Adding and taking away fractions

What this les­son is about#

Frac­tions turn up the moment any­thing is shared. Half a litre of milk, and then a quar­ter litre more. Three quar­ters of an hour on home­work and two thirds of an hour on cricket. A recipe that asks for one and a half cups of rice when you only want to cook for fewer peo­ple. Every time, you are qui­etly adding or tak­ing away frac­tions.

So how is it done? We'll begin with the easy case, where the pieces are already the same size, and then see what to do when they aren't. By the end you'll be com­fort­able adding and tak­ing away any two or three frac­tions, whole num­bers included, and you'll even be able to write the rule in let­ters.

Pic­ture a cake cut into three equal pieces. Each piece is $\displaystyle \frac{1}{3}$ of the cake. Two pieces are $\displaystyle \frac{2}{3}$. Adding frac­tions is really just count­ing pieces.

You met ratio­nal num­bers in Ratio­nal num­bers. A ratio­nal num­ber is any num­ber you can write as $\displaystyle \frac{p}{q}$, where $p$ and $q$ are inte­gers and $q \neq 0$. So $\displaystyle \frac{2}{3}$ is one, and so is $\displaystyle -\frac{7}{8}$. Frac­tion is the every­day word for it, so that is the word we'll use here.

When the bot­tom num­bers match#

Every frac­tion has a top num­ber and a bot­tom num­ber. The bot­tom num­ber tells you the size of each piece. The top num­ber tells you how many pieces you have.

These two num­bers have proper names: the top one is the numer­a­tor and the bot­tom one is the denom­i­na­tor. Mostly, though, we'll just say top and bot­tom.

You can have more pieces than one cake holds. Put a sec­ond cake beside the first. Cut it into three as well. Now there are six pieces to take from. Five of them is $\displaystyle \frac{5}{3}$, which is one whole cake and two pieces over.

In $\displaystyle \frac{2}{3}$ and $\displaystyle \frac{5}{3}$ the bot­tom num­ber is 3, so both are made of thirds. The pieces are the same size, and that is exactly why you can count them together.

Two thirds and five thirds make seven thirds. You add the top num­bers and leave the bot­tom num­ber as it is.

Two thirds plus five thirds

$$\displaystyle \begin{aligned}\frac{2}{3} + \frac{5}{3} &= \frac{2 + 5}{3} \\ &= \frac{7}{3}\end{aligned}$$

Seven thirds is more than one cake, and that is per­fectly fine. It just means two whole cakes and one piece over.

Why did­n't the bot­tom num­ber change? Because you never cut the cake again. The pieces are still thirds; you sim­ply have more of them.

Tak­ing away works the same way. You take one top num­ber from the other.

Some­times you are asked to give away more than you have. You hand over every­thing, and you still owe the rest, a bit like bor­row­ing from a friend. We write what you owe with a minus sign in front, so owing one whole cake is writ­ten $-1$.

Five thirds taken from two thirds

$$\displaystyle \begin{aligned}\frac{2}{3} - \frac{5}{3} &= \frac{2 - 5}{3} \\ &= \frac{-3}{3} \\ &= -1\end{aligned}$$

You had two thirds. You gave both away, and you still owe three more thirds. Three thirds make one whole cake. So you owe one whole cake, and we write that as $-1$.

The minus sign can sit on the top num­ber, or in front of the whole frac­tion. $\displaystyle \frac{-3}{3}$ and $\displaystyle -\frac{3}{3}$ mean the same amount. Fin­ished answers in this les­son put the minus in front. It tells you the whole amount is owed.

Remem­ber. Same bot­tom num­ber? Add or take away the top num­bers. Leave the bot­tom num­ber alone.

When the bot­tom num­bers are dif­fer­ent#

Now pic­ture two cakes of the same size, one cut into quar­ters and the other into sixths. The pieces are no longer the same size.

You can't count pieces of two dif­fer­ent sizes together, so you cut both cakes again until every piece is the same size.

The new size has to fit both cakes exactly. The small­est num­ber that both bot­tom num­bers divide into is the one you want. Its short name is the L.C.M, or low­est com­mon mul­ti­ple.

