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This pdf includes the following topics:-Distributive Property

Speeding Things Up a Bit

Simplify

Alternate solution

Applications

Examples

Speeding Things Up a Bit

Simplify

Alternate solution

Applications

Examples

1.
3.4. COMBINING LIKE TERMS 197

3.4 Combining Like Terms

We begin our discussion with the definition of a term.

Term. A term is a single number or variable, or it can be the product of a

number (called its coefficient ) and one or more variables (called its variable

part ). The terms in an algebraic expression are separated by addition symbols.

You Try It!

EXAMPLE 1. Identify the terms in the algebraic expression How many terms are in the

algebraic expression

3x2 + 5xy + 9y 2 + 12. 3x2 + 2xy − 3y 2 ?

For each term, identify its coefficient and variable part.

Solution. In tabular form, we list each term of the expression 3x2 + 5xy +

9y 2 + 12, its coefficient, and its variable part.

Term Coefficient Variable Part

3x2 3 x2

5xy 5 xy

2

9y 9 y2

12 12 None

Answer: 3

You Try It!

EXAMPLE 2. Identify the terms in the algebraic expression How many terms are in the

algebraic expression

a3 − 3a2 b + 3ab2 − b3 . 11 − a2 − 2ab + 3b2 ?

For each term, identify its coefficient and variable part.

Solution. The first step is to write each difference as a sum, because the terms

of an expression are defined above to be those items separated by addition

a3 + (−3a2 b) + 3ab2 + (−b3 )

In tabular form, we list each term of the expression a3 +(−3a2 b)+3ab2 +(−b3 ),

its coefficient, and its variable part.

Version: Fall 2010

3.4 Combining Like Terms

We begin our discussion with the definition of a term.

Term. A term is a single number or variable, or it can be the product of a

number (called its coefficient ) and one or more variables (called its variable

part ). The terms in an algebraic expression are separated by addition symbols.

You Try It!

EXAMPLE 1. Identify the terms in the algebraic expression How many terms are in the

algebraic expression

3x2 + 5xy + 9y 2 + 12. 3x2 + 2xy − 3y 2 ?

For each term, identify its coefficient and variable part.

Solution. In tabular form, we list each term of the expression 3x2 + 5xy +

9y 2 + 12, its coefficient, and its variable part.

Term Coefficient Variable Part

3x2 3 x2

5xy 5 xy

2

9y 9 y2

12 12 None

Answer: 3

You Try It!

EXAMPLE 2. Identify the terms in the algebraic expression How many terms are in the

algebraic expression

a3 − 3a2 b + 3ab2 − b3 . 11 − a2 − 2ab + 3b2 ?

For each term, identify its coefficient and variable part.

Solution. The first step is to write each difference as a sum, because the terms

of an expression are defined above to be those items separated by addition

a3 + (−3a2 b) + 3ab2 + (−b3 )

In tabular form, we list each term of the expression a3 +(−3a2 b)+3ab2 +(−b3 ),

its coefficient, and its variable part.

Version: Fall 2010

2.
198 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA

Term Coefficient Variable Part

a3 1 a3

−3a2 b −3 a2 b

3ab2 3 ab2

−b3 −1 b3

Answer: 4

Like Terms

We define what is meant by “like terms” and “unlike terms.”

Like and Unlike Terms. The variable parts of two terms determine whether

the terms are like terms or unlike terms.

Like Terms. Two terms are called like terms if they have identical variable

parts, which means that the terms must contain the same variables raised to

the same exponential powers.

Unlike Terms. Two terms are called unlike terms if their variable parts are

different.

You Try It!

Are −3xy and 11xy like or EXAMPLE 3. Classify each of the following pairs as either like terms or

unlike terms? unlike terms: (a) 3x and −7x, (b) 2y and 3y 2 , (c) −3t and 5u, and (d) −4a3

and 3a3 .

Solution. Like terms must have identical variable parts.

a) 3x and −7x have identical variable parts. They are “like terms.”

b) 2y and 3y 2 do not have identical variable parts (the exponents differ). They

are “unlike terms.”

c) −3t and 5u do not have identical variable parts (different variables). They

are “unlike terms.”

d) −4a3 and 3a3 have identical variable parts. They are “like terms.”

