s i m p l i f y i n g a l g e b r a simplifying...

Sim

plif

ying

Alg

ebra SIMPLIFYING

ALGEBRA

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Simplifying Algebra

100% Simplifying Algebra

Mathletics 100% © 3P Learning SERIES TOPIC

J 15

Algebra is mathematics with more than just numbers. Numbers have a fixed value, but algebra introduces variables – whose values can change. These are represented by letters. Although they work similarly to numbers it is important to be aware of how to add, subtract, multiply and divide expressions containing variables.

SIMPLIFYING ALGEBRA

What do I know now that I didn’t know before?

Answer these questions, before working through the chapter.

I used to think:

Answer these questions, after working through the chapter.

But now I think:

What is the difference between like and unlike terms?

Write an example explaining the distribution property

What is a binomial product?

What is the difference between like and unlike terms?

Write an example explaining the distribution property

What is a binomial product?

Simplifying Algebra


Mathletics 100% © 3P LearningSERIES TOPIC

J 52

Basics

Like Terms

It’s easy to see that 2 3 6# = and 3 2 6# = , so it doesn't matter in which order 2 and 3 are multiplied. This is always true! In any situation a b b a# #= . This is even true if more than two terms are multiplied: a b c a c b b a c# # # # # #= = .

This means xy = yx and they are like terms. abc = acb = bac are all equal and are also like terms.

The times sign #^ h is dropped in multiplcation of different terms. So x x4 4# = , a b ab# = and x y z xyz2 2# # =

The coefficients are always multiplied separately. Here are some more examples.

Terms with the same variables and indices are called like terms. If indices or variables differ they are called unlike terms.

Examples of like and unlike terms

Like Terms Unlike Terms

and

and

p p

xy xy

2 4

3 7

2 2

2 2

Different vairables

Different indices

andm n5 4

and4 4a b ab2 2

Only like terms can be added or subtracted.

Simplify the following as much as possible

Write the following products using algebra

2 3 7 2 9a ab a ab a ab2 2 2+ + - = +

Like terms

Like terms

This can't be simplified anymore because these are unlike terms.

Multiplying Terms

a b c3m n

mn3

#

=

p q r

p q r

pqr

2 4

2 4

8

# #

# # # #=

=

^ h

xy x y

x x y y

x y

6 3

6 3

18

2 3

2 3

3 4

#

# # # #=

=

^ ^ ^h h h

Coefficients multiplied separately Coefficients multiplied separately

Simplifying Algebra



J 35

Basics

Rules for Multiplication

When terms have the same sign their product is positive, when they have different signs their product is negative.

• #+ + = +

• #+ - = -

• #- + = -

• #- - = +

Positive # Positive = Positive

Positive # Negative = Negative

Negative # Positive = Negative

Negative # Negative = Positive

Here are some examples of multiplying with signs

Simplify the following

a

c

a b

b

d

a b

a b

ab

7 2

7 2

14

#

# # #

-

= -

=-

^ ^h h

m n p

m n p

mnp

2 2 5

2 2 5

20

# #

# # # # #

- -

= - -

=

^ ^h h

x y

x y

xy

3 5

3 5

15

#

# # #

- -

= - -

=

^ ^h h

t u v

t u v

tuv

5 4

5 4 1

20

# #

# # # # #

- - -

= - - -

=-

^ ^h h

The middle steps of c and d can be skipped if you remember this rule:

• If there are an even amount of negatives then the product is positive.

• If there are an odd amount of negatives then the product is negative.

Order of Operations

Remember that brackets and multiplication (and division) are always done before addition and subtraction.

