Adding and subtracting expressions

Alge­bra lets you write a rule once and use it for every num­ber. Before long you meet two expres­sions that need to be put together: the perime­ter of a field whose sides are $2x + 3$ and $x + 5$, the total cost of two orders, or the dif­fer­ence between two peo­ple's sav­ings. Adding and sub­tract­ing expres­sions is the skill behind all of these, and it rests on one sim­ple idea: you can only com­bine things that count the same thing.

An expres­sion is a piece of alge­bra made of terms joined by plus or minus signs, like $3a + 2$.

Two expres­sions can be put together to make one. You can add them. You can also take one away from the other.

This les­son shows you both jobs. It also shows the one step that catches most peo­ple out. That step is a minus sign in front of a bracket.

Only like terms join

Terms, co-effi­cients, like terms and unlike terms are taught in the Early Years course Intro­duc­tion to Alge­bra. Read that course again if they feel shaky.

Look at $3a$ and $2a$. Both of them count the same thing, $a$. So $3$ of them and $2$ more make $5$ of them. You can write $3a + 2a = 5a$.

Terms that count the same let­ter are called like terms. Terms that count dif­fer­ent let­ters are unlike terms.

The num­ber in front of a term is its co-effi­cient. In $3a$ the co-effi­cient is $3$.

Remem­ber. Only like terms add or sub­tract. Unlike terms stay side by side in the answer.

A good way to pic­ture this is with tiles. Let a large blue tile stand for $x$ and a small orange tile stand for $y$. You can count blue tiles with blue tiles, and orange with orange, but a blue tile and an orange tile never merge into a new kind of tile.

The hor­i­zon­tal method

Hor­i­zon­tal means you write every­thing along one line.

Add $3a$ and $2a$

$$\begin{aligned}&3a + 2a \\ &= (3 + 2)a \\ &= 5a\end{aligned}$$

The mid­dle step shows you why the answer is $5a$. The $a$ was taken out­side, and $3$ and $2$ were added.

That is the whole method. You gather the like terms, then you add the num­bers in front of them.

The next sum has two let­ters in it.

Add $4x + 2y$ and $3x + 6y$

$$\begin{aligned}&4x + 2y + 3x + 6y \\ &= 4x + 3x + 2y + 6y \\ &= (4 + 3)x + (2 + 6)y \\ &= 7x + 8y\end{aligned}$$

There were two piles this time. The $x$ terms made $7x$ and the $y$ terms made $8y$.

Those two can­not join, so the answer is $7x + 8y$.

A longer exam­ple with a minus sign inside

Add $5x - 3y + 2$ and $2x + y - 7$. Each term car­ries the sign writ­ten in front of it, so $-3y$ is a neg­a­tive term and $-7$ is a neg­a­tive num­ber.

$$\begin{aligned}&5x - 3y + 2 + 2x + y - 7 \\ &= 5x + 2x - 3y + y + 2 - 7 \\ &= (5 + 2)x + (-3 + 1)y + (2 - 7) \\ &= 7x - 2y - 5\end{aligned}$$

When you move a term to sit beside its part­ners, its sign trav­els with it. That is why $-3y$ stays neg­a­tive in the sec­ond line.

A printed line worth check­ing

The chap­ter works one more sum along a line. This is how the book prints it.

Add $3a + 2$ and $5a + 7$, printed as the chap­ter has it

$$\begin{aligned}&3a + 2 + 5a + 7 \\ &= 7a + 9\end{aligned}$$

That printed answer does not check out. Gather the $a$ terms first. Here $3 + 5 = 8$, so they give $8a$, not $7a$.

The plain num­bers are fine, because $2 + 7 = 9$.

Work the sum your­self and you get $8a + 9$. The acad­emy is being asked which line is right.

Check­ing a printed answer is a good habit. Do the same with your own work.

The ver­ti­cal method

Ver­ti­cal means you write one expres­sion under the other.

The chap­ter files this next sum under the ver­ti­cal method, so it is done that way here.

Add $2a + 3b$ and $5a + 4b$. Set them out in two columns, one for $a$ and one for $b$.

In the $a$ col­umn, $2 + 5 = 7$. In the $b$ col­umn, $3 + 4 = 7$.

Like terms sat in the same col­umn. Each col­umn was then added on its own.

The answer is $7a + 7b$. The two parts can­not join, so both stay.

