CHAPTER 5

EQUALITIES AND INEQUALITIES IN ONE VARIABLE

Lesson 5.1 Properties of Real Numbers

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Name the property that is being used or illustrated.

Basic Properties of Real Numbers

Recall several basic properties in addition and multiplication of real
numbers.

A. Closure Property
Each pair of real number has a unique (one and only one) sum which is
also a real number.
8 + 5 = 13 6 + 4 = 10

Each pair of real numbers has a unique product which is also a real
number.
8(5) = 40 6(4) = 24

B. Commutative Property
Adding two real numbers will give the same sum no matter in what order
the numbers are added.
8+5=5+8 12.4 + 0.8 = 0.8 + 12.4

Multiplying two real numbers will give the same product no matter in what
order the numbers are multiplied.
1 3 3 1
x = x
8(5) = 5(8) 4 5 5 4

C. Associative Property
Adding three or more real numbers will give the same sum no matter how
the numbers are grouped.
(6 + 4) + 8 = 6 + (4 + 8)

Multiplying three or more real numbers will give the same product no matter
how the numbers are grouped.
1 1
(8 x 2 ) x 20 = 8 x ( 2 x 20)

D. Identity Property
The sum of any given number and 0 is identical to the given number. The
number 0 is called the additive identity.
0+7=7 21 + 0 = 21

The product of any given number and 1 is identical to the given number.
The number 1 is called the multiplicative identity.
1x7=7 21 x 1 = 21

E. Inverse Property
The sum of a real number and its opposite is 0. The number opposite the
given real number is called the additive inverse.
2 −2
3
+ ( )
3
=0
-5 + 5 = 0

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 The product of a real number and its reciprocal is 1. The reciprocal of the
given number is called the multiplicative inverse.
2 3 −1
⋅ =1 ¿ =1
3 2 -5 5

F. Multiplicative Property of Zero

The product of any given number and zero is zero.
0 x 13 = 0 27 x 0 = 0

G. Distributive Axioms
Multiplication is distributive with respect to addition.
7(8 + 9) = 7(8) + 7(9)

All the axioms mentioned are very useful in simplifying expressions.

Self-Check 5.1
A. Name the property illustrated.
1. 8 + 9 = 17
2. 7 + 14 = 14 + 7
3. 0.4 x 0.6 = 0.24
4. 0.5(3 + 4) = 1.5 + 2.0
3 1 1 3
+ = +
5. 4 2 2 4

B. Fill in each blank and name the property used.

6. (15 + 5) + 8 = ___
7. 23 + 9 = 9 + ___
8. 37 + 53 = ___
9. 18 x ___ = 18

( + )
1 1
10. 8 2 4 = ___ + 2

C. Write each expression in a different form using the distributive property.

11. 3(4 + 5)
12. (6 + 5)7
13. 5(a+b)
14. 8m – 15m
15. 1.6(3 + 4)

Lesson 5.2 Properties of Equality

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Name the property that is being used or illustrated.

A. Reflexive Property
Think about the statements:
3 = 3, 7 = 7, 10.5 = 10.5
Obviously, any number is equal to itself.
Thus, for any real number a, a = a

B. Symmetric property
Consider the statements:
If 3 + 5 = 8, then 8 = 3 + 5
If 15 = 6 + 9, then 6 + 9 = 15
If 20 = 4(5), then 4(5) = 20
If the left and the right members of an equation are interchanged, their
values remain the same.
Thus, for any real number a and b, if a = b, then b = a.

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 C. Transitive Property
Look at these statements:
If 8 + 5 = 13 and 13 = 6 + 7, then 8 + 5 = 6 + 7
If 8(5) = 40 and 40 = 4(10), then 8(5) = 4(10)
If the left and the right members of an equation are equal to the same
quantity, then the two quantities are equal.
Thus, for any real numbers a, b and c, if a = b, and b = c, then a = c.

D. Addition and Subtraction Properties

Consider again these statements:
3=3 3+5=8
1. If 3 + 5 = 8, then (3 + 5) + 3 = 8 + 3, and 11 = 11.
If the same number is added to the left and right members of an
equation, the sums are equal.

2. If 3 + 5 = 8, then (3 + 5) – 3 = 8 – 3, and 5 = 5.
If the same number is subtracted from the left and right members of
an equation, the differences are equal.
Thus, for any real numbers a, b and c, if a =b, then a + c = b + c and
a–c=b–c

E. Multiplication and Division Properties

Look at these statements:
3=3 4(6) = 24
1. If 4(6) = 24, then 4(6)(3) = 24(3), and 72 =72
If the same number is multiplied to the left and the right members of
an equation, the products are equal.