Answer: Like terms

Version: Fall 2010

Term Coefficient Variable Part

a3 1 a3

−3a2 b −3 a2 b

3ab2 3 ab2

−b3 −1 b3

Answer: 4

Like Terms

We define what is meant by “like terms” and “unlike terms.”

Like and Unlike Terms. The variable parts of two terms determine whether

the terms are like terms or unlike terms.

Like Terms. Two terms are called like terms if they have identical variable

parts, which means that the terms must contain the same variables raised to

the same exponential powers.

Unlike Terms. Two terms are called unlike terms if their variable parts are

different.

You Try It!

Are −3xy and 11xy like or EXAMPLE 3. Classify each of the following pairs as either like terms or

unlike terms? unlike terms: (a) 3x and −7x, (b) 2y and 3y 2 , (c) −3t and 5u, and (d) −4a3

and 3a3 .

Solution. Like terms must have identical variable parts.

a) 3x and −7x have identical variable parts. They are “like terms.”

b) 2y and 3y 2 do not have identical variable parts (the exponents differ). They

are “unlike terms.”

c) −3t and 5u do not have identical variable parts (different variables). They

are “unlike terms.”

d) −4a3 and 3a3 have identical variable parts. They are “like terms.”

Answer: Like terms

Version: Fall 2010

3.
3.4. COMBINING LIKE TERMS 199

Combining Like Terms

When using the distributive property, it makes no difference whether the mul-

tiplication is on the left or the right, one still distributes the multiplication

times each term in the parentheses.

Distributive Property. If a, b, and c are integers, then

a(b + c) = ab + ac and (b + c)a = ba + ca.

In either case, you distribute a times each term of the sum.

“Like terms” can be combined and simplified. The tool used for combining

like terms is the distributive property. For example, consider the expression

3y + 7y, composed of two “like terms” with a common variable part. We can

use the distributive property and write

3y + 7y = (3 + 7)y.

Note that we are using the distributive property in reverse, “factoring out”

the common variable part of each term. Checking our work, note that if we

redistribute the variable part y times each term in the parentheses, we are

returned to the original expression 3y + 7y.

You Try It!

EXAMPLE 4. Use the distributive property to combine like terms (if pos- Simplify: −8z − 11z

sible) in each of the following expressions: (a) −5x2 − 9x2 , (b) −5ab + 7ab, (c)

4y 3 − 7y 2 , and (d) 3xy 2 − 7xy 2 .

Solution. If the terms are “like terms,” you can use the distributive property

to “factor out” the common variable part.

a) Factor out the common variable part x2 .

−5x2 − 9x2 = (−5 − 9)x2 Use the distributive property.

2

= −14x Simplify: −5 − 9 = −5 + (−9) = −14.

b) Factor out the common variable part ab.

−5ab + 7ab = (−5 + 7)ab Use the distributive property.

= 2ab Simplify: −5 + 7 = 2.

c) The terms in the expression 4y 3 − 7y 2 have different variable parts (the

exponents are different). These are “unlike terms” and cannot be combined.

Version: Fall 2010

Combining Like Terms

When using the distributive property, it makes no difference whether the mul-

tiplication is on the left or the right, one still distributes the multiplication

times each term in the parentheses.

Distributive Property. If a, b, and c are integers, then

a(b + c) = ab + ac and (b + c)a = ba + ca.

In either case, you distribute a times each term of the sum.

“Like terms” can be combined and simplified. The tool used for combining

like terms is the distributive property. For example, consider the expression

3y + 7y, composed of two “like terms” with a common variable part. We can

use the distributive property and write

3y + 7y = (3 + 7)y.

Note that we are using the distributive property in reverse, “factoring out”

the common variable part of each term. Checking our work, note that if we

redistribute the variable part y times each term in the parentheses, we are

returned to the original expression 3y + 7y.

You Try It!

EXAMPLE 4. Use the distributive property to combine like terms (if pos- Simplify: −8z − 11z

sible) in each of the following expressions: (a) −5x2 − 9x2 , (b) −5ab + 7ab, (c)

4y 3 − 7y 2 , and (d) 3xy 2 − 7xy 2 .

Solution. If the terms are “like terms,” you can use the distributive property

to “factor out” the common variable part.

a) Factor out the common variable part x2 .

−5x2 − 9x2 = (−5 − 9)x2 Use the distributive property.