Every $a$ term is in the $a$ col­umn and every $b$ term is in the $b$ col­umn. A term can­not drift into the wrong pile.

Which method to use

Both meth­ods give the same answer. You pick the one that is eas­ier to hold in your head.

Long expres­sions are easy to mud­dle. Columns keep every term in its place.

Sub­tract­ing

Sub­trac­tion fol­lows the same rule. Only like terms can be taken away from each other.

Sub­tract $2a$ from $7a$

$$\begin{aligned}&7a - 2a \\ &= (7 - 2)a \\ &= 5a\end{aligned}$$

Read the order with care. Sub­tract $2a$ from $7a$ means you start at $7a$.

The expres­sion you are tak­ing away is writ­ten sec­ond. Swap­ping them gives $2a - 7a$, which is $-5a$, not $5a$.

Order mat­ters in sub­trac­tion in a way it does not in addi­tion.

A sub­trac­tion where the answer has a minus sign

Sub­tract $4p$ from $p$. Start at $p$, which is $1p$.

$$\begin{aligned}&p - 4p \\ &= (1 - 4)p \\ &= -3p\end{aligned}$$

A neg­a­tive answer is per­fectly nor­mal in alge­bra. It sim­ply means you took away more than you started with.

Tak­ing away a whole bracket

Look at $3(2a + 5)$. The $3$ mul­ti­plies the $2a$ to give $6a$. It also mul­ti­plies the $5$ to give $15$. So $3(2a + 5) = 6a + 15$.

Writ­ten in gen­eral, that is $a(b + c) = ab + ac$, which is how the chap­ter prints it.

It says that the thing out­side a bracket mul­ti­plies every term inside it. Not just the first term. Every one.

This rule has a name: the dis­trib­u­tive rule.

A minus sign in front of a bracket is a $-1$ sit­ting out­side it.

That $-1$ mul­ti­plies every term inside, just like any other num­ber out­side a bracket. And $-1$ times a pos­i­tive term gives a neg­a­tive one. So $-(b + c)$ is $-b - c$. Every sign inside flips over.

This is the hard­est step in the les­son. The safe habit is to write the brack­ets in first.

Sub­tract $2a + 3$ from $9a + 8$

$$\begin{aligned}&(9a + 8) - (2a + 3) \\ &= 9a + 8 - 2a - 3 \\ &= 9a - 2a + 8 - 3 \\ &= (9 - 2)a + (8 - 3) \\ &= 7a + 5\end{aligned}$$

Look at step two. The $+3$ turned into $-3$ when the bracket came off.

For­get that flip and you add the $3$ instead. Then $8 + 3 = 11$, and you get $7a + 11$, which is wrong.

A bracket with a minus sign already inside

Sub­tract $4p - 2q$ from $p + 5q$. This time the bracket holds a minus sign, and it flips to a plus.

$$\begin{aligned}&(p + 5q) - (4p - 2q) \\ &= p + 5q - 4p + 2q \\ &= (1 - 4)p + (5 + 2)q \\ &= -3p + 7q\end{aligned}$$

Tak­ing away a neg­a­tive amount is the same as adding it, so $-(-2q)$ became $+2q$.

Remem­ber. Take away a whole bracket and every sign inside it changes. A plus becomes a minus, and a minus becomes a plus.

Let­ters that stand for whole expres­sions

A cap­i­tal let­ter can stand for a whole expres­sion. $A$ might mean $3a + 2b$, all of it.

The chap­ter asks you to find $2A - 3B$ and $(A + B) - (A - B)$.

Its own $A$ and $B$ were drawn on the page and are lost. Take these two instead.

Take $A = 3a + 2b$ and $B = a + b$.

Put the expres­sion in place of each cap­i­tal let­ter. Keep the brack­ets while you do it.

Putting an expres­sion in place of a let­ter like this is called sub­sti­tu­tion.

Find $2A - 3B$

$$\begin{aligned}&2A - 3B \\ &= 2(3a + 2b) - 3(a + b) \\ &= 6a + 4b - 3a - 3b \\ &= 6a - 3a + 4b - 3b \\ &= (6 - 3)a + (4 - 3)b \\ &= 3a + b\end{aligned}$$

The $3$ out­side the sec­ond bracket hit both terms inside. The minus sign flipped both of them.

The next one needs two small sums first. In $A + B$ the let­ters give $3 + 1 = 4$ lots of $a$ and $2 + 1 = 3$ lots of $b$.

