YEAR 7

ALGEBRAIC EXPRESSIONS
3(x – 8)
is
3x + (-24) or 3x - 24

-2x + 5x – 9 + 10
is
3x + 1
4x + 12
is
4(x + 3)

S Survey
P Parentheses
C Catch & Combine
= ...in equations...
A Clear Add/Subtract
D/M Clear Division/Multiplication

Name
Period
 Simplifying Algebraic Expressions by Combining Like Terms + C
Objective: Students will identify like terms.
Students will simplify algebraic expressions by combining like terms.

Term Definition Picture/Example

Terms Quantities that you There are 3 terms in
ADD to form an
algebraic expression
are called terms. 4n + 6b – 8

The terms are:

Like Terms terms with the

same variable raised to
the same power

You can COMBINE Like You CAN add/subtract

Terms like terms.
**COMBINE means
add, so use the
addition rules
(SSS, DSD)
Unlike Terms terms whose variables
are not the same, or
who have the same
variable, but it’s raised
to a different power

You CANNOT
add/subtract unlike
terms.

1
2
For each algebraic expression, identify the number of terms. Then list the
coefficients and any constant terms.

6a + 3 6a – 3 0.2x – y + 8z ½n
Expression
Number of Terms

Coefficient(s)

Constant(s)

Identify the number of terms, the coefficients, and the constant term of the
expressions below.

1. 7p – 6pc + 3c - 2

Number of terms:

Coefficients:

Constant terms:

2. 8 + 4ab - 5b

Number of terms:

Coefficients:

Constant terms:
3
To simplify by combining like terms: + C
1. Search for like terms (same variable raised to the same power; and constants
with other constants).

2. Catch the first term and any like terms.

3. Combine them using the addition rules. (SSS, DSD)

4. Continue with other like terms.

*Remember that an “invisible 1” lurks in front of variables that appear to have no

coefficient attached to them.

1) 4x + 5x + 7 + x + 2 2) 2n + 3 – 5n + 6

3) - 9b + 2n – 4 + 2b 4) -7g + 3 – 8 – 3g + 7h

5) -8 + 2d – 7 – 5d + 12 6) 5b + 7 – 3b – 4

4
Identify the number of terms, the coefficient(s), and the constant HOMEWORK
term(s) of the expressions below.

1. 6p – 7pc + 9c – 4 2. 3 + 4ab - 5b + m
+ C

Number of terms: Number of terms:

Coefficients: Coefficients:

Constant terms: Constant terms:

3. Sarah was asked to identify all coefficients and constants of the

expression 4 + n + 7m. She said that 4 is a constant, and 7 is a coefficient.

What is her error?

a. She did not include the constant 1.

b. She said 4 is a constant. It is actually a coefficient.
c. She did not include the coefficient 1.
d. She said 7 is a coefficient. It is actually a constant.

4. Add. 2a + 8 + 4b + 5 5. Add. 8x – 7 + 6x + 8

6. Find the sum. 8x + 2 -9x + 7 7. Find the sum. 3n + 4 – 8n -1

5
 Variable A symbol used to represent an
unknown amount.

The symbol is usually a letter

of the alphabet.

Coefficient The number being multiplied by

a variable.

It is the number attached to

the variable, and is usually in
front.

A variable with no coefficient

*Special
has an “INVISIBLE 1”
note! attached to it!

Constant A number that doesn’t change.

There is no variable attached
to a constant.

Algebraic An expression that contains

Expression variables.

6
Expanding Algebraic Expressions (The Distributive Property) day 1

Objective: Students will simplify algebraic expression using the distributive property.

Term Definition Example

Distributive The distributive property
combines multiplication with
Property addition and subtraction

You can multiply constants and algebraic terms simply by

multiplying the constant and the coefficient.
The variable remains the same.

Remember, if the variable has no coefficient, it’s an invisible 1.

a. 2(3x) = b. -2(3d) =

c. 5(n) = d. -3y(4) =

You can also multiply variables by one another.

e. a•t= f. b(y) =

g. 3c(b) = h. 2n(4x) =

But what happens when you have more than one term inside the
parentheses?

7
Examples: 2(n + 4) 3(x – 8)

8
 The Distributive Property x P (clear Parentheses)

KISS YOUR PARENTHESES GOODBYE!

Step 1: Catch the number touching the parentheses (on the

outside) and any number inside that has a sign.