Simplify the following as much as possible

xy x y3 2 4#+ -

xy xy3 8= -

xy5=-

Simplify like terms -(-42) = 42

Multiplication FIRST Multiplication FIRST Multiplication FIRST

a b b c5 2 6 7# #- - -

( )ab ac10 42=- - -

10 42ab ac=- +

Simplifying Algebra



J 54

Basics

2. Complete the following

1. Say if these are like or unlike terms

3. Use addition and subtraction to simplify the following as much as possible

a and 2a anda a2

pq and qp andcd d c2 2

a

a

a

b

b

b

c

c

c

c

d

d

d

d

3 4 4# #= yz =

ghi = abcd =

4 5 3 1x x x+ - + 11 20 8 7a b b a- + +

2 3 7 5 6jk j k jk k- + + + + 6 2 3 6ab ba ab ba+ - +

4 5 7 2def edf de ef ed- + + + - 4 2 3 8 4x y xy x xy yx2 2 2 2 2+ - + +

Simplifying Algebra



J 55

Basics

4. Use multiplication and the order of operations to simplify the following as much as possible

ba

dc

fe

hg

ji

lk

a4 3# x y2 6#

5 5t u#- 10p q#-

7 4 2b d c# #- 3 6g h i# #- -

8 2 3ux v w# #- - - 2 3 2p q r s# # #- -

x x y4 2 2#+ 3 2 4x y xy#- +

a b b a5 2 3 10# #+ - x w w z wz zw7 4 2 2# #- - - + +

Simplifying Algebra



J 56

Knowing More

When algebraic terms are divided, algebraic fractions are formed. These can be simplified by cancelling like terms.Write the division as a fraction. Always simplify the coefficients and cancel the variables where necessary.

Algebraic Fractions (Dividing Terms)

Simplify the following as much as possible:

a bm m

mm

mm

3 6

63

63

21

'

=

#=

=

Write as fraction Write as fraction

y is only in the numerator, so it is not cancelled. x is in the numerator and denominator, so x is cancelled

m is in both the numerator and denominator, so it can be cancelled

x xy12 4'

xyx

xyx

y

412

412

3

#

=

=

=

Rules for Division

Algebraic division with signs have the same rules as multiplication: If the terms have the same sign their quotient is positive, when they have different signs their quotient is negative.

• '+ + = + or ++ =+

• '+ - = - or -+ =-

• '- + = - or +- =-

• '- - = + or -- =+

Positive ' Positive = Positive

Positive ' Negative = Negative

Negative ' Positive = Negative

Negative ' Negative = Positive

Remember, always write the division as a fraction first. Then simplify the coefficients and cancel the variables.

a b

Simplify the following as much as possible:

Write as fraction Write as fraction

• The coefficients have been simplified • x is in both the numerator and

denominator, so it can be cancelled• The answer is negative '+ -^ h

• The coefficients have been simplified • a and c are in both the numerator

and denominator, so they can be cancelled

• b cannot be cancelled because it only appears in the numerator

• The answer is positive '- -^ h

x x15 10'- abc ac20 2'- -

xx

1015

23

=-

=-

acabc

b

220

10

=-

-

=

Simplifying Algebra



J 75

Knowing More

a b c

Order of Operations

Remember that brackets and division (and multiplication) are always done before addition and subtraction.

Simplify

Simplify

a bx xy y

yxy

x

6 4 2

24

4

'-

=

x

x x

6

6 2

= -

= -

division has been done firstmultiplication and division have been done first

Revising Index Laws

Here is a reminder of index laws.

1. Multiplication with indices: a a am n m n# = +

3. Raising an index to an index: a am n mn=^ h

5. Quotients raised to indices: ba

b

am

m

m

=` j

7. Negative indices: oraa b

aab1n

n

n n= =- -

` `j j

2. Division with indices: a aa

a am nn

mm n

' = = -

4. Products raised to indices: ab a bm m m=^ h

6. The zero index: for1 0a a0 !=

8. Fractional indices: a a anm mn n m

= = ^ h

pqr r p q

rpqr

pq

36 12 7 4

12

36

31

' #+

=

pq

pq pq

28

3 28

= +

= +

Index laws are used to simplify algebraic fractions. More than one law might be necessary to simplify a fraction.

pq p q q

p qq

q

p q

p q

1 1

2 3 3 8 0 4

3 64

4

3 6

3 2

' #

' #=

=

=

-^ ^h h b b

b b

b

b

34 4 7

43

47

48

2

#

#=

=

=

^ hx y x y

x y

x y

xy

y

x

1

2 3 5 7

2 5 7 3

7 4

7

4

4

7

# '

=

=

=

=

+ -

-

c m

Simplifying Algebra



J 58

Knowing More

Combining Index Laws and Algebraic Fractions

The next logical step is to mix indices and algebraic fractions. Algebraic fractions with indices work with the same two steps.

• Step 1: Simplify the numerator and denominator separately.

