Adding and subtracting expressions

Before long in alge­bra 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. It rests on one idea you met in the first les­son: only like terms can be com­bined. In this les­son you will add and sub­tract expres­sions in two ways, along a line and in columns, and learn the step that catches most peo­ple out, a minus sign in front of a bracket.

Only like terms join#

Like terms have the same lit­eral fac­tors, such as $3a$ and $2a$, or $5x^2$ and $-7x^2$. They count the same thing, so their co-effi­cients can be added or sub­tracted. Unlike terms, such as $3a$ and $2b$, stay side by side in the answer.

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

There are two ways of set­ting out the work: the hor­i­zon­tal method and the ver­ti­cal method.

Adding: the hor­i­zon­tal method#

Write every­thing along one line, then gather each group of like terms and add their co-effi­cients.

Exam­ple 1#

Add $3a$ and $2a$; then add $3a + 2$ and $5a + 7$.

$$3a + 2a = (3 + 2)a = 5a$$

$$3a + 2 + 5a + 7 = (3 + 5)a + (2 + 7) = 8a + 9$$

Exam­ple 2#

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

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

Exam­ple 3#

Add $x^2 + y^2 + 3xy$ and $2x^2 + 5y^2 - 7xy$.

$$\begin{aligned} & x^2 + y^2 + 3xy + 2x^2 + 5y^2 - 7xy \\ &= (1 + 2)x^2 + (1 + 5)y^2 + (3 - 7)xy \\ &= 3x^2 + 6y^2 - 4xy \end{aligned}$$

When a term moves to sit beside its part­ners, its sign trav­els with it. That is why $-7xy$ stays neg­a­tive.

Exam­ple 4: Frac­tional co-effi­cients#

Add $\displaystyle \frac{3}{2}a^2 + \frac{5}{3}ab + \frac{1}{4}b^2$ and $\displaystyle \frac{5}{6}a^2 - \frac{3}{4}ab + \frac{3}{4}b^2$.

Work out each group of co-effi­cients sep­a­rately:

$$\displaystyle \frac{3}{2} + \frac{5}{6} = \frac{9 + 5}{6} = \frac{14}{6} = \frac{7}{3} \qquad \frac{5}{3} - \frac{3}{4} = \frac{20 - 9}{12} = \frac{11}{12} \qquad \frac{1}{4} + \frac{3}{4} = 1$$

So the sum is $\displaystyle \frac{7}{3}a^2 + \frac{11}{12}ab + b^2$.

Adding: the ver­ti­cal method#

Write one expres­sion under the other, with like terms in the same col­umn, then add each col­umn on its own. The columns do the match­ing for you, which helps when there are many terms.

Exam­ple 5#

Add $\displaystyle 2x^2 + 3y^2 + \frac{1}{2}xy$ and $\displaystyle \frac{5}{2}x^2 + xy - 7y^2$.

Put the sec­ond expres­sion in the same order as the first, so that like terms line up.

Col­umn by col­umn: $\displaystyle 2 + \frac{5}{2} = \frac{4 + 5}{2} = \frac{9}{2}$, then $3 - 7 = -4$, then $\displaystyle \frac{1}{2} + 1 = \frac{3}{2}$. The sum is $\displaystyle \frac{9}{2}x^2 - 4y^2 + \frac{3}{2}xy$.

Exam­ple 6#

Add $2u + 3v$, $-5v + 6w$ and $7u - 7w$.

Leave a gap where an expres­sion has no term of that kind.

The sum is $9u - 2v - w$.

Sub­tract­ing: the hor­i­zon­tal method#

Read the order with care. "Sub­tract $2a$ from $7a$" means start at $7a$ and take $2a$ away: $7a - 2a = 5a$. The expres­sion after "from" is writ­ten first.

When the expres­sion being taken away is neg­a­tive, or has more than one term, put it in a bracket first. A minus sign in front of a bracket changes the sign of every term inside it.