Find­ing the L.C.M with a lad­der#

A lad­der is a quick way to find it. You write the num­bers in rows going down, like the steps of a lad­der, and each row comes from the one above it.

A fac­tor of a num­ber is a num­ber that goes into it exactly. The fac­tors of 6 are 1, 2, 3 and 6.

Here is how you climb down. Divide by a num­ber that goes into at least two of them, and copy down again any num­ber it won't divide. Stop when no num­ber goes into two or more of them.

Then mul­ti­ply together the num­bers you divided by, along with the num­bers left at the bot­tom. You'll see this hap­pen in the sums below.

Here is why it works. Each time you divide, you take out a part the num­bers share. When noth­ing is shared any more, you have found every part just once. Mul­ti­ply them all back together and you get the small­est num­ber both bot­tom num­bers fit into.

Adding five quar­ters and seven sixths#

First build the lad­der for 4 and 6.

Noth­ing goes into both 2 and 3, so you stop. Mul­ti­ply the 2 you used by the 2 and 3 left at the bot­tom. That gives $2 \times 2 \times 3 = 12$.

So every piece will be a twelfth. Four goes into twelve three times. Six goes into twelve two times.

Mul­ti­ply the top and bot­tom of $\displaystyle \frac{5}{4}$ by 3. Mul­ti­ply the top and bot­tom of $\displaystyle \frac{7}{6}$ by 2.

Is this allowed? Yes, com­pletely. You have only cut each piece into smaller pieces. There are more of them now, but each one is smaller, so the amount of cake stays the same.

Five quar­ters plus seven sixths

$$\displaystyle \begin{aligned}\frac{5}{4} + \frac{7}{6} &= \frac{5 \times 3}{4 \times 3} + \frac{7 \times 2}{6 \times 2} \\ &= \frac{15}{12} + \frac{14}{12} \\ &= \frac{15 + 14}{12} \\ &= \frac{29}{12}\end{aligned}$$

Both are twelfths now, so you can count them. Fif­teen pieces and four­teen pieces make twenty-nine pieces.

Tak­ing seven twelfths from five eighths#

Build the lad­der for 8 and 12. This one takes two rows.

You divided by 2 twice. The num­bers 2 and 3 are left at the bot­tom. So the L.C.M is $2 \times 2 \times 2 \times 3 = 24$.

Eight goes into twenty-four three times. Twelve goes into twenty-four two times.

Five eighths minus seven twelfths

$$\displaystyle \begin{aligned}\frac{5}{8} - \frac{7}{12} &= \frac{5 \times 3}{8 \times 3} - \frac{7 \times 2}{12 \times 2} \\ &= \frac{15}{24} - \frac{14}{24} \\ &= \frac{15 - 14}{24} \\ &= \frac{1}{24}\end{aligned}$$

The two amounts turn out to be very close; the gap between them is just one twenty-fourth.

Adding five thirds and eight sev­enths#

Three and seven share no fac­tor; only 1 goes into both. With noth­ing to take out, the lad­der has no work to do.

The size that fits both is sim­ply the two bot­tom num­bers mul­ti­plied together, here $3 \times 7 = 21$. Twenty-one pieces split evenly into three parts, and just as evenly into seven.

Five thirds plus eight sev­enths

$$\displaystyle \begin{aligned}\frac{5}{3} + \frac{8}{7} &= \frac{5 \times 7}{3 \times 7} + \frac{8 \times 3}{7 \times 3} \\ &= \frac{35}{21} + \frac{24}{21} \\ &= \frac{35 + 24}{21} \\ &= \frac{59}{21}\end{aligned}$$

Tak­ing eight fifths from two sev­enths#

Seven and five share no fac­tor either. So the L.C.M is $7 \times 5 = 35$.

Two sev­enths minus eight fifths

$$\displaystyle \begin{aligned}\frac{2}{7} - \frac{8}{5} &= \frac{2 \times 5}{7 \times 5} - \frac{8 \times 7}{5 \times 7} \\ &= \frac{10}{35} - \frac{56}{35} \\ &= \frac{10 - 56}{35} \\ &= -\frac{46}{35}\end{aligned}$$

$\displaystyle \frac{2}{7}$ is a small part of one whole, while $\displaystyle \frac{8}{5}$ is more than one whole. You can't give away that much, so you end up owing, and that is why the answer car­ries a minus sign.