Step 2: Multiply the number on the outside of the parentheses

by the FIRST number inside the parentheses.

I “times” over the rainbow ☺

Step 3: Multiply the number on the outside by the SECOND

number that’s inside.

Examples:

1. 3(4x + 2) 2. -3(4x + 2) 3. -3(4x - 2)

9
 xP
1. 5(x + 3) b. 2(x + 1) c. 4(x + 5)

2. -3(x + 4) b. -6(x + 5) c. -1(x + 4)

3. -4(x - 4) b. -8(x - 3) c. -1(x - 7)

4. -3(2x - 5) b. -2(4x - 7) c. -2(6x - 8)

5. a(b - 4) b. n(d + 1) c. y(5 - z)

6. 2a(3p - 5) b. 4n(6d + k) c. 5y(6h - w)

9
 HOMEWORK xP

a. 3(x + 2) b. 5(2y – 7) c. -2(n + 9)

d. -3(k – 1) e. -4(1 + a) f. 3(d – 4)

g. -1(3x + 4) h. -3(b – 9 + 2y) i. -5(2 – m)

j. 3(n – 4 + 5y) k. -6(j – 2 + 3k) l. -1(3 - h)

pg. 10
 Expanding Algebraic Expressions (The Distributive Property) day 2

Let’s review using the distributive property:

xP
1. 3(x – 4) 2. 4(n + 1) 3. -5(x – 5)

1 1 1
4. (10) 5. (12) 6. (9)
2 2 3

1 1 1
7. (16) 8. (6x + 10) 9. (8x − 4)
4 2 2

1 1
10. (12x + 9) 11. (15x − 3) 12. 5(2x + 1 – n)
3 3

Objective: Students will simplify algebraic expression using the distributive

property. Students will recognize that a problem can be written in different
forms. Students will recognize that some problems that have parentheses, may
NOT require the dist. property.

Remember!

The distributive property is multiplication over addition or subtraction. This

means that in order to distribute, you must have a term (constant or variable)
that is touching parentheses that contain more than one term. Those terms must
be separated by addition or subtraction signs, NOT multiplication or division signs.

pg. 11
 ERROR ALERT!

Some addition and subtraction problems look very similar to distributive property
problems.

Ex. -3(x + 7) This expression requires the distributive property.

(5 – x)(-2) This expression requires the distributive property.

(5 – x) – 8 and 2(n • 4) These expressions do not!

1. Circle the problems that require the distributive property. Put an X through
those that do NOT require the distributive property.

-a(3 + b) (-a)(3) + (b) (3 + b) - a (3 + b)(-a)

4(2 • n) (3 – g)(-5) (-4)(6) + n (8 + h) - 2

SHOULD I USE THE DISTRIBUTIVE PROPERTY??

YES NO
-3(x + 1) -3 + (x + 1)

(2x – 4)(3) (2x – 4) + 3

(3 + y)(-2) (3 y)(−2)

pg. 12
 Use the Distributive Property to expand each expression.
xP

Ex. -4(2n + 5)

a. 3(4x + 2) b. -5( y – 7) c. -2(n + 9)

d. 3(-1 – 5c) e. x(3 + 4y) f. -1(a – 1)

1 3
− (8x + 12) (8x - 12)
g. ½(4x – 6) h. 4 i. 4

pg. 13
 HOMEWORK
Expand the expressions that require the distributive property.
Put an X through the expressions that do not require the distributive property.
REMEMBER—you can only distribute (multiply) over addition or subtraction!

1. -6(a + 8) 2. 4(1 + 8x + a) 3. (-5n + 7)6

4. 2 + (4k – 3) 5. (9m + 10) • 2 6. -8(-b – 4)

7. (3)(y)(-2) 8. 5(3 • y) 9. -4(-6p + 7)

1 1
(9n + 15) (6b − 10)
10. -4(n  5) 11. 3 12. 2

pg. 14
 Simplifying Algebraic Expressions by Distributing
and Combining Like Terms

pg. 15
 S
Simplifying Algebraic Expressions by Distributing P
and Combining Like Terms C
=
A
Objective: Students will simplify algebraic expressions by combining like D/M
terms.

Students will recognize when the distributive property is required to

simplify an expression and when it is not.

Before simplifying an expression that contains parentheses,

you must determine whether or not you need to use the distributive
property. If so, DO IT FIRST!