• Step 2: Simplify the fraction and cancel variables where necessary.

Here are some examples of algebraic fraction containing indices.

a

c d

b

Simplify as much as possible:

y y yy y y

yy

y

y

4 2 24 2 2

816

2

2

3

3 1

2

# #

+ +

=

=

=

-

Simplify numerator and denominator separately



Simplify fraction using index laws for division


x is in the denominator because its index is negative




Fractional indices

x y z

x y z

x y z

x y z

x y z

x y z

xy z

18

6

18

36

2

2

2

5 2

2 3 2

5 2

4 6 2

4 5 6 2 2 1

1 4

4

=

=

=

=

- - -

-

^ h

x y

x y

x y

x y

x y

x y

x y

x y

16

12

16

12

4

12

3

3

6 8

6 8

6 8 21

6 8

3 4

6 8

6 3 8 4

3 4

=

=

=

=

- -

^ h

p q

p q p q

p q

p q p q

p q

p q

p q

p q

2 4

8 16

128

128

128

3 3

2 3 2 2 2

3 3

6 3 4 4

3 3

10 7

10 3 7 3

7 4

#

#

-

=-

=-

=-

=-

- -

^ ^h h

Simplifying Algebra



J 95

Questions Knowing More

1. Find the quotient of the following:

a

a

e

c

g

d

d

b

f

d

b

e

c

f

a a8 2' a a2' pqr pr4 2'-

100 20amn n'- - wx xy18 9'- st stu5 20'-

2. Answer these questions about order of operations:

Simplify a b4 2#

Simplify a b4 2'

Simplify b a2 4#

Simplify 2b a4'

Is this statement true or false: a b b a4 2 2 4# #=

Is this statement true or false: a b b a4 2 2 4' '=

Is the order of the terms in multiplication important?

Is the order of the terms in division important?

Simplifying Algebra



J 510

Knowing MoreQuestions

3. Use the order of operations to simplify these as much as possible:

a

d

b

e

c

f

a ab b10 40 10'- xyz xz y26 2 6' +

xz xyz xz24 4# ' xy x yz z20 5 30 6' '+ d f def e15 3 200 10# '-

14 2 6mn n m' -

Simplifying Algebra



J 115

Questions Knowing More

a

c

e

b

d

f

4. Simplify these algebraic fractions as much as possible

y y yy y y2 33 4 5

# #

+ +

x x

x

2 2

42 4

3 2

#^

^

h

h

x y

xy x y

2

5 22 2 3

2 2 2 3#

^

^ ^

h

h h

pq

pq p q32

2 3 3#

-

-^ ^h h

f g f g

f g fg

3 2

9 22 2 4

2 3 3

#

#

^

^

h

h

q r

qr q r

3 2 2

2 3 2#-

^e

ho

Simplifying Algebra



J 512

Knowing MoreQuestions

5. Answer these questions

a

b

Use the index law aa

1nn

=- to prove ba

abn n

=-

` `j j (Hint: )ba

ban n1

=- -

` ``j j j

Use this result to simplify this fraction x y

xy

3

32 3

2 2 2-

^

^e

h

ho

Simplifying Algebra



J 135

Using Our Knowledge

p q p

pq p p

3 2 6 9

6 18 272

- + -

=- - +

^ h

Expanding Brackets

A term outside a bracket is multipled with all terms inside a bracket, so that

a b c a b a c ab ac# #+ = + = +^ h

This property of multiplication with brackets is called the 'distribution property'. Here are some examples:

Expand the following



a

a

a

b

b

b

x

x

3 2

6 3

+

= +

^ h

( )y

y

4 5

4 20

- -

=- +

y4 # 4 5#- - p q3 2#- 3p p6#- 3p 9#- -

x x y

x x xy

5 2 3

5 2 32

+ +

= + +

^ h

5x # x x2# x y3#3 2#- 3 x#

There is no difference if the number on the outside is negative. Always multiply the term outside with all the terms inside.

Sometimes like terms can be simplified after expanding brackets.