4(6 ) 24
=
2. If 4(6) = 24, then 3 3 , and 8 = 8
If the same nonzero number is divided into the left and the right
members of an equation, the quotients are equal.
Thus, for any real numbers a, b and c, if a = b, then ac = bc and
a b
=
c c where c ¿ 0

Self-Check 5.2
A. Tell what property is illustrated in each of the statements below.
1. 13 = 13
2. If 5 = 2 + 3, then 2 + 3 = 5
3. If 8 + 9 = 9 + 8, then 9 + 8 = 8 + 9
4. 16 – 5 = 16 – 5
5. If 6 + 2 = 8 and 8 = 5 + 3, then 6 + 2 = 5 + 3
6. 18(0) = 0

B. Give the correct expressions to make a true statement.

7. 12 = ___, Reflexive Property
8. If a = 14, then ___ = 10, Addition Prop.
9. (13 – 5)6 = 8(__), Subtraction Prop.
10. 0 + __ = 38, Additive Identity

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 Problem Set #1
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________
Instructions: 1. Use one-half (crosswise) sheet of paper.
2. Write the answers only.

A. Name the property illustrated.

1. (8 + 10) + 6 = 8 + (10 + 6)
2. 30.56 + 0 = 30.56
3. 2.5 x 3.5 = 3.5 x 2.5
4. 12(0.5 + 0.8) = 6.0 + 9.6
3 2 4
( )( )
3 2 4
+ + = + +
5. 5 5 5 5 5 5

B. Fill in each blank and name the property used.

6. (12 + 8) + 25 = ___ + (12 + 8)
7. 15 x (3 x 6) = (___ x 3) x 6
8. 5(0.4 + 0.25) = 2.0 + ___
9. (12 x 0.8)(___) = 0
10. (___) + 62 = 18 + (8 + 62)

C. Write each expression in a different form using the distributive property.

11. 4(3x + 5y)
12. 6r – nr
13. (a – k)11
14. 12(a + b – c)
15. 45x + 9y

D. Tell what property is illustrated in each of the statements below.

16. If 5 = 9 – 4, then 9 – 4 = 5
17. If 3(5) = 15, then 3(5)(4) = 15(4)
18. If 5(6) = 30 and 30 = 3(10), then 5(6) = 3(10)
19. 9 + 7 = (3 + 6) + 7
5 (6) 30
=
20. If 5(6) = 30, then 5 5
21. (25 + 8) + 0 = (25 + 8)

E. Give the correct expressions to make a true statement.

6 (5) ( __ )
=
22.If 6(5) = 30, then 3 3 Division Prop.
23. If 8(7) = 56 and 56 = 28(2), then ___ Transitive property
24. 4(5)(6) = (__)6, Multiplication Prop.
25. 8(15)(_) =0, Multiplication Prop. Of Zero

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Lesson 5.3 Solving First Degree Equations in One Variable
Intended Learning Outcome:
At the end of this lesson, you are expected to:
1. Find the solution set of simple equations in one variable.

By using the properties of equality, first degree equations in one variable

can be solved.

A. Addition Property of Equality

If the same number is added to the left and right members of an equation,
the equation remains unchanged.

Examples:
A. x – 5 = 8 B. x – 12 = -18 C. x – 3.1 = 5.7
x–5+5=8+5 x – 12 + 12 = -18 + 12 x – 3.1 + 3.1 = 5.7 + 3.1
x + 0 = 13 x + 0 = -6 x + 0 = 8.8
x = 13 x = -6 x = 8.8

B. Subtraction Property of Equality

If the same number is subtracted from the left and right members of an
equation, the equation remains unchanged.

Examples:
A. x+4=6 B. x + 12 = 25 C. x + 1.9 = 2.6
x+4–4=6–4 x + 12 – 12 = 25 – 12 x + 1.9 – 1.9 = 2.6 – 1.9
x+0=2 x + 0 = 13 x + 0 = 0.7
x=2 x = 13 x = 0.7

C. Multiplication Property of Equality

If both members of an equation are multiplied by the same number, the
equation remains unchanged.