One more worked exam­ple: tenths and fif­teenths#

Here is a pair whose L.C.M is not sim­ply the two bot­tom num­bers mul­ti­plied. Add $\displaystyle \frac{3}{10}$ and $\displaystyle \frac{7}{15}$.

Both 10 and 15 divide by 5. Divide once by 5 and you are left with 2 and 3, which share noth­ing. So the L.C.M is $5 \times 2 \times 3 = 30$. Ten goes into thirty three times. Fif­teen goes into thirty two times.

$$\displaystyle \begin{aligned}\frac{3}{10} + \frac{7}{15} &= \frac{3 \times 3}{10 \times 3} + \frac{7 \times 2}{15 \times 2} \\ &= \frac{9}{30} + \frac{14}{30} \\ &= \frac{23}{30}\end{aligned}$$

Mul­ti­ply­ing the bot­tom num­bers would have given 150. That works too, but you would then have to can­cel $\displaystyle \frac{115}{150}$ by 5 to reach the same answer. The lad­der saves that step.

A whole num­ber and a frac­tion#

The num­bers in this sec­tion, like $1$, $2$, $3$ and $5$, have no pieces left over. Think of a whole cake, not a slice of one. Even so, any of them can be writ­ten as a frac­tion.

The trick is to put the whole num­ber over 1. So $\displaystyle 5 = \frac{5}{1}$ and $\displaystyle 1 = \frac{1}{1}$.

This works because of what the bot­tom num­ber means: how many pieces one whole is cut into. Cut a cake into one piece and, well, it is still whole.

Once it looks like a frac­tion, you work exactly as before.

Five thirds plus one

$$\displaystyle \begin{aligned}\frac{5}{3} + 1 &= \frac{5}{3} + \frac{1}{1} \\ &= \frac{5}{3} + \frac{3}{3} \\ &= \frac{5 + 3}{3} \\ &= \frac{8}{3}\end{aligned}$$

Five thirds minus one

$$\displaystyle \begin{aligned}\frac{5}{3} - 1 &= \frac{5}{3} - \frac{1}{1} \\ &= \frac{5}{3} - \frac{3}{3} \\ &= \frac{5 - 3}{3} \\ &= \frac{2}{3}\end{aligned}$$

One whole is three thirds. That is why the 1 turned into $\displaystyle \frac{3}{3}$.

Five plus seven sixths

$$\displaystyle \begin{aligned}5 + \frac{7}{6} &= \frac{5}{1} + \frac{7}{6} \\ &= \frac{30}{6} + \frac{7}{6} \\ &= \frac{30 + 7}{6} \\ &= \frac{37}{6}\end{aligned}$$

Five minus seven sixths

$$\displaystyle \begin{aligned}5 - \frac{7}{6} &= \frac{5}{1} - \frac{7}{6} \\ &= \frac{30}{6} - \frac{7}{6} \\ &= \frac{30 - 7}{6} \\ &= \frac{23}{6}\end{aligned}$$

Five wholes is thirty sixths. Each whole holds six sixths, and $5 \times 6 = 30$.

Three frac­tions at once#

The same steps work just as well for three frac­tions, and you can do the whole line in one go.

Three like frac­tions in one line

$$\displaystyle \begin{aligned}\frac{5}{3} + \frac{7}{3} - \frac{4}{3} &= \frac{5 + 7 - 4}{3} \\ &= \frac{8}{3}\end{aligned}$$

Those bot­tom num­bers already matched. When they do not, take the L.C.M of all three.

Here is the lad­der for 6, 8 and 12. A num­ber that will not divide just waits. Copy it down again and try the next row.

Noth­ing now goes into two of them, so you stop. Mul­ti­ply the num­bers down the side, and the 2 left at the bot­tom. That gives $2 \times 2 \times 3 \times 2 = 24$.

Six goes into twenty-four four times. Eight goes into it three times. Twelve goes into it two times.