Exam­ple 7#

Sub­tract $-2a$ from $7a$.

$$(7a) - (-2a) = 7a + 2a = 9a$$

Tak­ing away a neg­a­tive amount is the same as adding it.

Exam­ple 8#

Sub­tract $4x - 3y$ from $8x - 7y$.

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

The $-3y$ inside the bracket became $+3y$ when the bracket came off.

Exam­ple 9#

Sub­tract $2a + 3$ from $9a + 8$.

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

For­get to change the $+3$ and you would get $7a + 11$, which is wrong.

Exam­ple 10: Let­ters for whole expres­sions#

If $A = 2x^2 + x + 5$ and $B = 3x^2 - 2x + 7$, find (i) $2A - 3B$ and (ii) $(A + B) - (A - B)$.

A cap­i­tal let­ter here stands for a whole expres­sion, so keep it in brack­ets when you put it in.

(i)
$$\begin{aligned} 2A - 3B &= 2(2x^2 + x + 5) - 3(3x^2 - 2x + 7) \\ &= 4x^2 + 2x + 10 - 9x^2 + 6x - 21 \\ &= -5x^2 + 8x - 11 \end{aligned}$$

(ii) First $A + B = 5x^2 - x + 12$ and $A - B = -x^2 + 3x - 2$.

$$\begin{aligned} (A + B) - (A - B) &= (5x^2 - x + 12) - (-x^2 + 3x - 2) \\ &= 5x^2 - x + 12 + x^2 - 3x + 2 \\ &= 6x^2 - 4x + 14 \end{aligned}$$

Notice that $6x^2 - 4x + 14$ is exactly $2B$. That is no acci­dent: $(A + B) - (A - B) = A + B - A + B = 2B$, what­ever $A$ and $B$ are.

Sub­tract­ing: the ver­ti­cal method#

Write the expres­sion to be taken away under the first one, with like terms in the same col­umn. Then change the sign of every term in the bot­tom row, writ­ing the new sign under the old one, and add down the columns.

Exam­ple 11#

Sub­tract $3x + 4y$ from $4x - 3y$.

In the $x$ col­umn, $4 - 3 = 1$. In the $y$ col­umn, $-3 - 4 = -7$. The answer is $x - 7y$.

Exam­ple 12#

Sub­tract $a^3 - b^3 + 3ab + 1$ from $a^3 - 2b^3 + 5$.

The $a^3$ terms can­cel, $-2b^3 + b^3 = -b^3$, $5 - 1 = 4$, and $0 - 3ab = -3ab$. The answer is $-b^3 + 4 - 3ab$.

Prac­tice#

1. Add $5a - 2b$ and $a + 7b$.
2. Add $2a + 3b - 4c$ and $\displaystyle 7a - \frac{2}{3}b + 5c$.
3. Add $3p^2 - 4pq + q^2$, $-p^2 + 2pq$ and $5pq - 3q^2$ by the ver­ti­cal method.
4. Sub­tract $5y$ from $12y$, and $-6xy$ from $-2xy$.
5. Sub­tract $\displaystyle 2a + \frac{3}{2}b$ from $7a - 5b$.
6. Sub­tract $3x - 4y + z$ from $5x + 2y - 3z$.
7. Sub­tract $\displaystyle \frac{3}{4}x^2 - \frac{1}{12}xy + 5y^2$ from $\displaystyle -5x^2 + 3xy - \frac{1}{2}y^2$.
8. If $\displaystyle A = 5x^2 - \frac{3}{4}xy + 7y^2$ and $\displaystyle B = 3x^2 + \frac{1}{2}xy - \frac{5}{3}y^2$, find $5B - A$.
9. With the same $A$ and $B$, find $3A - 5B$.
10. Sub­tract $2m^2 - 3m + 4$ from $m^2 + m - 1$ by the ver­ti­cal method.

Answers#