Examples:
x x x
=3 =−5 =8
A. 2 B. 7 C. 2.6
x x x
2⋅ =3⋅2 7⋅ =−5⋅7 (2. 6 ) =8(2. 6)
2 7 2 .6
x=6 x = -35 x = 20.8

D. Division Property of Equality

If the left and right members of an equation are divided by the same
number, the equation remains unchanged.

Examples:
A. 5x = 35 B. 12y = -72 C. 3.6x = 180
5 x 35 12 y −72 3 . 6 x 180
= = =
5 5 12 12 3. 6 3 . 6
x=7 x = -6 x = 50

Solving Other Types of First Degree Equations

Other types of equations in one variable are solved using the properties of
real numbers and the properties of equality.

Examples:
A. 2x + 3 = 9
2x + 3 – 3 = 9 – 3 Subtraction property of equality
2x = 6

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 2x 6
=
2 2 Division property of equality
x=3

B. 5y – 4 = 12 – y
5y – 4 + 4 = 12 + 4 – y Addition property of equality
5y + y = 16 – y + y Addition property of equality
6y = 16
6 y 16
=
6 6
2
2
y= 3

C. 2(x – 4) = 10 + 5x
2x – 8 = 10 + 5x Distributive property
2x – 8 + 8 = 10 + 8 + 5x Addition property of equality
2x = 18 + 5x
2x – 5x = 18 + 5x – 5x Subtraction property of equality
-3x = 18
−3 x 18
=
−3 −3 Division property of equality
x = -6

Self-Check 5.3
Solve for x.
1. x – 18 = 32 9. 4 + 3y = 16
2. x – 28 = 25 10. 5(x + 2) = 25
3. x + 20 = 11 11. 4 + 3x = 4(2x + 6)
4. x + 43 = -16 12. 6x – 5 = 2x + 3
x
=5
5. −8 13. 3(3x – 2) = 4x + 9
x 3
=
6. 3 4 14. 2(x – 3) = 4(3 + x)

7. -7x = 105 15. a – (3a + 5) = 7 – 5a

8. 3.2x = -96

Lesson 5.4 Solving First Degree Inequalities in One Variable

Intended Learning Outcome:
At the end of this lesson, you are expected to:
1. Find the solution set of simple inequalities in one variable.

As in equation, inequalities are solved using the properties of addition,

subtraction, multiplication and division. This time, these are called properties of
inequalities.

Examples:
A. For addition B. For subtraction
x–2>6 x + 15 > -7
x–2+2>6+2 x + 15 – 15 > -7 – 15
x+0>8 x + 0 > -22
x>8 x > -22

C. For multiplication D. For division

x
≥7
6 2x ¿ 8
x 2x 8
6⋅ ≥7⋅6 ≤
6 2 2
x ¿ 42 x ¿ 4

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 Take note that when you divide/multiply an inequality by a negative number, the
sign of inequality (> or <) is reversed.

Example.

-3x > 6
−3 x 6
>
−3 −3
x < -2

Self-Check 5.4
Find the solution set using the properties of inequalities.
1. x + 7 < 1 6. 0.04x > -2.8
x
≤2. 8
2. x – 11 > 14 7. −3
2 1 x 1
x+ > <
3. 5 2 8. −6 2
x
>3 . 2
4. x – 25 > 28 9. 1 . 4
5. x + 3.2 ¿ 6.4 10. 4x ¿ 3.6

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 Problem Set #2
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________
Instructions: 1. Use one whole sheet of paper.
2. Show your solution.

A. Solve the equation by finding the value of the variable.

1. 8x – 7 + 2x = 6 + 5x + 2 6. 9 – 5y = 3(y – 5)
2. 5(3x – 2) – 6x = -1 7. 11x – 9 = 2x + 9
3. x + 4 = 2(2x – 5) 8. 4 + 3x = x – 6
4. 10(x – 1) = 2(2x – 5) 9. 7x – 1 = 9 – 5x
5. 3x + 7x = 16 + 4x 10. 6y = 12 – 2y + 4

B. Solve each inequality

11. 3 x−2<4 16. 10−2 x >8
12. 4 x +9> x−5 17. 7−5 x <2 x+1
13. 6 x+4<3 x +13 18. 4−2( x +3 )<10
14. 7 x−3( x−4 )>9 19. 2(3 x−2)<7
15. 3( x−5 )<2 x +3 20. 9−2 x <3 x−6

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Lesson 5.5 Word Problems
Intended Learning Outcome:
At the end of this lesson, you are expected to:
1. Solve problems involving first degree equations.