2

= −14x Simplify: −5 − 9 = −5 + (−9) = −14.

b) Factor out the common variable part ab.

−5ab + 7ab = (−5 + 7)ab Use the distributive property.

= 2ab Simplify: −5 + 7 = 2.

c) The terms in the expression 4y 3 − 7y 2 have different variable parts (the

exponents are different). These are “unlike terms” and cannot be combined.

Version: Fall 2010

4.
200 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA

d) Factor out the common variable part xy 2 .

3xy 2 − 7xy 2 = (3 − 7)xy 2 Use the distributive property.

2

= −4xy Simplify: 3 − 7 = 3 + (−7) = −4.

Answer: −19z

Speeding Things Up a Bit

Once you’ve written out all the steps for combining like terms, like those shown

in Example 4, you can speed things up a bit by following this rule:

Combining Like Terms. To combine like terms, simply add their coefficients

and keep the common variable part.

Thus for example, when presented with the sum of two like terms, such as in

5x + 8x, simply add the coefficients and repeat the common variable part; that

is, 5x + 8x = 13x.

You Try It!

Combine: −3x2 − 4x2 EXAMPLE 5. Combine like terms: (a) −9y − 8y, (b) −3y 5 + 4y 5 , and (c)

−3u2 + 2u2 .

Solution.

a) Add the coefficients and repeat the common variable part. Therefore,

−9y − 8y = −17y.

b) Add the coefficients and repeat the common variable part. Therefore,

−3y 5 + 4y 5 = 1y 5 .

However, note that 1y 5 = y 5 . Following the rule that the final answer

should use as few symbols as possible, a better answer is −3y 5 + 4y 5 = y 5 .

c) Add the coefficients and repeat the common variable part. Therefore,

−3u2 + 2u2 = (−1)u2 .

However, note that (−1)u2 = −u2 . Following the rule that the final answer

should use as few symbols as possible, a better answer is −3u2 + 2u2 = −u2 .

Answer: −7x2

Version: Fall 2010

d) Factor out the common variable part xy 2 .

3xy 2 − 7xy 2 = (3 − 7)xy 2 Use the distributive property.

2

= −4xy Simplify: 3 − 7 = 3 + (−7) = −4.

Answer: −19z

Speeding Things Up a Bit

Once you’ve written out all the steps for combining like terms, like those shown

in Example 4, you can speed things up a bit by following this rule:

Combining Like Terms. To combine like terms, simply add their coefficients

and keep the common variable part.

Thus for example, when presented with the sum of two like terms, such as in

5x + 8x, simply add the coefficients and repeat the common variable part; that

is, 5x + 8x = 13x.

You Try It!

Combine: −3x2 − 4x2 EXAMPLE 5. Combine like terms: (a) −9y − 8y, (b) −3y 5 + 4y 5 , and (c)

−3u2 + 2u2 .

Solution.

a) Add the coefficients and repeat the common variable part. Therefore,

−9y − 8y = −17y.

b) Add the coefficients and repeat the common variable part. Therefore,

−3y 5 + 4y 5 = 1y 5 .

However, note that 1y 5 = y 5 . Following the rule that the final answer

should use as few symbols as possible, a better answer is −3y 5 + 4y 5 = y 5 .

c) Add the coefficients and repeat the common variable part. Therefore,

−3u2 + 2u2 = (−1)u2 .

However, note that (−1)u2 = −u2 . Following the rule that the final answer

should use as few symbols as possible, a better answer is −3u2 + 2u2 = −u2 .

Answer: −7x2

Version: Fall 2010

5.
3.4. COMBINING LIKE TERMS 201

A frequently occurring instruction asks the reader to simplify an expression.

Simplify. The instruction simplify is a generic term that means “try to write

the expression in its most compact form, using the fewest symbols possible.”

One way you can accomplish this goal is by combining like terms when they

are present.

You Try It!

EXAMPLE 6. Simplify: 2x + 3y − 5x + 8y. Simplify: −3a + 4b − 7a − 9b

Solution. Use the commutative property to reorder terms and the associative

and distributive properties to regroup and combine like terms.

2x + 3y − 5x + 8y = (2x − 5x) + (3y + 8y) Reorder and regroup.

= −3x + 11y Combine like terms:

2x − 5x = −3x and 3y + 8y = 11y.

Alternate solution. Of course, you do not need to show the regrouping step.

