the whole from the wholes and the sixths from the sixths:

\begin{aligned} 4\tfrac{5}{6} - 1\tfrac{2}{6} &= (4 - 1) + \left(\tfrac{5}{6} - \tfrac{2}{6}\right) \\ &= 3 + \tfrac{3}{6} = 3\tfrac{3}{6} \end{aligned}

She has 3\tfrac{3}{6} miles left.

When the fraction part taken is bigger. Agata’s practice loop is 4\tfrac{1}{6} miles. She has run 2\tfrac{5}{6} miles of it. How far is left?

4\tfrac{1}{6} - 2\tfrac{5}{6}

The wholes are fine: 2 can be taken from 4. The sixths are not. There is only 1 sixth, and 5 sixths must be taken away (Adding and subtracting fractions with like denominators showed that you cannot take away more pieces than there are). But the wholes hold more sixths. Every mile is \tfrac{6}{6} of a mile. So trade one of the 4 wholes for 6 sixths:

\begin{aligned} 4\tfrac{1}{6} &= 3 + 1 + \tfrac{1}{6} \\ &= 3 + \tfrac{6}{6} + \tfrac{1}{6} \\ &= 3\tfrac{7}{6} \end{aligned}

Figure 3Trading one whole for \tfrac{6}{6} changes the name, not the amount. 4\tfrac{1}{6} and 3\tfrac{7}{6} are the same distance.

The trade does not change the distance. It only renames it: three whole miles and seven sixths of a mile is the same as four miles and one sixth. Now there are enough sixths to take 5 of them:

\begin{aligned} 4\tfrac{1}{6} - 2\tfrac{5}{6} &= 3\tfrac{7}{6} - 2\tfrac{5}{6} \\ &= (3 - 2) + \left(\tfrac{7}{6} - \tfrac{5}{6}\right) \\ &= 1\tfrac{2}{6} \end{aligned}

Agata has 1\tfrac{2}{6} miles left. This is the same trade as regrouping in whole-number subtraction, where one ten is traded for ten ones. Here one whole is traded for \tfrac{6}{6}, because the pieces are sixths.

Check by adding. What is left plus what was taken must give back the start (Subtraction as an unknown addend): 2\tfrac{5}{6} + 1\tfrac{2}{6} = 3\tfrac{7}{6}, and \tfrac{7}{6} = 1\tfrac{1}{6}, so the sum is 4\tfrac{1}{6}. It checks.

The improper-fraction route. Turn both into sixths: 4\tfrac{1}{6} = \tfrac{25}{6}, since 4 \times 6 + 1 = 25, and 2\tfrac{5}{6} = \tfrac{17}{6}, since 2 \times 6 + 5 = 17. Then

\tfrac{25}{6} - \tfrac{17}{6} = \tfrac{8}{6} = 1\tfrac{2}{6}

No trade is needed, because the improper fraction already counts every sixth, the ones inside the wholes too.

Simpler names are optional. The answer 1\tfrac{2}{6} could also be written 1\tfrac{1}{3}, since \tfrac{2}{6} and \tfrac{1}{3} are the same amount (Simplifying a fraction to simplest form). Keeping the sixths the problem gave is fully correct.

Word problems. First check that every amount is measured in the same whole, here miles. Then read what happens:

Distances are joined, or the question asks for a total: add.
Part of a distance is done and the question asks how much is left: subtract.
Two distances are compared, and the question asks how much farther or how much less: subtract.

A story with two steps, such as two runs and then the distance still to go, takes one step at a time. Worked examples 5 to 7 are word problems of this kind.

Worked examples

Examples 1 and 2 belong to Part 1, Examples 3 and 4 to Part 2, and Examples 5, 6 and 7 to Part 3.

Example 1

Two warm-up runs

Habiba runs 1\tfrac{3}{8} miles before lunch and 2\tfrac{4}{8} miles after lunch. How far does she run?

Solution

The question asks for a total, so add. Both fraction parts are eighths, so group wholes with wholes and eighths with eighths, which the commutative and associative properties allow:

\begin{aligned} 1\tfrac{3}{8} + 2\tfrac{4}{8} &= (1 + 2) + \left(\tfrac{3}{8} + \tfrac{4}{8}\right) \\ &= 3 + \tfrac{7}{8} = 3\tfrac{7}{8} \end{aligned}

The eighths make 3 + 4 = 7 eighths, less than the 8 in a whole, so no trade is needed. Habiba runs 3\tfrac{7}{8} miles. A size check: about 1 and a bit plus 2 and a half is a bit less than 4, and 3\tfrac{7}{8} is just under 4.