To solve a word problem, one must know how to write equations. There are
steps to follow when writing the equation and finally finding its solution.
1. Read and explore the problem. Choose the variable to represent the
unknown number in the problem. This is called defining the variable.
2. Use this variable in writing expressions for other unknown numbers in
the problem.
3. Read the problem again and decide how the unknown number relates
to other information in the problem.
4. Write an equation to represent the relationship.
5. Solve for the unknown.
6. Check by going back to the original statement.

Examples:
A. Number Relation
One number is 3 less than another number. If their sum is 49, find the two
numbers.
Step 1: Let x be the first number.
Step 2: x – 3 is the second number
Step 3: The sum of the two numbers is 49
Step 4: x + (x – 3) = 49
Step 5: x + x – 3 = 49
2x – 3 = 49
2x = 52
x = 26 – first number
(x - 3) = (26 – 3) = 23 – second number
Step 6: Check. x + (x – 3) = 49
26 + 23 = 49
49 = 49

B. Consecutive numbers
The sum of three consecutive even integers is 96. Find the largest of these
numbers.
Step 1: Let x be the first even integer.
Step 2: x + 2 is the second even integer.
x + 4 is the third and largest even integer.
Step 3: The sum of the three consecutive integers is 96.
Step 4: (x) + (x +2) + (x + 4) = 96
Step 5: x + x + 2 + x + 4 = 96
3x + 6 = 96
3x = 90
x = 30 – first even integer
(x + 2) = (30 + 2) = 32 – second even integer
(x + 4) = (30 + 4) = 34 – third even integer
Step 6: Check. (x) + (x +2) + (x + 4) = 96
30 + 32 + 34 = 96
96 = 96

C. Coin Problem
Mark has some 1-peso coins and 4 more 5-peso coins than 1-peso coins,
making a total of 22 coins. How much money does he have?
Step 1: Let x be the number of 1-peso coins.
x + 4 is the number of 5-peso coins.
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 Step 2: 1.00x is the amount of 1-peso coins.
5.00(x + 4) is the amount of 5-peso coins.
Step 3: The total number of 1-peso coins and 5-peso coins is 22.
Step 4: x + (x + 4) = 22
Step 5: x + x + 4 = 22
2x + 4 = 22
2x = 18
x = 9 – number of 1-peso coins
(x + 4) = (9 + 4) = 13 – number of 5-peso coins
Step 6: Check. x + (x + 4) = 22
9 + 13 = 22
22 = 22
How much money does he have?
1.00x is the amount of 1-peso coins.
1.00(9) = 9.00
5.00(x + 4) is the amount of 5-peso coins.
5.00(13) = 65.00
9.00 + 65.00 = 74.00 – total amount of money.

D. Age Problem
Six years ago, Mrs. Dela Rosa was 5 times as old as her daughter, Leila.
How old is Leila now if her age is one-third of her mother’s present age?
Step 1: Let x be Leila’s age now.
3x is Mrs. Dela Rosa’s age now.
Step 2: x – 6 is the age of Leila six years ago.
3x – 6 is Mrs. Dela Rosa’s age six years ago.
Step 3: Leila’s age is one-third of her mother’s present age.
Step 4: 5(x – 6) = 3x – 6
Step 5: 5(x – 6) = 3x – 6
5x – 30 = 3x – 6
5x – 3x = 30 – 6
2x = 24
x = 12 - Leila’s age now
3x = 3(12) = 36 - Mrs. Dela Rosa’s age now
Step 6: Check.

Self- Check 5.5

Represent each problem by an equation and find its solution.

1. The sum of two numbers is 120. If the greater number is four times the less,
what are the numbers?

2. Find a number such that 13 less than twice the number is 137.

3. Find three odd consecutive integers whose sum is 33.

4. Find four consecutive even integers such that the 6 times the first is equal to
the sum of the other three.

5. Mario has P8.00 in 25-centavo coins and 10-centavo coins. He has 10 more
10-centavo coins than 25-centavo coins. Find the number of each coin.

6. Ken has 11 more 1-peso coins than 5-peso coins. How many coins does he
have if the total value of his coins is PhP263?

7. Anita is 2 years younger than her sister Fe. Eight times Anita’s age increased
by 2 years is six times Fe’s age diminished by 4 years. Find their ages.

8. Ariel is now 4 times as old as Ben. Five years ago, the sum of their ages was
45 years. How old is each?

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 Problem Set #3
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________
Instructions: 1. Use one whole sheet of paper.
2. Show your solution.