If you are more comfortable combining like terms in your head, you are free to

present your work as follows:

2x + 3y − 5x + 8y = −3x + 11y.

Answer: −10a − 5b

You Try It!

EXAMPLE 7. Simplify: −2x − 3 − (3x + 4). Simplify: −9a − 4 − (4a − 8)

Solution. First, distribute the negative sign.

−2x − 3 − (3x + 4) = −2x − 3 − 3x − 4 −(3x + 4) = −3x − 4.

Next, use the commutative property to reorder, then the associative property

to regroup. Then combine like terms.

= (−2x − 3x) + (−3 − 4) Reorder and regroup.

= −5x + (−7) Combine like terms:

−2x − 3x = −5x.

= −5x − 7 Simplify:

−5x + (−7) = −5x − 7.

Version: Fall 2010

A frequently occurring instruction asks the reader to simplify an expression.

Simplify. The instruction simplify is a generic term that means “try to write

the expression in its most compact form, using the fewest symbols possible.”

One way you can accomplish this goal is by combining like terms when they

are present.

You Try It!

EXAMPLE 6. Simplify: 2x + 3y − 5x + 8y. Simplify: −3a + 4b − 7a − 9b

Solution. Use the commutative property to reorder terms and the associative

and distributive properties to regroup and combine like terms.

2x + 3y − 5x + 8y = (2x − 5x) + (3y + 8y) Reorder and regroup.

= −3x + 11y Combine like terms:

2x − 5x = −3x and 3y + 8y = 11y.

Alternate solution. Of course, you do not need to show the regrouping step.

If you are more comfortable combining like terms in your head, you are free to

present your work as follows:

2x + 3y − 5x + 8y = −3x + 11y.

Answer: −10a − 5b

You Try It!

EXAMPLE 7. Simplify: −2x − 3 − (3x + 4). Simplify: −9a − 4 − (4a − 8)

Solution. First, distribute the negative sign.

−2x − 3 − (3x + 4) = −2x − 3 − 3x − 4 −(3x + 4) = −3x − 4.

Next, use the commutative property to reorder, then the associative property

to regroup. Then combine like terms.

= (−2x − 3x) + (−3 − 4) Reorder and regroup.

= −5x + (−7) Combine like terms:

−2x − 3x = −5x.

= −5x − 7 Simplify:

−5x + (−7) = −5x − 7.

Version: Fall 2010

6.
202 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA

Alternate solution. You may skip the second step if you wish, simply com-

bining like terms mentally. That is, it is entirely possible to order your work

as follows:

−2x − 3 − (3x + 4) = −2x − 3 − 3x − 4 Distribute negative sign.

= −5x − 7 Combine like terms.

Answer: −13a + 4

You Try It!

Simplify: EXAMPLE 8. Simplify: 2(5 − 3x) − 4(x + 3).

−2(3a − 4) − 2(5 − a) Solution. Use the distributive property to expand, then use the commutative

and associative properties to group the like terms and combine them.

2(5 − 3x) − 4(x + 3) = 10 − 6x − 4x − 12 Use the distributive property.

= (−6x − 4x) + (10 − 12) Group like terms.

= −10x − 2 Combine like terms:

−6x − 4x = −10x and

10 − 12 = −2.

Alternate solution. You may skip the second step if you wish, simply com-

bining like terms mentally. That is, it is entirely possible to order your work

as follows:

2(5 − 3x) − 4(x + 3) = 10 − 6x − 4x − 12 Distribute.

= −10x − 2 Combine like terms.

Answer: −4a − 2

You Try It!

Simplify: EXAMPLE 9. Simplify: −8(3x2 y − 9xy) − 8(−7x2 y − 8xy).

(a2 − 2ab) − 2(3ab + a2 )

Solution. We will proceed a bit quicker with this solution, using the distribu-

tive property to expand, then combining like terms mentally.

−8(3x2 y − 9xy) − 8(−7x2 y − 8xy) = −24x2 y + 72xy + 56x2 y + 64xy

= 32x2 y + 136xy

Answer: −a2 − 8ab

Version: Fall 2010

Alternate solution. You may skip the second step if you wish, simply com-

bining like terms mentally. That is, it is entirely possible to order your work

as follows:

−2x − 3 − (3x + 4) = −2x − 3 − 3x − 4 Distribute negative sign.