4 3x x x x2 5 22- + -^ ^h h

x x23 10 2= -

x x x x8 4 15 62 2= - + -

Like terms

Like terms Like terms

Like terms

2q p pq q p pq1 4 3 2 5- + - + -^ ^h h

q pq pq pq pq2 8 6 2 102 2=- - + + -

q pq pq2 6 4 2=- - -

Simplifying Algebra



J 514

Using Our Knowledge

Multiplying Brackets

Brackets can also be multiplied together. Both terms in the first bracket are multiplied with both terms in the second bracket.

( )( )a b c d+ +

( )( )x x3 2+ + ( )( )y y2 3 3 4+ -

1

1

3

3

4

4

2

2

ac ad bc bd= + + +

The product of two brackets can also be thought of in this way

( )( ) ( ) ( )a b c d a c d b c d

ac ad bc bd

+ + = + + +

= + + +

Find the following products:

a

c

b

d

1 1

1 1

3 3

3 3

4 4

4 4

2 2

2 2

( ) ( ) ( ) ( )x x x x2 3 3 2# # #= + + + + ( ) ( ) ( ) ( )y y y y2 3 2 4 3 3 3 4# # # #= + - + + -

x x x2 3 62= + + + y y y6 8 9 122

= - + -

x x5 62= + + y y6 122

= + -

Like terms

Like terms

p q p q

p p p q q p q q

p pq pq q

p pq q

2 3 4

2 2 4 3 3 4

2 8 3 12

2 5 12

2 2

2 2

# # # #

+ -

= + - + + -

= - + -

= - -

^ ^

^ ^ ^ ^

h h

h h h h

a b c d

a c a d b c b d

ac ad bc bd

3 5 3 2

3 3 3 2 5 3 5 2

9 6 15 10

# # # #

- +

= + + - + -

= + - -

^ ^

^ ^ ^ ^

h h

h h h h

Simplifying Algebra



J 155

Questions Using Our Knowledge

1. Expand these brackets:

a

c

e

g

i

b

d

f

h

j

( )y5 2 3+ ( )t6 4 3-

( )m4 3 5- - -

( )x y21 8 2-

( )x3 9 4- -

( 4 )n m41 16 2- -

2x x y4 2-^ h x xy y4 32-^ h

5p q pq p2 3+ +^ h 3xy x y2 2 1- +^ h

Simplifying Algebra



J 516

Using Our KnowledgeQuestions

2. Expand these brackets:

a x x y xy2 3 4 4+ - +^ h b 2abc ab ac a b bc c3 4 2 5 32 2- - - + - +^ h

4 2x x x x3 7 52- + -^ ^h h

3. Expand then simplify using like terms:

a

c

b

d

d c cd cd d cd1 2 3 5 3 22 2- + + - -^ ^h h

5 4p q pq q p q p3 2 2 2 32 2- - + -^ ^h h 3 (8 2 4 )ab a b b a b a b ab a a b2 62 2 3 2 2 2

- - + - + - +^ h

Simplifying Algebra



J 175


4. Expand and simplify:

a

c

e

g

b

d

f

h

x x2 2 1+ -^ ^h h st t t s2 3 4 2+ -^ ^h h

k k4 1 3 2- -^ ^h h

x y x y+ -^ ^h h

ab a b ab2 3 62- +^ ^h h

ab cd ab cd- +^ ^h h

x y xy xy x3 4 22 2- +^ ^h h ab c a bc abc a b c3 4 5 23 2 2 2 2 2

- -^ ^h h

Simplifying Algebra



J 518

Using Our Knowledge

Step 1: Find a common denominator

Step 2: Write both fractions over this denominator

Step 1: Find a common denominator

Step 2: Write both fractions over this denominator

Adding and Subtracting Algebraic Fractions

Like all fractions, algebriac fractions can only be added or subtracted if they have the same denominator. If the fractions already have a common denominator, then simplify the numerators only.

Find the following products:

a

c

a b

d

b

These fractions already have a common denominator

12 is the lowest common denominator of 3, 4 and 6

30 is the lowest common denominator of 15, 30 and 10

ab is the lowest common denominator of a and ab

This is the product of the original denominators

This is the product of the original denominators

Use the distributive property to expand the bracket

Multiply the brackets to find this denominator

Sometimes its difficult to find a common denominator. In these situations, the common denominator can be found by multiplying the denominators together.