Three unlike frac­tions in one line

$$\displaystyle \begin{aligned}\frac{5}{6} + \frac{7}{8} - \frac{3}{12} &= \frac{5 \times 4}{6 \times 4} + \frac{7 \times 3}{8 \times 3} - \frac{3 \times 2}{12 \times 2} \\ &= \frac{20}{24} + \frac{21}{24} - \frac{6}{24} \\ &= \frac{20 + 21 - 6}{24} \\ &= \frac{35}{24}\end{aligned}$$

The next sum has owed amounts in it, so think about money for a moment. If some­one can­cels a debt you owe, you are bet­ter off, so tak­ing away an owed amount turns into adding. Pile on another debt and you are worse off, so adding an owed amount turns into tak­ing away.

First the lad­der for 9, 12 and 18. This one starts by divid­ing by 3.

Noth­ing goes into two of them now, so you stop. Mul­ti­ply the num­bers down the side, and the 2 left at the bot­tom. That gives $3 \times 2 \times 3 \times 2 = 36$.

Nine goes into thirty-six four times. Twelve goes into it three times. Eigh­teen goes into it two times.

Six­teen ninths, with two owed amounts

$$\displaystyle \begin{aligned}\frac{16}{9} - \left( -\frac{5}{12} \right) + \left( -\frac{7}{18} \right) &= \frac{16}{9} + \frac{5}{12} - \frac{7}{18} \\ &= \frac{16 \times 4}{9 \times 4} + \frac{5 \times 3}{12 \times 3} - \frac{7 \times 2}{18 \times 2} \\ &= \frac{64}{36} + \frac{15}{36} - \frac{14}{36} \\ &= \frac{64 + 15 - 14}{36} \\ &= \frac{65}{36}\end{aligned}$$

Notice that both signs were sorted out on the very first line. After that it is the same job as before.

Two more worked exam­ples#

A whole num­ber minus a small frac­tion. Take $\displaystyle \frac{3}{8}$ away from 2. Two wholes are six­teen eighths, because $2 \times 8 = 16$.

$$\displaystyle \begin{aligned}2 - \frac{3}{8} &= \frac{16}{8} - \frac{3}{8} \\ &= \frac{13}{8}\end{aligned}$$

Two owed amounts added. Add $\displaystyle -\frac{2}{5}$ and $\displaystyle -\frac{3}{10}$. Both amounts are owed, so the total owed grows. The L.C.M of 5 and 10 is 10, since 5 already goes into 10.

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

You owed four tenths and then three tenths more, so you owe seven tenths alto­gether.

Three frac­tions with a mix of signs. Work out $\displaystyle \frac{3}{4} - \frac{5}{6} + \frac{1}{3}$. The L.C.M of 4, 6 and 3 is 12.

$$\displaystyle \begin{aligned}\frac{3}{4} - \frac{5}{6} + \frac{1}{3} &= \frac{9}{12} - \frac{10}{12} + \frac{4}{12} \\ &= \frac{9 - 10 + 4}{12} \\ &= \frac{3}{12} \\ &= \frac{1}{4}\end{aligned}$$

This answer can­cels, because 3 and 12 both divide by 3. Make it a habit: glance for a com­mon fac­tor before you write the final answer.

Prac­tice#

Try these on paper. The answers come after them.

The first answer can be writ­ten more sim­ply. Both 12 and 9 divide by 3, so $\displaystyle \frac{12}{9}$ becomes $\displaystyle \frac{4}{3}$. Mak­ing a frac­tion smaller like this is called can­celling. The amount does not change, only the way you write it.

The same rules in let­ters#

Let­ters stand for any num­bers you like. These eight rules say what you have just done.

Look at the third rule first. $\displaystyle \frac{5}{3}$ and $\displaystyle \frac{8}{7}$ shared no fac­tor, so you mul­ti­plied 3 by 7. You mul­ti­plied the top of the first by the other bot­tom. Then you mul­ti­plied the top of the sec­ond by the first bot­tom. In let­ters that is $a \times d$ and $c \times b$.

Rules 3 and 4 use $b \times d$ as the new bot­tom num­ber. That always works. Both bot­tom num­bers divide into it.