= −5x − 7 Combine like terms.

Answer: −13a + 4

You Try It!

Simplify: EXAMPLE 8. Simplify: 2(5 − 3x) − 4(x + 3).

−2(3a − 4) − 2(5 − a) Solution. Use the distributive property to expand, then use the commutative

and associative properties to group the like terms and combine them.

2(5 − 3x) − 4(x + 3) = 10 − 6x − 4x − 12 Use the distributive property.

= (−6x − 4x) + (10 − 12) Group like terms.

= −10x − 2 Combine like terms:

−6x − 4x = −10x and

10 − 12 = −2.

Alternate solution. You may skip the second step if you wish, simply com-

bining like terms mentally. That is, it is entirely possible to order your work

as follows:

2(5 − 3x) − 4(x + 3) = 10 − 6x − 4x − 12 Distribute.

= −10x − 2 Combine like terms.

Answer: −4a − 2

You Try It!

Simplify: EXAMPLE 9. Simplify: −8(3x2 y − 9xy) − 8(−7x2 y − 8xy).

(a2 − 2ab) − 2(3ab + a2 )

Solution. We will proceed a bit quicker with this solution, using the distribu-

tive property to expand, then combining like terms mentally.

−8(3x2 y − 9xy) − 8(−7x2 y − 8xy) = −24x2 y + 72xy + 56x2 y + 64xy

= 32x2 y + 136xy

Answer: −a2 − 8ab

Version: Fall 2010

7.
3.4. COMBINING LIKE TERMS 203

We can simplify a number of useful formulas by combining like terms.

You Try It!

EXAMPLE 10. Find the perimeter P of the (a) rectangle and (b) square A regular hexagon has six

pictured below. Simplify your answer as much as possible. equal sides, each with length

L s x. Find its perimeter in

terms of x.

W W s s

L s

Solution. The perimeter of any polygonal figure is the sum of the lengths of

its sides.

a) To find the perimeter P of the rectangle, sum its four sides.

P = L + W + L + W.

Combine like terms.

P = 2L + 2W.

b) To find the perimeter P of the square, sum its four sides.

P = s + s + s + s.

Combine like terms.

P = 4s.

Answer: P = 6x

Sometimes it is useful to replace a variable with an expression containing

another variable.

You Try It!

EXAMPLE 11. The length of a rectangle is three feet longer than twice its The length L of a rectangle

width. Find the perimeter P of the rectangle in terms of its width alone. is 5 meters longer than twice

Solution. From the previous problem, the perimeter of the rectangle is given its width W . Find the

by perimeter P of the rectangle

P = 2L + 2W, (3.1) in terms of its width W .

where L and W are the length and width of the rectangle, respectively. This

equation gives the perimeter in terms of its length and width, but we’re asked

to get the perimeter in terms of the width alone.

However, we’re also given the fact that the length is three feet longer than

twice the width.

Version: Fall 2010

We can simplify a number of useful formulas by combining like terms.

You Try It!

EXAMPLE 10. Find the perimeter P of the (a) rectangle and (b) square A regular hexagon has six

pictured below. Simplify your answer as much as possible. equal sides, each with length

L s x. Find its perimeter in

terms of x.

W W s s

L s

Solution. The perimeter of any polygonal figure is the sum of the lengths of

its sides.

a) To find the perimeter P of the rectangle, sum its four sides.

P = L + W + L + W.

Combine like terms.

P = 2L + 2W.

b) To find the perimeter P of the square, sum its four sides.

P = s + s + s + s.

Combine like terms.

P = 4s.

Answer: P = 6x

Sometimes it is useful to replace a variable with an expression containing

another variable.

You Try It!

EXAMPLE 11. The length of a rectangle is three feet longer than twice its The length L of a rectangle

width. Find the perimeter P of the rectangle in terms of its width alone. is 5 meters longer than twice

Solution. From the previous problem, the perimeter of the rectangle is given its width W . Find the

by perimeter P of the rectangle

P = 2L + 2W, (3.1) in terms of its width W .

where L and W are the length and width of the rectangle, respectively. This

equation gives the perimeter in terms of its length and width, but we’re asked

to get the perimeter in terms of the width alone.

However, we’re also given the fact that the length is three feet longer than

twice the width.