Example 2

Three runs, one of them whole miles

Over three days Ufuk runs 2\tfrac{1}{10} miles, then 3 miles, then 1\tfrac{6}{10} miles. How far does he run in all?

Solution

A whole number is a mixed number with no fraction part, so it joins the other wholes. Reorder and regroup so all the wholes sit together and all the tenths sit together:

\begin{aligned} 2\tfrac{1}{10} + 3 + 1\tfrac{6}{10} &= (2 + 3 + 1) + \left(\tfrac{1}{10} + \tfrac{6}{10}\right) \\ &= 6 + \tfrac{7}{10} = 6\tfrac{7}{10} \end{aligned}

Ufuk runs 6\tfrac{7}{10} miles. The order the runs happened in does not matter to the total, which is what the commutative property says.

Example 3

Eighths that pass a whole

Agata runs 2\tfrac{5}{8} miles on Monday and 3\tfrac{7}{8} miles on Tuesday. How far does she run on the two days? Find the answer two ways.

Solution

First way, wholes and parts:

\begin{aligned} 2\tfrac{5}{8} + 3\tfrac{7}{8} &= (2 + 3) + \left(\tfrac{5}{8} + \tfrac{7}{8}\right) \\ &= 5 + \tfrac{12}{8} \end{aligned}

Twelve eighths is more than one mile, since a mile is 8 eighths. Split 12 = 8 + 4, so \tfrac{12}{8} = \tfrac{8}{8} + \tfrac{4}{8} = 1\tfrac{4}{8}. Give the whole mile to the wholes: 5 + 1\tfrac{4}{8} = 6\tfrac{4}{8}.

Second way, improper fractions. 2\tfrac{5}{8} = \tfrac{21}{8}, since 2 \times 8 + 5 = 21, and 3\tfrac{7}{8} = \tfrac{31}{8}, since 3 \times 8 + 7 = 31. Then

\tfrac{21}{8} + \tfrac{31}{8} = \tfrac{52}{8}

and 52 \div 8 = 6 \text{ R } 4, so \tfrac{52}{8} = 6\tfrac{4}{8}. Both ways give 6\tfrac{4}{8} miles.

Example 4

Fraction parts that make two wholes

Habiba runs 1\tfrac{2}{3} miles on each of two days and 2\tfrac{2}{3} miles on a third day. How far is that?

Solution

Add all three. Group the wholes and the thirds:

\begin{aligned} &1\tfrac{2}{3} + 2\tfrac{2}{3} + 1\tfrac{2}{3} \\ &= (1 + 2 + 1) + \left(\tfrac{2}{3} + \tfrac{2}{3} + \tfrac{2}{3}\right) \\ &= 4 + \tfrac{6}{3} \end{aligned}

A mile is 3 thirds, and 6 = 3 + 3, so the six thirds fill exactly two whole miles: \tfrac{6}{3} = \tfrac{3}{3} + \tfrac{3}{3} = 2. Then 4 + 2 = 6. Habiba runs exactly 6 miles, with no fraction part left over.

Example 5

How much farther

In one week Habiba runs 5\tfrac{7}{10} miles and Agata runs 2\tfrac{3}{10} miles. How much farther does Habiba run?

Solution

“How much farther” compares two distances, so subtract the smaller from the larger. The tenths to take, 3, are fewer than the tenths there, 7, so no trade is needed:

\begin{aligned} 5\tfrac{7}{10} - 2\tfrac{3}{10} &= (5 - 2) + \left(\tfrac{7}{10} - \tfrac{3}{10}\right) \\ &= 3\tfrac{4}{10} \end{aligned}

Habiba runs 3\tfrac{4}{10} miles farther. Check by adding to the smaller distance: 2\tfrac{3}{10} + 3\tfrac{4}{10} = 5\tfrac{7}{10}.

Example 6

The rest of the course

The long-distance course at the meet is 5\tfrac{3}{8} miles. Ufuk has run 2\tfrac{6}{8} miles of it. How far does he still have to run?