But it is not always the small­est one. So the answer may still need can­celling. Watch what rule 3 does to a sum you have already met.

The same sum done by rule three

$$\displaystyle \begin{aligned}\frac{5}{6} + \frac{3}{8} &= \frac{5 \times 8 + 6 \times 3}{6 \times 8} \\ &= \frac{40 + 18}{48} \\ &= \frac{58}{48} \\ &= \frac{29}{24}\end{aligned}$$

That matches the prac­tice answer. Both 58 and 48 divide by 2. So $\displaystyle \frac{58}{48}$ can­cels down to $\displaystyle \frac{29}{24}$.

The L.C.M route gave 24 straight away, which is exactly why it is worth the effort of learn­ing.

Remem­ber. You can only add frac­tions when the pieces are the same size. Mak­ing the bot­tom num­bers match is the whole job.

Com­mon mis­takes#

• Adding the bot­tom num­bers as well as the top ones. $\displaystyle \frac{1}{3} + \frac{1}{3}$ is $\displaystyle \frac{2}{3}$, not $\displaystyle \frac{2}{6}$. The size of the pieces does not change when you count them.
• Mul­ti­ply­ing only the bot­tom num­ber when chang­ing a frac­tion. If the bot­tom is mul­ti­plied by 3, the top must be mul­ti­plied by 3 too, or the amount changes.
• For­get­ting that a whole num­ber has a bot­tom num­ber of 1, and adding it straight to the top: $\displaystyle \frac{5}{3} + 1$ is not $\displaystyle \frac{6}{3}$.
• Los­ing a minus sign. Tak­ing away an owed amount turns into adding. Adding an owed amount turns into tak­ing away.
• Leav­ing an answer that can still be can­celled, such as $\displaystyle \frac{12}{9}$ instead of $\displaystyle \frac{4}{3}$.

Key terms#

Ratio­nal num­ber

A num­ber that can be writ­ten as $\displaystyle \frac{p}{q}$, where $p$ and $q$ are inte­gers and $q \neq 0$.

Numer­a­tor

The top num­ber of a frac­tion. It tells you how many pieces you have.

Denom­i­na­tor

The bot­tom num­ber of a frac­tion. It tells you how many equal pieces one whole is cut into.

Like frac­tions

Frac­tions with the same bot­tom num­ber, so their pieces are the same size.

L.C.M

The low­est com­mon mul­ti­ple: the small­est num­ber that each of the given num­bers divides into exactly.

Lad­der

A table of repeated divi­sion used to find the L.C.M of two or more num­bers.

Can­celling

Divid­ing the top and bot­tom of a frac­tion by the same num­ber, so it is writ­ten more sim­ply with­out chang­ing its value.

Answers#

These answers match the prac­tice table above, with the work­ing short­ened.

1. $\displaystyle \frac{7}{9} + \frac{5}{9} = \frac{12}{9} = \frac{4}{3}$
2. $\displaystyle \frac{7}{9} - \frac{5}{9} = \frac{2}{9}$
3. $\displaystyle \frac{5}{6} + \frac{3}{8} = \frac{20}{24} + \frac{9}{24} = \frac{29}{24}$
4. $\displaystyle \frac{5}{6} - \frac{3}{8} = \frac{20}{24} - \frac{9}{24} = \frac{11}{24}$
5. $\displaystyle \frac{5}{3} + \frac{7}{8} = \frac{40}{24} + \frac{21}{24} = \frac{61}{24}$
6. $\displaystyle \frac{5}{3} - \frac{7}{8} = \frac{40}{24} - \frac{21}{24} = \frac{19}{24}$
7. $\displaystyle \frac{5}{3} + 6 = \frac{5}{3} + \frac{18}{3} = \frac{23}{3}$
8. $\displaystyle \frac{5}{3} - 6 = \frac{5}{3} - \frac{18}{3} = -\frac{13}{3}$
9. $\displaystyle 3 + \frac{5}{7} = \frac{21}{7} + \frac{5}{7} = \frac{26}{7}$
10. $\displaystyle 3 - \frac{5}{7} = \frac{21}{7} - \frac{5}{7} = \frac{16}{7}$