Version: Fall 2010

8.
204 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA

Three Twice the

Length is longer than

Feet Width

L = 3 + 2W

Because L = 3+2W , we can replace L with 3+2W in the perimeter equation 3.1.

P = 2L + 2W

P = 2(3 + 2W ) + 2W

Use the distributive property, then combine like terms.

P = 6 + 4W + 2W

P = 6 + 6W.

Answer: P = 6W + 10 This last equation gives the perimeter P in terms of the width W alone.

You Try It!

The width W of a rectangle EXAMPLE 12. The width of a rectangle is two feet less than its length.

is 5 feet less than twice its Find the perimeter P of the rectangle in terms of its length alone.

width L. Find the perimeter Solution. Again, the perimeter of a rectangle is given by the equation

P of the rectangle in terms

of its length L. P = 2L + 2W, (3.2)

where L and W are the length and width of the rectangle, respectively. This

equation gives the perimeter in terms of its length and width, but we’re asked

to get the perimeter in terms of the length alone.

However, we’re also given the fact that the width is two feet less than the

length.

Width is Length minus Two feet

W = L − 2

Because W = L−2, we can replace W with L−2 in the perimeter equation 3.2.

P = 2L + 2W

P = 2L + 2(L − 2)

Use the distributive property, then combine like terms.

P = 2L + 2L − 4

P = 4L − 4.

Answer: P = 6L − 10 This last equation gives the perimeter P in terms of the length L alone.

Version: Fall 2010

Three Twice the

Length is longer than

Feet Width

L = 3 + 2W

Because L = 3+2W , we can replace L with 3+2W in the perimeter equation 3.1.

P = 2L + 2W

P = 2(3 + 2W ) + 2W

Use the distributive property, then combine like terms.

P = 6 + 4W + 2W

P = 6 + 6W.

Answer: P = 6W + 10 This last equation gives the perimeter P in terms of the width W alone.

You Try It!

The width W of a rectangle EXAMPLE 12. The width of a rectangle is two feet less than its length.

is 5 feet less than twice its Find the perimeter P of the rectangle in terms of its length alone.

width L. Find the perimeter Solution. Again, the perimeter of a rectangle is given by the equation

P of the rectangle in terms

of its length L. P = 2L + 2W, (3.2)

where L and W are the length and width of the rectangle, respectively. This

equation gives the perimeter in terms of its length and width, but we’re asked

to get the perimeter in terms of the length alone.

However, we’re also given the fact that the width is two feet less than the

length.

Width is Length minus Two feet

W = L − 2

Because W = L−2, we can replace W with L−2 in the perimeter equation 3.2.

P = 2L + 2W

P = 2L + 2(L − 2)

Use the distributive property, then combine like terms.

P = 2L + 2L − 4

P = 4L − 4.

Answer: P = 6L − 10 This last equation gives the perimeter P in terms of the length L alone.

Version: Fall 2010

9.
3.4. COMBINING LIKE TERMS 205

§ § § Exercises § § §

In Exercises 1-16, combine like terms by first using the distributive property to factor out the common

variable part, and then simplifying.

1. 17xy 2 + 18xy 2 + 20xy 2 9. −11x − 13x + 8x

2. 13xy − 3xy + xy 10. −9r − 10r + 3r

3. −8xy 2 − 3xy 2 − 10xy 2 11. −5q + 7q

4. −12xy − 2xy + 10xy 12. 17n + 15n

5. 4xy − 20xy 13. r − 13r − 7r

3 3

6. −7y + 15y 14. 19m + m + 15m

7. 12r − 12r 15. 3x3 − 18x3

8. 16s − 5s 16. 13x2 y + 2x2 y

In Exercises 17-32, combine like terms by first rearranging the terms, then using the distributive

property to factor out the common variable part, and then simplifying.

17. −8 + 17n + 10 + 8n 25. −14x2 y − 2xy 2 + 8x2 y + 18xy 2

18. 11 + 16s − 14 − 6s 26. −19y 2 + 18y 3 − 5y 2 − 17y 3

19. −2x3 − 19x2 y − 15x2 y + 11x3 27. −14x3 + 16xy + 5x3 + 8xy

20. −9x2 y − 10y 3 − 10y 3 + 17x2 y 28. −16xy + 16y 2 + 7xy + 17y 2

21. −14xy − 2x3 − 2x3 − 4xy 29. 9n + 10 + 7 + 15n

3 3

22. −4x + 12xy + 4xy − 12x 30. −12r + 5 + 17 + 17r

23. −13 + 16m + m + 16 31. 3y + 1 + 6y + 3

24. 9 − 11x − 8x + 15 32. 19p + 6 + 8p + 13

In Exercises 33-56, simplify the expression by first using the distributive property to expand the ex-

pression, and then rearranging and combining like terms mentally.