Solution

Part of the course is done and the question asks what is left, so subtract: 5\tfrac{3}{8} - 2\tfrac{6}{8}. There are 3 eighths and 6 must be taken, so trade one of the 5 wholes for \tfrac{8}{8}:

5\tfrac{3}{8} = 4 + \tfrac{8}{8} + \tfrac{3}{8} = 4\tfrac{11}{8}

Now subtract wholes from wholes and eighths from eighths:

4\tfrac{11}{8} - 2\tfrac{6}{8} = 2\tfrac{5}{8}

since 4 - 2 = 2 and 11 - 6 = 5. Ufuk has 2\tfrac{5}{8} miles to go. Check by adding: 2\tfrac{6}{8} + 2\tfrac{5}{8} = 4\tfrac{11}{8} = 5\tfrac{3}{8}. The improper route agrees: \tfrac{43}{8} - \tfrac{22}{8} = \tfrac{21}{8} = 2\tfrac{5}{8}.

Example 7

A whole number of miles, two steps

Agata wants to run 7 miles during the meet week. She runs 1\tfrac{3}{4} miles on the first day and 2\tfrac{3}{4} miles on the second. How far does she still have to run?

Solution

Step one, the total so far. Add the two runs:

\begin{aligned} 1\tfrac{3}{4} + 2\tfrac{3}{4} &= 3 + \tfrac{6}{4} \\ &= 3 + 1\tfrac{2}{4} = 4\tfrac{2}{4} \end{aligned}

since \tfrac{6}{4} = \tfrac{4}{4} + \tfrac{2}{4}. Step two, what is left of the 7 miles. The whole number 7 has no quarters at all, so trade one whole for \tfrac{4}{4}: 7 = 6\tfrac{4}{4}. Then

6\tfrac{4}{4} - 4\tfrac{2}{4} = 2\tfrac{2}{4}

Agata has 2\tfrac{2}{4} miles still to run (also called 2\tfrac{1}{2} miles). Check: the three amounts should make the whole goal, and 1\tfrac{3}{4} + 2\tfrac{3}{4} + 2\tfrac{2}{4} = 5\tfrac{8}{4} = 7.

Boundaries

The denominators must be the same. Fifths add to fifths because the pieces are the same size. A mixed number in halves and one in thirds cannot be added part by part until their pieces are renamed to one size, which is the work of Adding mixed numbers and Subtracting mixed numbers.
The amounts must be of the same whole. Miles add to miles. A fraction of a mile and a fraction of a lap of the track are different wholes, and adding their numbers would name no real distance.
A finished answer has a fraction part less than 1. 4\tfrac{6}{5} is a true name for 5\tfrac{1}{5}, but its fraction part still holds a whole. Trading it to the wholes finishes the answer.
A trade moves exactly one whole, written in the problem’s pieces. In sixths, one whole is \tfrac{6}{6}; in eighths, \tfrac{8}{8}. Trading changes the name of the amount, never the amount.
You cannot take away more than there is. With distances run, 2\tfrac{1}{5} - 3\tfrac{4}{5} has no answer here: the second amount is bigger than the first, so no trade can help. Every difference in this lesson is 0 or more.
Simplifying is a choice. 1\tfrac{2}{6} and 1\tfrac{1}{3} are the same number, and either is correct.
The denominator is never 0. A whole cut into 0 pieces has no pieces to count.

Where this goes next

The next step is mixed numbers whose fraction parts have different denominators. Adding 1\tfrac{1}{2} miles and 2\tfrac{1}{4} miles first renames the halves as fourths, then goes on exactly as here, in Adding mixed numbers. Subtracting, with the same trading of a whole, is in Subtracting mixed numbers. Stories that mix joining, taking away and comparing, with any fractions, grow into Word problems that add or subtract fractions.

Exercises

Part 1: wholes with wholes, fraction parts with fraction parts

Find 3\tfrac{1}{4} + 2\tfrac{2}{4}.
Find 4\tfrac{2}{7} + 1\tfrac{3}{7}.
Find 6\tfrac{1}{9} + 5.
Find 2\tfrac{3}{10} + 4\tfrac{5}{10}.
Find 1\tfrac{1}{6} + 2\tfrac{2}{6} + 3\tfrac{1}{6}.
Ufuk walks 1\tfrac{2}{5} miles to the track meet and 1\tfrac{1}{5} miles home. How far does he walk?
Find the missing mixed number: 3\tfrac{2}{8} + ? = 7\tfrac{7}{8}.