A. Simplify the expressions below (Hint: ONE of them needs dist. property).

a. 2x + 5 + (6x + 1) b. 2x + 5 + 6(x + 1)

B. Sometimes you will need to CATCH the term (including a subtraction sign)
before you distribute.

1. x – 2(x + 3) 2. 3n – 2(n + 5) 3. 4h – 3(2h + 5)

pg. 16
 Simplify. (Ask… Do I need to distribute? If so, do it FIRST!)
S
1. 3(x + 6y – 7) P
C
=
A
D/M

2. -7(2x – 4)

3. 4(3x + 7) - 5x

4. 3x - 2(-4x + 5)

pg. 17
 S
Extra Practice
P
Rewrite the expressions without parentheses. C
(Hint: check for dist. property first!) Then simplify. =
A
1. 8 + (2x – 1)7 2. -3(3x – 5) + 8x D/M

3. 9x + 2 + (3x – 7) 4. 12x + (-4x + 1)3

5. 2y + 7(2y – 3) 6. -8y + 4(3y + 3) - 5

7. -4 + (3n – 5) – 7n Notice the difference between #7 and #8. 8. -4(3n – 5) – 7n

pg. 18
 S
HOMEWORK
P
Simplify. C
=
1a. -3(2x – 3y - 5) 1b. -4(2y - 3) + 7y
A
D/M

2a. (b – 7) + (3b – 9) 2b. (-6x – 11) + (5x + 12)

3a. –2(-4x + 5) + 6x 3b. 7x + 3(-2x + 3)

4. For each algebraic expression, identify the number of terms. Then list the
coefficients and any constant terms.

Expression 8x - 3 4x + y + 11

Number of Terms
Coefficient(s)
Constant(s)

pg. 19
 Simplifying Algebraic Expressions by Distributing
and Combining Like Terms
x P
1.a 3(x + 6y – 7) 1.b. -7(2x – 4)

Distribute first…. then catch and combine like terms. x P

2a. 4(3x + 7) - 5x 2.b. -2(-4x + 5) + 3x
+ C

Ask… Do I need to distribute? If not, Catch and Combine Like Terms.

3a. 5x + 6y – 4x + 3y – 9 3.b. - 9x + 7 –x + 5x - 4y

Ask… Do I need to distribute? If so, do it first. Then “Bring Down the

Junk.” If not, drop the parenthesis and then catch and combine like
terms.

4a. (x – 5) + (4x – 2) 4b. 6x + 4(-2x + 3)

pg. 20
 HOMEWORK

Rewrite the expression without parentheses. Ask…

Do I need to distribute? If so, do it first. Then x P
“Bring Down the Junk.” If not, drop the parenthesis and
then catch and combine like terms.
+ C

1. 7(2x – 1) + 8 2. -3(3x – 5) + 8x

3. 9x + 2(2x – 7) 4. 12x + 3(-4x + 1)

5. 12y + 7(2y – 3)

pg. 21
 Adding and Subtracting Algebraic Expressions

Objective: Students will add and subtract algebraic expressions.

Recall that only like terms can be added or subtracted.

Simplify the following problem by combining like terms.

Ex 1: (2n + 3) + (4n + 5) Ex 2: (-3h + 2) + 3(4h - 2)

***Subtraction of expressions can be especially difficult!

Note the difference between the two problems below.

Which problem requires the distributive property to simplify it?

Rewrite using the distributive property where necessary.

(8x – 3) + 2(3x + 1) and (8x – 3) – 2(3x + 1)

pg. 22
 Note the difference between the two problems below.
Which problem requires the distributive property to simplify it?

Rewrite using the distributive property where necessary.

(7x – 3) + 1(4x + 1) and (7x – 3) – 1(4x + 1)

Recall that often times in math the 1 is “invisible”, as is a -1.

Here are the same problems rewritten with an “invisible” 1.

(7x – 3) + (4x + 1) and (7x – 3) – (4x + 1)

HOWEVER, that means when there is a subtraction sign between expressions,

you must think of it is as distributing a -1.

a. Distribute the -1 b. Distribute the -1

(6n – 5) – (2n + 1) (8a + 1) – (2a + 4)

pg. 23
 c. Distribute the -1 d. Distribute the -1

(5h – 4) – (2h + 1) (2n + 4) – (n - 1)

e. Distribute the -1 f. Distribute the -1

(-2x – 5) – (6x + 3) (-3y + 1) – (4y + 8)

pg. 24
 HOMEWORK
1. Distribute the -1

(6h + 4) – (2h + 3) x P
+ C

2. Distribute the -1

(6h + 4) – (2h - 3)

3. Distribute the -1

(6h - 4) – (2h - 3)

pg. 25
 Factoring Algebraic Expressions
Objective:

Students will factor algebraic expressions by REVERSING the distributive property.