33. −4(9x2 y + 8) + 6(10x2 y − 6) 37. −s + 7 − (−1 − 3s)

3 3

34. −4(−4xy + 5y ) + 6(−5xy − 9y ) 38. 10y − 6 − (−10 − 10y)

35. 3(−4x2 + 10y 2 ) + 10(4y 2 − x2 ) 39. −10q − 10 − (−3q + 5)

36. −7(−7x3 + 6x2 ) − 7(−10x2 − 7x3 ) 40. −2n + 10 − (7n − 1)

Version: Fall 2010

§ § § Exercises § § §

In Exercises 1-16, combine like terms by first using the distributive property to factor out the common

variable part, and then simplifying.

1. 17xy 2 + 18xy 2 + 20xy 2 9. −11x − 13x + 8x

2. 13xy − 3xy + xy 10. −9r − 10r + 3r

3. −8xy 2 − 3xy 2 − 10xy 2 11. −5q + 7q

4. −12xy − 2xy + 10xy 12. 17n + 15n

5. 4xy − 20xy 13. r − 13r − 7r

3 3

6. −7y + 15y 14. 19m + m + 15m

7. 12r − 12r 15. 3x3 − 18x3

8. 16s − 5s 16. 13x2 y + 2x2 y

In Exercises 17-32, combine like terms by first rearranging the terms, then using the distributive

property to factor out the common variable part, and then simplifying.

17. −8 + 17n + 10 + 8n 25. −14x2 y − 2xy 2 + 8x2 y + 18xy 2

18. 11 + 16s − 14 − 6s 26. −19y 2 + 18y 3 − 5y 2 − 17y 3

19. −2x3 − 19x2 y − 15x2 y + 11x3 27. −14x3 + 16xy + 5x3 + 8xy

20. −9x2 y − 10y 3 − 10y 3 + 17x2 y 28. −16xy + 16y 2 + 7xy + 17y 2

21. −14xy − 2x3 − 2x3 − 4xy 29. 9n + 10 + 7 + 15n

3 3

22. −4x + 12xy + 4xy − 12x 30. −12r + 5 + 17 + 17r

23. −13 + 16m + m + 16 31. 3y + 1 + 6y + 3

24. 9 − 11x − 8x + 15 32. 19p + 6 + 8p + 13

In Exercises 33-56, simplify the expression by first using the distributive property to expand the ex-

pression, and then rearranging and combining like terms mentally.

33. −4(9x2 y + 8) + 6(10x2 y − 6) 37. −s + 7 − (−1 − 3s)

3 3

34. −4(−4xy + 5y ) + 6(−5xy − 9y ) 38. 10y − 6 − (−10 − 10y)

35. 3(−4x2 + 10y 2 ) + 10(4y 2 − x2 ) 39. −10q − 10 − (−3q + 5)

36. −7(−7x3 + 6x2 ) − 7(−10x2 − 7x3 ) 40. −2n + 10 − (7n − 1)

Version: Fall 2010

10.
206 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA

41. 7(8y + 7) − 6(8 − 7y) 49. −8(−n + 4) − 10(−4n + 3)

42. −6(−5n − 4) − 9(3 + 4n) 50. 3(8r − 7) − 3(2r − 2)

43. 7(10x2 − 8xy 2 ) − 7(9xy 2 + 9x2 ) 51. −5 − (10p + 5)

44. 10(8x2 y − 10xy 2 ) + 3(8xy 2 + 2x2 y) 52. −1 − (2p − 8)

45. −2(6 + 4n) + 4(−n − 7) 53. 7(1 + 7r) + 2(4 − 5r)

46. −6(−2 − 6m) + 5(−9m + 7) 54. (5 − s) + 10(9 + 5s)

47. 8 − (4 + 8y) 55. −2(−5 − 8x2 ) − 6(6)

48. −1 − (8 + s) 56. 8(10y 2 + 3x3 ) − 5(−7y 2 − 7x3 )

57. The length L of a rectangle is 2 feet 60. The width W of a rectangle is 9 feet

longer than 6 times its width W . Find shorter than its length L. Find the

the perimeter of the rectangle in terms of perimeter of the rectangle in terms of its

its width alone. length alone.