Part 2: when the fraction parts make more than a whole

Find 2\tfrac{3}{4} + 1\tfrac{3}{4}.
Find 5\tfrac{4}{6} + 2\tfrac{5}{6}.
Find 1\tfrac{7}{10} + 3\tfrac{3}{10}.
Find 4\tfrac{2}{3} + 2\tfrac{2}{3}.
Write 3\tfrac{9}{5} as a finished mixed number.
Find 2\tfrac{4}{5} + 1\tfrac{3}{5} by first writing both as improper fractions. Then check with the wholes-and-parts route.
Agata runs 1\tfrac{5}{8} miles on Monday, 2\tfrac{6}{8} miles on Tuesday and 1\tfrac{7}{8} miles on Wednesday. How far does she run in the three days?

Part 3: subtracting, and word problems

Find 6\tfrac{5}{8} - 2\tfrac{3}{8}.
Find 5\tfrac{1}{4} - 1\tfrac{3}{4}.
Find 7\tfrac{2}{10} - 3\tfrac{9}{10}.
Find 4 - 1\tfrac{2}{3}.
Find 9\tfrac{4}{6} - 5\tfrac{4}{6}.
The fun run is 6\tfrac{1}{4} miles long. Habiba stops at a water table after 3\tfrac{3}{4} miles. How much farther does she have to go?
In one week Ufuk runs 4\tfrac{3}{10} miles and Agata runs 2\tfrac{7}{10} miles. How much farther does Ufuk run?
On meet day Habiba covers 5 miles in all. She walks 1\tfrac{4}{10} miles and runs 2\tfrac{8}{10} miles in races. The rest is a slow cool-down jog after the last race. How far does she jog to cool down?

Reasoning

To find 2\tfrac{3}{7} + 4\tfrac{1}{7}, Agata adds 2 + 4 and then \tfrac{3}{7} + \tfrac{1}{7}. Name the properties of addition that allow her to add the parts in that order, and explain what each one does here. Give the sum.
In 5\tfrac{2}{6} - 1\tfrac{5}{6}, explain why the fraction parts cannot be subtracted as they stand, and what trading one whole does. Give the answer.
Habiba says: “When I take a mixed number away from a whole number, I always have to trade a whole first, unless the mixed number has no fraction part.” Is she right? Explain, and show it with 6 - 2\tfrac{3}{8}.