Adding and subtracting fractions with complicated denominators

x 32

75

++

7 x 3+^ h

x x

x

x

x

7 314

7 3

5 3

7 3

14 5 3

=+

++

+

=+

+ +

^ ^^

^^

h hh

hh

Step 3: Simplify Step 3: Simplify

xx

xx

7 2114 5 15

7 215 29

=+

+ +

=++

m n m n2 2+

--

( )( )m n m n+ -

m n m n

m n

m n m n

m n

m n m n

m n m n

2 2

2 2

=+ -

--

+ -

+

=+ -

- - +

^ ^^

^ ^^

^ ^^ ^

h hh

h hh

h hh h

m mn mn n

m n m n

m n

n

2 2 2 2

4

2 2

2 2

=+ - -

- - -

=--

y y y

y y y

y

3 43

6

124 9 2

1211

+ -

=+ -

=

p q p q

p q p q

q q

15

8

30

2

10

2

30

16 2 6 3

30

25

6

5

- ++

=- + +

= =

ab a

abb

5 5

5 5

+

= +

h h

h h

h

199

194

199 4

195

-

= -

=

Simplifying Algebra



J 195


5. Simplify these fractions:

a

c

e

g

b

d

f

h

x x47

45+

x x32

125-

mn km3 2+

x65

1 22+-

a a2 13 4-

-y y3 13

52

++

y y64

11

+-

- p p p

325 2

2+ -

Simplifying Algebra



J 520

Using Our Knowledge

Multiplying and dividing algebraic fractions is easy because a common denominator is not necessary. To multiply algebraic fractions, simply multiply the numerators and the denominators separately.

Multiplying and Dividing Algebraic Fractions

Find these products:

a

c

a b

b

d

xy

xy

xy

34

43

43

#

#

#=

=

pqp

qp p

qp

3

2

3

2

3

2 2

#

#

#=

=

t

t

t

t

t

t

t

t

510

5

10

5

10

2

2

4

2

2

4

2

4

4 2

2

#

#

#=

=

=

=

-

a b

a b

a b

a b

a b

a b

a b

b

a

2 2

24

2 2

24

4

24

6

6

7 2

4 4

4 4

7 2

4 4

7 2

7 4 2 4

2

3

# #

#

#=

=

=

=

- -

To divide algebraic fractions, simply find the reciprocal of the second fraction and multiply.

Find these quotients:

x y

xy

yx

yx

yx

82

16

82 16

82 16

832

4

'

#

##

=

=

=

=

cab

cb

cab

b

c

c b

ab c

b c

abc

ba

43

2412

53

12

24

5 12

3 24

60

72

56

2

2

2

2

'

#

#

#

-

= -

= -

= -

=-

Multiply numerators and denominators separately


'flip' the second fraction to find the reciprocal.

'flip' the second fraction to find the reciprocal.



Simplifying Algebra



J 215


6. Write these in simplest form:

a

c

e

g

b

d

f

h

p p2 10#

mnn510

#

mn

m n5252 3

#p q

p

q7

3 62

4#

ps t

t

p s9

8

16

273 2

4

3 2

#ba

cb

dc

# #

e f

d

d

e f d ef

5

3

3

443 2 4

4 5 2 3

# #a

wx ywxya b

2

52510

2

2 3 4

#-

Simplifying Algebra



J 522

Using Our KnowledgeQuestions

7. Write these in simplest form:

a

c

e

b

d

f

p q2

3

6'

ba a

2

2

#

a b a b3

49

23 5 2 2

#f

d e

d f

e

8

3

16

92

2

2

3

'

x y

s t

x y

s t5 203 4

3 2

6 2

5

'p q r

a b c

p q r

a bc

7

6

21

24 3 2

2 3 2

2 5

3 4

'

Simplifying Algebra



J 235

Thinking More

p q

p p q q

p pq q

4 2

4 2 4 2 2

16 16 4

2

2 2

2 2

-

= + - + -

= - +

^

^ ^ ^ ^

h

h h h h

Shortcuts for Multiplying Brackets

A product of two brackets it called a ‘binomial product’. Some binomial products can be found more easily than others. Look at this example:

a b a b

a ab ab b

a b

2 2

2 2

+ -

= - + -

= -

^ ^h h

The middle terms cancel each other away

Can you see the shortcut? Their product is the square of the first term minus the square of the second term. Here are some examples using this shortcut:


x x

x

x

2 2

2

4

2 2

2

+ -

= -

= -

^ ^

^ ^

h h

h h

a b y y

y

y

3 5 3 5

3 5

9 25

2 2

2

- +

= -

= -

^ ^

^ ^

h h

h h

What about the perfect square of a bracket? Look at this example:

2

a b a b a b

a ab ab b

a ab b

2

2 2

2 2

+ = + +

= + + +

= + +

^ ^ ^h h h

1st termsquared

1st termsquared

2nd termsquared

2nd termsquared

2nd termsquared

2 # product of terms

1st termsquared

2nd termsquared


1st termsquared

2nd termsquared


Can you see the shortcut? Here are some examples using this shortcut:


x

x x

x x

4

2 4 4

8 16

2

2 2

2

+

= + +

= + +

^

^ ^ ^ ^

h

h h h h

a b

1st termsquared

Simplifying Algebra



J 524

Thinking More

These shortcuts may seem silly for these easier examples, but they are especially useful for more complicated examples.


a

c d

bx

xx

x

xx

xx

4 4

4

16

2 2

22

+ -

= -

= -

` `

` ^

j j

j h

x y pq x y pq

x y pq

x y p q

3 4 3 4

3 4

9 16

2 3 2 3

2 2 3 2

4 2 2 6

+ -

= -

= -

^ ^

^ ^

h h

h h

2

25 2

25

yy

y yy y

yyy

y

yy

523

5 523

23

215

4

9

154

9

2

22

22

22

-

= + - + -

= + - +

= - +

c

^ ^ c c

c

m

h h m m

m

2

4 2 9

4 2 9

m n n

m n m n n n

m n m n n

m n m n n

2 3

2 2 3 3

6

1

3 2 2

3 2 2 3 2 2

6 4 3 3 2

6 4 3 3 2

-

= + - + -

= + - +

= - +

^

^ ^ ^ ^

^

h

h h h h

h

We can even use these rules in calculations using actual numbers.

a b11 9

11 9 11 9

20 2

40

2 2-

= + -

=

=

^ ^

^ ^

h h

h h

36

30 2 6

900 360 36

30 6

30 6

1296

2

2

2 2

= +

= + +

= + +

=

^

^ ^

h

h h

Simplifying Algebra



J 255

Questions Thinking More

1. Prove that x y x y x y2 2+ - = -^ ^h h

2. Simplify these perfect squares

a

c

e

b

d

f

p2 1 2+^ h x y5 7 2-^ h

m n3 2 2- -^ h p pq3 2-^ h

xx1 2

+` j yy

2 322

-c m

Simplifying Algebra



J 526

Thinking MoreQuestions

3. Simplify these binomial products:

a

c

e

g

b

d

f

h

x x3 3+ -^ ^h h m m5 5- +^ ^h h

b c b c2 3 2 3+ -^ ^h h x y x y7 3 7 3+ -^ ^h h

xy x y xy x y2 2 2 2+ -^ ^h h a b ab a b ab10 4 10 42 3 2 2 3 2

- +^ ^h h

mm

mm

21

212 2

+ -` `j jx

yx

y52 3

52 3- +` `j j

Simplifying Algebra



J 275

Questions Thinking More

4. Find 9 and 4 in each of the following

a

c

a

e

c

b

f

d

b

d

9x x x3 2 2+ = + +9^ h 4 9x x xy y2 122 2 2

+ = + +9^ h

5 16 40 25q p pq q2 2 2- = - +9^ h 4 16m m mn2 2 2 2

+ = + +9 4^ h

5. Find these values using expanded form:

16 62 2- 64 362 2

-

292 372

77 272 2- .12 62 (Hint: 12.6 = 12 + 0.6)

Simplifying Algebra



J 528

Thinking MoreQuestions

6. A stage in a stadium is made up of a rectangle and a square in the following diagram. The dimensions are in terms of x.