Term Definition Example

A number that is List the factors of 18 and of 12.
Factor
(as a noun) multiplied by another
number to get a
product.

Factor When you factor an Factor: 18n + 12

(as a verb) expression, you reverse
Skeleton:
the distributive
( + )
property.
1. Find the GCF of the numbers.
The number on the 2. Find the GCF of the variables.
outside of the 3. The GCF’s go on the outside.
4. The leftovers go on the inside.
parentheses is the GCF.

Greatest The largest factors

that is shared by two
Common
or more numbers. The GCF of 18 and 12 is .
Factor

(See top example.)

Factored Form An expression in
factored form has the The factored form of 18n + 12
GCF on the outside of is .
the parentheses, and a
sum or difference on
the inside.

(See 2nd example.)

pg. 26
 Ex 1. Factor. 14xy + 21x

To factor an algebraic expression you use division to undo the distributive property.

1. Make the “skeleton” of a distributive property problem under the given one.
It will look like this: ( + ) or ( - ) this.

2. Look for a variable that is shared by the terms (they may share more
than one); circle it and then place the shared variable on the outside,
next to the parentheses.

3. Bring down any variables not circled.

4. Find the GCF of the coefficients. (Check to see if the smallest # you see is a
factor of the others. If so, it’s the GCF. Otherwise, use the “rainbow” method.)

5. Place the GCF on the outside of the parentheses. If there are any
letters already there, the GCF will become their coefficient.

To fill the inside:

6. Divide the 1st term by the GCF you found. Place the quotient on the line.

7. Divide the 2nd term by the GCF you found. Place the quotient on the line.

8. Repeat until all terms have been divided by the GCF.

9. Check your work by applying the distributive property to your answer to

see if it matches the original expression.

Ex 2. 12x + 6
pg. 27
 Ex. 3 5b – 15n Ex. 4 3n – nj

Use the distributive property to check.

1. 10n + 15 Check:

2. 8y - 12y Check:

3. 8n – 2 Check:

4. 14z + 21 Check:

Check:
5. 4h - 12

pg. 28
 6. 8a + 4 Check:

7. 12a + 16b – 10c Check:

8. 12x – 24y - 3 Check:

9. 9n + 7n Check:

10. -5xy + 25x Check:

11. 18b + 3 Check:

pg. 29
 HOMEWORK
1. Write the factored form of each expression.
2. Check using distributive property.

a. 18a + 3 Check:

Check:
b. -2c + 6d

Check:
c. 20xy + 10x

Error analysis: Ali factored the above problem, c, and got 2x(10y + 5). When
she checked it using the distributive property, she got the original problem! Since
her check worked, she thinks she has the correct answer. How can that be?
Explain her mistake.

3. Error analysis: Jamie incorrectly factored 15x – 20xy. She got 5x(3 – 4xy).
Factor the expression correctly.

What error did she likely make?

a. She did not have the correct operation inside parentheses.

b. She did not factor the variable from the first term.
c. She did not factor the variable from the second term.
d. She did not simplify the terms inside the parentheses.
pg. 30
 HOMEWORK
More factoring practice.

1. 24xy + 10x

2. 15m – 18mn

3. 4x – xy

4. 42x + 7y

5. 4x – 2xy

6. 32xyz + 12xy

pg. 31
pg. 32
 Indices part 1
In this chapter, you are going to

•
• Learn about negative indices
• Learn about fractional indices
• Use fractional indices to evaluate square roots and cube roots

Recap
1. Numbers written in index form

81 = = 34 Index form
3×3×3×3

Expanded form

34 is read as 3 to the power or index of 4 where 𝟑 is called the base.