Answers

5\tfrac{3}{4}. Wholes: 3 + 2 = 5. Quarters: 1 + 2 = 3.
5\tfrac{5}{7}. Wholes: 4 + 1 = 5. Sevenths: 2 + 3 = 5.
11\tfrac{1}{9}. Wholes: 6 + 5 = 11; the only ninths are the \tfrac{1}{9}.
6\tfrac{8}{10}. Wholes: 2 + 4 = 6. Tenths: 3 + 5 = 8.
6\tfrac{4}{6}. Wholes: 1 + 2 + 3 = 6. Sixths: 1 + 2 + 1 = 4.
2\tfrac{3}{5} miles. 1\tfrac{2}{5} + 1\tfrac{1}{5} = (1 + 1) + \tfrac{3}{5}.
4\tfrac{5}{8}. The missing amount is 7\tfrac{7}{8} - 3\tfrac{2}{8}: wholes 7 - 3 = 4, eighths 7 - 2 = 5. Check: 3\tfrac{2}{8} + 4\tfrac{5}{8} = 7\tfrac{7}{8}.
4\tfrac{2}{4}. Wholes and parts give 3\tfrac{6}{4}, and \tfrac{6}{4} = 1\tfrac{2}{4}, so 3 + 1\tfrac{2}{4} = 4\tfrac{2}{4}.
8\tfrac{3}{6}. Wholes and parts give 7\tfrac{9}{6}, and \tfrac{9}{6} = 1\tfrac{3}{6}, so 7 + 1\tfrac{3}{6} = 8\tfrac{3}{6}.
5. Wholes and parts give 4\tfrac{10}{10}, and \tfrac{10}{10} = 1, so the sum is exactly 5.
7\tfrac{1}{3}. Wholes and parts give 6\tfrac{4}{3}, and \tfrac{4}{3} = 1\tfrac{1}{3}, so 6 + 1\tfrac{1}{3} = 7\tfrac{1}{3}.
4\tfrac{4}{5}. \tfrac{9}{5} = \tfrac{5}{5} + \tfrac{4}{5} = 1\tfrac{4}{5}, so 3\tfrac{9}{5} = 3 + 1\tfrac{4}{5} = 4\tfrac{4}{5}.
4\tfrac{2}{5}. 2\tfrac{4}{5} = \tfrac{14}{5} and 1\tfrac{3}{5} = \tfrac{8}{5}, so the sum is \tfrac{22}{5}, and 22 \div 5 = 4 \text{ R } 2. Check: wholes and parts give 3\tfrac{7}{5}, and \tfrac{7}{5} = 1\tfrac{2}{5}, so 4\tfrac{2}{5} again.
6\tfrac{2}{8} miles. Wholes: 1 + 2 + 1 = 4. Eighths: 5 + 6 + 7 = 18, and \tfrac{18}{8} = \tfrac{8}{8} + \tfrac{8}{8} + \tfrac{2}{8} = 2\tfrac{2}{8}. So 4 + 2\tfrac{2}{8} = 6\tfrac{2}{8}.
4\tfrac{2}{8}. Wholes: 6 - 2 = 4. Eighths: 5 - 3 = 2. No trade needed.
3\tfrac{2}{4}. Trade one whole: 5\tfrac{1}{4} = 4\tfrac{5}{4}. Then 4\tfrac{5}{4} - 1\tfrac{3}{4} = 3\tfrac{2}{4}.
3\tfrac{3}{10}. Trade one whole: 7\tfrac{2}{10} = 6\tfrac{12}{10}. Then 6\tfrac{12}{10} - 3\tfrac{9}{10} = 3\tfrac{3}{10}.
2\tfrac{1}{3}. Write 4 = 3\tfrac{3}{3}. Then 3\tfrac{3}{3} - 1\tfrac{2}{3} = 2\tfrac{1}{3}.
4. Wholes: 9 - 5 = 4. Sixths: 4 - 4 = 0, so there is no fraction part.
2\tfrac{2}{4} miles. Trade one whole: 6\tfrac{1}{4} = 5\tfrac{5}{4}, and 5\tfrac{5}{4} - 3\tfrac{3}{4} = 2\tfrac{2}{4}. Check: 3\tfrac{3}{4} + 2\tfrac{2}{4} = 5\tfrac{5}{4} = 6\tfrac{1}{4}.
1\tfrac{6}{10} miles farther. Trade one whole: 4\tfrac{3}{10} = 3\tfrac{13}{10}, and 3\tfrac{13}{10} - 2\tfrac{7}{10} = 1\tfrac{6}{10}.
\tfrac{8}{10} of a mile of cool-down jogging. Walking and races: 1\tfrac{4}{10} + 2\tfrac{8}{10} = 3\tfrac{12}{10} = 4\tfrac{2}{10} miles. The rest: 5 = 4\tfrac{10}{10}, and 4\tfrac{10}{10} - 4\tfrac{2}{10} = \tfrac{8}{10}.
6\tfrac{4}{7}. The sum is 2 + \tfrac{3}{7} + 4 + \tfrac{1}{7}. The commutative property lets her swap \tfrac{3}{7} and 4, so the order becomes 2 + 4 + \tfrac{3}{7} + \tfrac{1}{7} without changing the total. The associative property lets her group (2 + 4) and \left(\tfrac{3}{7} + \tfrac{1}{7}\right) and add each group first. Then 6 + \tfrac{4}{7} = 6\tfrac{4}{7}.
3\tfrac{3}{6}. The first number has only 2 sixths in its fraction part, and 5 sixths must be taken, so there are not enough loose sixths. But each whole is \tfrac{6}{6}. Trading one whole gives 5\tfrac{2}{6} = 4 + \tfrac{6}{6} + \tfrac{2}{6} = 4\tfrac{8}{6}, the same amount with more sixths showing. Then 4\tfrac{8}{6} - 1\tfrac{5}{6} = 3\tfrac{3}{6}.
Yes. A whole number has 0 pieces in its fraction part, and any mixed number with a fraction part takes away at least one piece, so a whole must be traded for pieces first. With no fraction part, only wholes are taken and no trade is needed. Example: 6 = 5\tfrac{8}{8}, so 6 - 2\tfrac{3}{8} = 5\tfrac{8}{8} - 2\tfrac{3}{8} = 3\tfrac{5}{8}.