Find expression for the area of the square in the form of a binomial product

Find expression for the area of the rectangle in the form of a binomial product

Show that the total area of the stage is ( )( ) ( )x x x2 2 4 2 3 2- + + -

Simplify the expression for the area of the stage

Find the value of the area of the stage if 5x =

Rectangle

Square

2x + 4

2x - 3

a

b

c

d

e

x -

2


Mathletics 100% © 3P Learning

Simplifying Algebra

SERIES TOPIC

J 295

Answers

Basics: Knowing More:

Knowing More:

Using Our Knowledge:

a

a

a

b

b

b

c

c

c

e

d

d

d

d

3 4 4 3# #=

x6 1+ a b4 12-

j k jk3 13 3+ + ab11

edf ed f3 3 73+ +

x y xyz x8 10 32 2+ -

Like terms

Like terms Like terms

Unlike terms

yz z y#=

ghi g h i# #=

abcd a b c d# # #=

1.

2.

3.

4.

1.

2.

2.

3.

4.

5.

1.

2.

a

c

g

e

i

k

b

d

h

f

j

l

a12 xy12

tu25- pq10-

bcd56- ghi18

uvwx48- pqrs12

x xy4 4+ xy2-

ab20- zw xw11 7-

a

a

e

c

g

d

h

b

f

d

b

e

c

f

4a1 q2-

am5 yw2-

u41-

8ab

8ab

True, they are both 8ab.

False, they are not equal.

Yes, the answers are different.

No, the answers are the same.

ba2

ab

2

a

d

b

e

c

f

a6 y19 m

xyz6 y9 df25

a

c

b

d

y

22 x2

12

x y25 2 p q9 4 3

e ff

g43

3r81

2

a bba

abn n

=-

` `j jy

x92

8

a

c

e

g

i

b

d

f

h

j

y10 15+ t24 18-

m12 20+

x y4 -

x27 12+

n m4 6- +

x xy8 42- x y yx4 33 2

-

pq p q p5 10 152 2+ +

x y xy xy6 6 32 2- +

a x x xy x y2 3 4 42 2+ - +

b 6 2 8 4 10 6abc a b c a bc a b ab c abc2 2 2 2 3 2 2 2 3- + + - + -



Simplifying Algebra

SERIES TOPIC

J 530

Answers

x x2 62-a

c

b

d

cd cd d c d cd3 3 3 23 2 2 2 3+ + - -

pq q q7 12 182 2+ -

ab a b a b4 13 113 2 3 2- -

a

c

b

d

x2 22-

st s t t st8 4 12 62 2 2- + -

k k k8 14 3 22 2- + - +

ab a b ab a b6 6 83 2 2 2 2+ - -

e

f

x y2 2-

a b c d2 2 2 2-

g x y x y x y x y3 6 4 83 2 3 2 3 2 2+ - -

h a b c a b c a b c a b c15 6 20 82 4 2 3 5 3 3 3 3 4 3 4- - +

Using Our Knowledge: Using Our Knowledge:

Thinking More:

3.

4.

5.

6.

7.

2.

3.a

a

a

c

c

c

e

e

e

g

b

b

b

d

d

d

f

f

f

h

x3 x4

kmnk n3 2-

xx

6 1217 10

--

a aa2 1

11 4--

^ h y yy

5 3 121 2

++

^ h

y yy1 6

3 10

- +-

^ ^h h p

p

2

16 42

-

a

c

e

g

b

d

f

h

p20

2

m2

mn

52 p

q

7

182

2

t

p s

2

32

2 5

da

de f5

2 6

a bxy2 2-

a

c

e

b

d

f

qp9

ba2

3

a b21

8 5 7

fe

d

3

22

4

s y

tx

4 2 2

3

ac p r

b q92 2

2 2

p p4 4 13+ +

x xy y25 70 492 2- +

m mn n9 12 42 2+ +

p p q p q9 62 2 2 2+ +

xx1 2

2

2+ +

y yy

4 12 94

2- +

g

h

x 92- m25 2

-

b c4 92 2- x y49 92 2

-

x y x y2 2 4 4-

a b a b100 164 6 2 4-

mm

4

12

4-

x xy

xy25

456

56 9

2 - + -



Simplifying Algebra

SERIES TOPIC

J 315

Answers

Thinking More:

4.

5.

6.

a

c

b

d

6=9 y3=9

p4=9 n

n

4

16

2

4

=

=

9

4

a

e

c

b

f

d

220 2800

841 1369

5200 .158 76

a

e

b

d

x2 3 2-^ h

x x2 2 4- +^ ^h h

x x6 6 12- -

119

Simplifying Algebra



J 532

Notes

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