58. The length L of a rectangle is 7 feet 61. The length L of a rectangle is 9 feet

longer than 6 times its width W . Find shorter than 4 times its width W . Find

the perimeter of the rectangle in terms of the perimeter of the rectangle in terms of

its width alone. its width alone.

59. The width W of a rectangle is 8 feet 62. The length L of a rectangle is 2 feet

shorter than its length L. Find the shorter than 6 times its width W . Find

perimeter of the rectangle in terms of its the perimeter of the rectangle in terms of

length alone. its width alone.

§ § § Answers § § §

1. 55xy 2 15. −15x3

3. −21xy 2 17. 2 + 25n

5. −16xy 19. 9x3 − 34x2 y

7. 0 21. −18xy − 4x3

9. −16x 23. 3 + 17m

25. −6x2 y + 16xy 2

11. 2q

27. −9x3 + 24xy

13. −19r

Version: Fall 2010

41. 7(8y + 7) − 6(8 − 7y) 49. −8(−n + 4) − 10(−4n + 3)

42. −6(−5n − 4) − 9(3 + 4n) 50. 3(8r − 7) − 3(2r − 2)

43. 7(10x2 − 8xy 2 ) − 7(9xy 2 + 9x2 ) 51. −5 − (10p + 5)

44. 10(8x2 y − 10xy 2 ) + 3(8xy 2 + 2x2 y) 52. −1 − (2p − 8)

45. −2(6 + 4n) + 4(−n − 7) 53. 7(1 + 7r) + 2(4 − 5r)

46. −6(−2 − 6m) + 5(−9m + 7) 54. (5 − s) + 10(9 + 5s)

47. 8 − (4 + 8y) 55. −2(−5 − 8x2 ) − 6(6)

48. −1 − (8 + s) 56. 8(10y 2 + 3x3 ) − 5(−7y 2 − 7x3 )

57. The length L of a rectangle is 2 feet 60. The width W of a rectangle is 9 feet

longer than 6 times its width W . Find shorter than its length L. Find the

the perimeter of the rectangle in terms of perimeter of the rectangle in terms of its

its width alone. length alone.

58. The length L of a rectangle is 7 feet 61. The length L of a rectangle is 9 feet

longer than 6 times its width W . Find shorter than 4 times its width W . Find

the perimeter of the rectangle in terms of the perimeter of the rectangle in terms of

its width alone. its width alone.

59. The width W of a rectangle is 8 feet 62. The length L of a rectangle is 2 feet

shorter than its length L. Find the shorter than 6 times its width W . Find

perimeter of the rectangle in terms of its the perimeter of the rectangle in terms of

length alone. its width alone.

§ § § Answers § § §

1. 55xy 2 15. −15x3

3. −21xy 2 17. 2 + 25n

5. −16xy 19. 9x3 − 34x2 y

7. 0 21. −18xy − 4x3

9. −16x 23. 3 + 17m

25. −6x2 y + 16xy 2

11. 2q

27. −9x3 + 24xy

13. −19r

Version: Fall 2010

11.
3.4. COMBINING LIKE TERMS 207

29. 24n + 17 47. 4 − 8y

31. 9y + 4 49. 48n − 62

33. 24x2 y − 68 51. −10 − 10p

2 2

35. −22x + 70y

53. 15 + 39r

37. 2s + 8

55. −26 + 16x2

39. −7q − 15

57. 4 + 14W

41. 98y + 1

59. 4L − 16

43. 7x2 − 119xy 2

61. 10W − 18

45. −40 − 12n

Version: Fall 2010

29. 24n + 17 47. 4 − 8y

31. 9y + 4 49. 48n − 62

33. 24x2 y − 68 51. −10 − 10p

2 2

35. −22x + 70y

53. 15 + 39r

37. 2s + 8

55. −26 + 16x2

39. −7q − 15

57. 4 + 14W

41. 98y + 1

59. 4L − 16

43. 7x2 − 119xy 2

61. 10W − 18

45. −40 − 12n

Version: Fall 2010