How to Simplify Expressions with Many Variables

Everything from 2.5 still works when more than one letter is in view. Take $8 a b + 9 a b$ . The coefficients add the 2.4 way, $8 + 9 = 17$ , so $8 a b + 9 a b = 17 a b$ , and the $a b$ stays as it is. The one new question is what counts as the same kind of term, now that a single term can carry several letters.

Problem

Combine

7 s t + 6 s t

into one term. Type its coefficient.

Show a hint

$s t$ is one variable part, so this is a single family.
Add the coefficients, the 2.4 way.

Show the full solution

s t

is one variable part, so the coefficients add,

7 + 6 = 13

. Two letters in a term does not make it two families. The whole variable part is what has to match.

Same variable part means the same tile shape, so three

x y

tiles stack and their coefficients add to

3 x y

. An

x y

tile and an

x z

tile are different shapes, so they stay as two separate terms no matter how alike the letters look.

Problem

A classmate rewrote

5 c d + 5 c

10 c d

, adding the coefficients as if the terms matched. Evaluate the original

5 c d + 5 c

c = 2, d = 4

. Type the value.

Show a hint

Are $5 c d$ and $5 c$ really like terms? Compare their variable parts.
Put $c = 2, d = 4$ into the original $5 c d + 5 c$ , one letter at a time.

Show the full solution

Substitute into the original.

5 c d = 5 \cdot 2 \cdot 4 = 40

and

5 c = 5 \cdot 2 = 10

, so the value is

40 + 10 = 50

. The classmate's

10 c d

gives

80

at the same input, and one input with two different outputs proves

5 c d

and

5 c

are unlike.

Problem

Simplify

11 m n + 9 m - 5 n m

by combining like terms. Type the coefficient of the

m n

term.

Show a hint

$n m$ is the same variable part as $m n$ , so the order does not matter.
The minus travels with its term into the $m n$ family; the $9 m$ is a different family.

Show the full solution

Read

n m

m n

, so

11 m n - 5 m n = 6 m n

, and the

9 m

is a separate family. The coefficient is

6

. Order inside a term is free, so a backward-written term still joins its family, and its minus sign comes with it.

Problem

Distribute

8 r (2 s + t)

. Type the coefficient of the

r s

term.

Show a hint

The $8 r$ multiplies every term inside the parentheses.
You only need the term that has $r$ and $s$ .

Show the full solution

8 r \cdot 2 s = 16 r s

and

8 r \cdot t = 8 r t

, so the

r s

coefficient is

16

. The

8 r

multiplies both terms inside, not just the first.

Problem

Simplify

2 (4 u v + 5 v) + 7 u v

by distributing then combining like terms. Type the coefficient of the

u v

term.

Show a hint

Distribute the $2$ across both terms inside first.
Only the $u v$ terms combine; the plain $v$ term is a different family.

Show the full solution

Distribute first,

2 (4 u v + 5 v) = 8 u v + 10 v

. Only the

u v

terms combine,

8 u v + 7 u v = 15 u v

, so the coefficient is

15

. The

10 v

has a different variable part, so it never joins the

u v

count.

Problem

Find the greatest common factor of the terms in

8 p q r + 12 p q s

, typed like

2 x y

, the number first and the letters in alphabetical order.

Show a hint

Take the GCD of $8$ and $12$ first.
Which letters appear in BOTH terms? Only those come out front.

Show the full solution

The GCD of

8

and

12

4

, and

p

and

q

are the letters sitting in both terms, so the GCF is

4 p q

. The

r

and

s

each appear in only one term, so neither can come out front.

Problem

Evaluate

4 g h - 2 g + h

g = 5, h = 2

. Type the value.

Show a hint

$4 g h$ needs both letters, so $4 \cdot g \cdot h$ .
Handle the $- 2 g$ and $+ h$ after the $4 g h$ term.

Show the full solution

4 \cdot 5 \cdot 2 = 40

, then

- 2 \cdot 5 = - 10

and

+ 2

, so

40 - 10 + 2 = 32

. A term like

4 g h

needs both values, and reading it as

4 g

drops the

h

Problem

Simplify

2 d (4 e + f) + 9 d e

by distributing and combining, then evaluate the result at

d = 2, e = 1, f = 5

. Type the value.