2. Multiplication law

𝒂𝒎 × 𝒂𝒏 = 𝒂𝒎+𝒏

Examples:

(a) 𝒂𝟑 × 𝒂𝟐 = 𝒂𝟑+𝟐 = 𝒂𝟓
(b) 𝟒𝟓 × 𝟒𝟐 = 𝟒𝟓+𝟐 = 𝟒𝟕
(c) 𝟑𝒂𝟑 × 𝟐𝒂𝟐 = (𝟑 × 𝟐)𝒂𝟑+𝟐 = 𝟔𝒂𝟓

3. Division law

𝒂𝒎
𝒂𝒎 ÷ 𝒂𝒏 = = 𝒂𝒎−𝒏 , 𝒂 ≠ 𝟎
𝒂𝒏
Examples

(a) 𝒂𝟓 ÷ 𝒂𝟐 = 𝒂𝟓−𝟐 = 𝒂𝟑
𝟑𝟕
(b) 𝟑𝟐 = 𝟑𝟕−𝟐 = 𝟑𝟓

𝟏𝟐𝒎𝟓 𝟏𝟐 𝟓−𝟒 𝟏

(c) = ( 𝟑 )𝒎 = 𝟒𝒎 or simply 𝟒𝒎
𝟑𝒎𝟒

4. Power law

(𝑎𝑚)𝑛 = 𝑎𝑚𝑛
Examples

(a) (𝑚2)4 = 𝑚2×4 = 𝑚8

(b) (53)2 = 53×2 = 56
(c) (𝑚2)4 × (𝑚3)5 = 𝑚8 × 𝑚15 = 𝑚23
7 3
(𝑝3) ×(𝑝4) 𝑝21×𝑝12 𝑝33
(d) = = = 𝑝25
(𝑝2)4 𝑝8 𝑝8

5. Meaning of 𝒂𝟎

Any number raised to the power of zero is 1

𝒂𝟎 = 1

Examples

(a) 𝑚0 = 1
(b) 30 = 1
(c) 3𝑚0 = 3
(d) 𝑏 5 ÷ 𝑏5 = 𝑏 0 = 1
So far we have done a recap of the various laws of indices you have learned in
grade 7. Now we are going to learn some more laws and rules of indices in
grade 8.

1. Consider the following

(𝒂𝒃)𝟑 = (𝒂𝒃) × (𝒂𝒃) × (𝒂𝒃)
= 𝒂×𝒃×𝒂×𝒃×𝒂×𝒃
= 𝒂𝟑𝒃𝟑
So we conclude that

(𝒂𝒃)𝟑 = 𝒂𝟑𝒃𝟑
Similarly

𝒂 𝟐 𝒂 𝒂 𝒂×𝒂 𝒂𝟐
( ) = ( )×( ) = =
𝒃 𝒃 𝒃 𝒃 × 𝒃 𝒃𝟐

We conclude that

𝒂𝟐 𝒂𝟐
( ) =
𝒃 𝒃𝟐

Rule 1 :

Rule 2:
𝒂 𝒏 𝒂𝒏

Examples
Simplify the following:
(a) (2𝑚)3
(b) (3𝑝)2
(c) (2𝑚)0
(d) 2(𝑚𝑛)0
Solutions:
(a) (2𝑚)3 = 23𝑚3 = 8𝑚3 ( according to rule 1, the power 3 belongs to both 2 and 𝑚 )
(b) (3𝑝)2 = 32𝑝2 = 9𝑝2 ( according to rule 1, the power 2 belongs to both 3 and 𝑝 )
(c) (2𝑚)0 = 20𝑚0 = 1 × 1 = 1 since (2𝑚)0 = 1
(d) 2(𝑚𝑛)0 = 2 × 𝑚0 × 𝑛0 = 2 × 1 × 1 = 2 since 2(𝑚𝑛)0 = 2
2 2 22 4
(e) ( ) = 2 = 2 ( according to rule 2, the power 2 belongs to both 2 and 𝑝 )
𝑝 𝑝 𝑝
2𝑚 2 22𝑚2 4𝑚2

(f) ( ) = = ( according to rule 2, the power 2 belongs to 2 , 𝑚 and 𝑛 )

𝑛 𝑛2 𝑛2

2. Negative indices

𝟏
𝒂−𝒙 =
𝒂𝒙

Examples
Evaluate the following:
(a) 5−2
(b) 2−3
(c) 4𝑦−3
(d) 3𝑚−2
(e) 𝑎−3 × 𝑎−2
(f) (23)−2