In $A - B$ they give $3 - 1 = 2$ lots of $a$ and $2 - 1 = 1$ lot of $b$.

Find $(A + B) - (A - B)$

$$\begin{aligned}&(A + B) - (A - B) \\ &= (3a + 2b + a + b) - (3a + 2b - a - b) \\ &= (4a + 3b) - (2a + b) \\ &= 4a + 3b - 2a - b \\ &= (4 - 2)a + (3 - 1)b \\ &= 2a + 2b\end{aligned}$$

The answer $2a + 2b$ is just $B$ dou­bled.

That is no acci­dent. The sec­ond bracket held $-B$, and tak­ing away $-B$ adds $B$ back.

So there is one $B$ from the first bracket and one more from the sec­ond. Two lots of $B$ make $2B$.

The first bracket gave $+A$ and the sec­ond gave $-A$. A term and the same term with a minus sign make noth­ing. So the two lots of $A$ can­celled out.

This works for any $A$ and any $B$ you choose.

Sub­tract­ing in columns

The chap­ter now takes $3x + 4y$ away from another expres­sion. That expres­sion was drawn and is lost, so take $9x + 7y$.

The bot­tom row is the whole expres­sion you are tak­ing away. It is the same as a bracket with a minus sign in front.

Change the sign of every term in that row, then add down.

In the $x$ col­umn, $9 - 3 = 6$. In the $y$ col­umn, $7 - 4 = 3$.

The answer is $6x + 3y$.

Chang­ing the signs first is the same flip as before. The columns sim­ply keep it tidy.

Prac­tice

Try these. Write the brack­ets in first on every sub­trac­tion.

1. Add $4x$ and $9x$.
2. Add $2m + 5$ and $6m + 1$.
3. Add $3p + 2q$ and $4p + 6q$. Use the ver­ti­cal method.
4. Sub­tract $5y$ from $12y$.
5. Sub­tract $2a + 1$ from $6a + 9$.
6. Sub­tract $3x + 4y$ from $10x + 4y$.
7. Add $5a - 2b$ and $a + 7b$.
8. With $A = 3a + 2b$ and $B = a + b$, find $A + 2B$.

Remem­ber. Gather like terms, add or sub­tract their num­bers, and leave unlike terms alone. Off comes the bracket, over go the signs.

Com­mon mis­takes

• Join­ing unlike terms, for exam­ple writ­ing $3a + 2b = 5ab$. The answer stays $3a + 2b$.
• Chang­ing only the first sign when a bracket is taken away: $-(2a + 3)$ is $-2a - 3$, not $-2a + 3$.
• Read­ing "sub­tract $A$ from $B$" as $A - B$. The expres­sion after "from" is writ­ten first.
• Leav­ing a sign behind when terms are moved about. A term and its sign always travel together.
• For­get­ting that a let­ter on its own, such as $p$, has co-effi­cient $1$.
• In the ver­ti­cal method, putting a term in the wrong col­umn, or for­get­ting to change the signs of the bot­tom row before adding down.

Key terms

Expres­sion

Terms joined by plus or minus signs, such as $3a + 2$.

Term

One part of an expres­sion, together with the sign in front of it.

Co-effi­cient

The num­ber in front of a term; in $3a$ it is $3$.

Like terms

Terms that count the same let­ter, such as $3a$ and $2a$.

Unlike terms

Terms that count dif­fer­ent let­ters, such as $3a$ and $2b$.

Dis­trib­u­tive rule

$a(b + c) = ab + ac$: the num­ber out­side a bracket mul­ti­plies every term inside.

Sub­sti­tu­tion

Putting an expres­sion or num­ber in place of a let­ter.

Answers

These are the answers to the Prac­tice ques­tions, each checked by work­ing the sum in full.

1. $4x + 9x = 13x$
2. $(2m + 5) + (6m + 1) = 8m + 6$
3. $(3p + 2q) + (4p + 6q) = 7p + 8q$
4. $12y - 5y = 7y$
5. $(6a + 9) - (2a + 1) = 4a + 8$
6. $(10x + 4y) - (3x + 4y) = 7x$, because the $y$ terms can­cel.
7. $(5a - 2b) + (a + 7b) = 6a + 5b$
8. $A + 2B = (3a + 2b) + 2(a + b) = 5a + 4b$

The same answers, with the work­ing in brief:

Answers