Show a hint

Distribute $2 d$ first, keeping the $d$ on both new terms.
Combine the $d e$ terms, then substitute $d = 2, e = 1, f = 5$ .

Show the full solution

Distribute,

2 d (4 e + f) = 8 d e + 2 d f

, then combine,

8 d e + 9 d e = 17 d e

. Now substitute,

17 \cdot 2 \cdot 1 = 34

and

2 \cdot 2 \cdot 5 = 20

, so

34 + 20 = 54

. The

d

has to stay on both new terms, and

d e

and

d f

are different families, so they never merge.

That caps Chapter 2. Distribute, combine, factor, and evaluate are the whole toolkit, and none of them changed when several letters came into view. Only the like-term rule widened, since the variable part now has to match in full. Chapter 3 turns to exponents, the shorthand for a letter multiplied by itself over and over, along with the rules that come with them.

Practice these ideas

Practice

Combine

5 j k + 9 j k

. Type the coefficient.

Show the solution

j k

is one variable part, so the coefficients add,

5 + 9 = 14

Practice

Combine

7 g h + 8 h g

. Type the coefficient.

Show the solution

h g

and

g h

are the same variable part, so both terms sit in one family and

7 + 8 = 15

. Flipping the letters inside a term does not make a new family.

Practice

After combining like terms in

5 m p + 2 m + 4 m p

, how many terms remain?

Show the solution

The

m p

family gives

5 + 4 = 9 m p

, and

2 m

stands on its own, so

2

terms remain. The

2 m

has a different variable part, so it never joins the

m p

count.

Practice

Are

6 a b

and

6 a

like terms? Type yes or no.

Show the solution

The variable parts are

a b

and

a

, and those differ, so

no

. A shared coefficient and one shared letter are not enough, since the whole variable part has to match.

Practice

Evaluate

4 p q + 3 p

p = 2, q = 5

. Type the value.

Show the solution

4 \cdot 2 \cdot 5 = 40

and

3 \cdot 2 = 6

, so

40 + 6 = 46

. The

4 p q

term needs both values, and reading it as

4 p

drops the

q

Practice

Evaluate

2 m n + 4 m - n

m = 3, n = 1

. Type the value.

Show the solution

2 \cdot 3 \cdot 1 = 6

4 \cdot 3 = 12

, and the last term is

- 1

, so

6 + 12 - 1 = 17

. The

2 m n

term needs both values, not just the

m

Practice

Distribute

2 a (4 b + 3 c)

. Type the coefficient of the

a b

term.

Show the solution

2 a \cdot 4 b = 8 a b

, so the

a b

coefficient is

8

. The

2 a

multiplies both terms inside, not just the first.

Practice

Distribute

5 m (2 n + 4 p)

. Type the coefficient of the

m p

term.

Show the solution

The

m p

term comes from

5 m \cdot 4 p = 20 m p

, so the coefficient is

20

. The other product,

5 m \cdot 2 n = 10 m n

, is a different family.

Practice

Simplify

3 r (s + 2) + 4 r s

. Type the coefficient of the

r s

term.

Show the solution

Distribute,

3 r (s + 2) = 3 r s + 6 r

. Only the

r s

terms combine,

3 r s + 4 r s = 7 r s

, so the coefficient is

7

. The

6 r

has no

s

, so it stays out of the

r s

count.

Practice

Factor

8 a b + 12 a c

fully. Type the greatest common factor pulled out front, written like

5 x

(number first, letters alphabetical).

Show the solution

The GCD of

8

and

12

4

, and

a

is the only letter in both terms, so the GCF is

4 a

. The

b

and

c

each appear once, so neither comes out front.

Practice

Factor

6 c d h + 15 c d k

fully. Type the greatest common factor, written like

2 x y

(number first, letters alphabetical).

Show the solution

The GCD of

6

and

15

3

, and both

c

and

d

sit in every term, so the GCF is

3 c d

. The

h

and

k

each appear in only one term, so they stay inside.

Practice

Evaluate

5 s t - 2 s + t

s = 2, t = 3

. Type the value.

Show the solution

5 \cdot 2 \cdot 3 = 30

- 2 \cdot 2 = - 4

, and

+ 3

, so

30 - 4 + 3 = 29

. The

5 s t

term needs both values, and reading it as

5 s

drops the

t