Solutions
1 1
(a) 5−2 = =
52 25
1 1
(b) 2−3 = =
23 8
1 4
(c) 4𝑦−3 = 4( ) = 𝑦3
𝑦3
1 3
(d) 3𝑚−2 = 3 ( )=
𝑚2 𝑚2
1
(e) 𝑎−3 × 𝑎−2 = = 𝑎−5 = 𝑎−3+−2 𝑎5
34 1 1
(d) = 34−7 = 3−3 = =
37 33 27
1 1
(f) (23)−2 = 2−6 = =
26 64
3. Power law involving fractional indices.
𝟏 𝟏

Examples
Evaluate the following:
1
(a) (23)3
1
(b) (54)2
1
(c) 273
2
(d) 1253
2
(e) 80003
Solutions
1 1
(a) (23)3 = 23×3 = 21 = 2
1 1
(b) (54)2 = 54×2 = 52 = 25
1 1
(c) 27 =3 33×3 = 31 = 3
2 2
(d) 1253 = 53×3 = 52 = 25
2 2 2 2
(e) 80003 = (26 × 53)3 = (26)3 × (53)3 = 23×2 × 52 = 16 × 25 = 400
 Exercises for practice
1. Simplify the following.

(a) 𝑦4 × 𝑦2 (b) 5𝑦3 × 2𝑦2

(c) 𝑥𝑦3 × 𝑥2𝑦 (d) 5𝑎3𝑏5 × 2𝑎𝑏2
(e) 𝑦3 × 𝑥 × 𝑦2 × 𝑥4 (f) 3𝑚3 × 2𝑛 × 4𝑚2𝑛2 × 𝑚4

2. Simplify the following.

(a) 𝑦4 ÷ 𝑦2 (b) 𝑝7 ÷ 𝑝2
𝑚4 5 6
(c) (d) 𝑎 𝑏
𝑚2 𝑎𝑏3
𝑥8× 𝑥 5𝑚3×4𝑚3
(e) (f)
𝑥2× 𝑥4 2𝑚2× 𝑚4

3. Simplify the following, giving your answer as positive index.

(a) 𝑚−2 (b) 2𝑝−3

(c) 𝑦−3 × 𝑦2 (d) 3𝑚−2 × 𝑚−3
(e) 𝑎−3 ÷ 𝑎−2 (f) (2𝑎3)−2

4. Simplify the following.

(a) (3𝑚)4 (b) (𝑝𝑞4)2
(c) (𝑎2𝑏5)3 (d) (3𝑚3)4 × (2𝑚2)2
(e) 5(𝑝5𝑞)3 ÷ (𝑝3𝑞)3 (f) (𝑝3𝑞)7 × (𝑝5)3 × 𝑞5
𝑚 5 4 5
(g) ( ) (h) ( 𝑝2 )
𝑛 𝑞
𝑎 5
(i) 15(𝑎𝑏3)7 × 3( ) (j) 12(𝑚𝑛)7 × 3𝑚𝑛5 ÷ (3𝑚)4
𝑏

5. Evaluate the following

1 1
(a) (43)3 (b) (34)2
1 2
(c) (64𝑎3)3 (d) (125𝑝3)3
1 25 −21
16 (f) ( )
2
(e) ( )
81 16
Solutions
1. (a) 𝑦6 (b) 10𝑦5 (c) 𝑥 3 𝑦 4 (d) 10𝑎4𝑏7 (e) 𝑥 5 𝑦 5 (f) 24 𝑚9𝑛3

2. (a) 𝑦2 (b) 𝑝5 (c) 𝑚2 (d) 𝑎4𝑏3 (e) 𝑥3 (f) 10

1 2 1 3 1 1
3. (a) (b) (c) (d) (e) (f)
𝑚2 𝑝3 𝑦 𝑚5 𝑎 4𝑎6
4. (a) 81𝑚4 (b) 𝑝2𝑞8 (c) 𝑎6𝑏15 (d) 324𝑚16 (e) 5𝑝6 (f) 𝑝36𝑞12
𝑚5 𝑝20 4
(g) (h) (i) 45𝑎12𝑏16
𝑛5 𝑞10 9
4 4
5. (a) 4 (b) 9 (c) 4𝑎 (d) 25𝑝2 (e) (f)
9 5

Links for reference

1. http://mathematics.laerd.com/maths/indices-intro.php
2. https://revisionmaths.com/advanced-level-maths-revision/pure-
maths/algebra/indices
3. https://www.youtube.co
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