Solve the following problems.

1. One number is four more than a second number. If their sum is 38,
find the two numbers?

2. The sum of two consecutive integer is 99. Find the numbers.

3. If the sum of three consecutive odd integer is 219, find the numbers.

4. Arnel has 10 more 5-peso coins than 10-peso coins. How many coins does he
have if the total value of his coins is Php785?

5. Nazrat is 4 years older than Umaima. Two years ago, Nazrat was 5 times as
old as Umaima. How old is Nazrat today?

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 CHAPTER 6
SPECIAL PRODUCTS AND FACTORING

SPECIAL PRODUCTS

There are certain special products that occur frequently in Algebra. They are
here classified, stated and illustrated. The letters in the formula/pattern may
stand for any algebraic expression.

Lesson 6.1 Review in Multiplication by Monomials

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Perform multiplication by monomials.

Examples: Find the products of:

5 3
1. 2x y ( 3 x y ) = 6 x y
3 2 2

4 3 2 2
2. 3 x ( 4 x y−3 xy −7 y ) = 12 x y−9 x y−21x y
2 2 2

3. 3 a b ( 2 a b−5 ab+3 ab ) = 6 a b −15 a b +9 a b

2 2 3 2 5 3 3 3 3 4

4. 4 a b ( 3 a b−5 ab ) = 12 a b −20 a b
2 2 2 4 2 3 3

Self-Check 6.1
Find the product.
1. 7 ( 2 x+3 ) 6. 6 a b ( 2 ab−3 a b )
2 2 2

2. 4 (3 a−5 ) 7. x ( 3 x−2 y )
3. 3 x ( 7+ y ) 8. 3ab ( 5a−4 b )
4. 9 ( 2 x y −3 xy ) 9. 4 x ( 7 x −3 y )
2 2 2

5. 3 x ( 4 y−x ) 10. 5 xy ( 8 xy −7 x y )
2 2

Lesson 6.2 Product of the Sum and the Difference of the same
two terms
Intended Learning Outcome:
At the end of this lesson, you are expected to:
1. Use Special Products in finding the product of the sum and difference of the
same two terms.

( x+ y )( x− y ) =x2 − y 2
The product of the sum and difference of two terms is the square of the first
term minus the square of the second term.

The product of the sum and the difference of the same two terms is a special
case in the sense that its middle term is zero and leads to a very simple product
called the difference of two squares.

Examples:
1. ( 4 x +3 y ) ( 4 x−3 y ) =( 4 x )2 −( 3 y )2
2 2
= 16 x −9 y

2 2
2. ( 2 m4−5 n3 )( 2m4 +5 n3 )=( 2m4 ) −( 5 n3 )
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 8 6
=4 m −25 n

( 2 x +3 y ) ( 2 x−3 y ) ( 4 x2 +9 y 2 ) =[ ( 2 x )2 −( 3 y )2 ] ( 4 x 2 +9 y 2 )
3.
=( 4 x 2 −9 y 2 )( 4 x2 +9 y 2 )
2 2
=( 4 x ) −( 9 y )
2 2
4 4
=16 x −81 y

Self-Check 6.2
Find the following products.
1. ( x+4 )( x−4 ) 6. ( 3 a−b ) ( 3 a+b )
2. ( a−5 ) ( a+5 ) 7. ( 5 a2 +2b )( 5 a2 −2 b )
3. ( 2 a+7 ) ( 2 a−7 ) 8. ( 4 x+5 y ) ( 4 x−5 y )
4. ( c +d ) ( c−d ) 9. ( 2 a2 b 2 +c 3 )( 2a 2 b2−c 3 )
( 2x− y )( 2 x+ y ) 10. ( 4 x y −3 z )( 4 x y +3 z )
2 2
5.

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 Problem Set #4
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________
Instructions: 1. Use one-half (crosswise) sheet of paper.
2. Write the answers only.

A. Find the product.

1. 5 r ( 6 r −2rt +t ) 6. 2 a b ( a −3 ab+5 b )
2 2 4 2 2 2

2. 3 cd ( 2 a c−3 cb +4 bd ) 7. 5 x y ( 6 x y +2 x y −3 xy )
2 2 2 3 2 2 3

3. 7 ab ( 8 a −4 ab +5 b ) 8. 10 a b ( 3 a b−4 a b +5 ab )
2 2 2 2 3 2 2 3

4. 4 x y ( 3 x y−2 xy −5 y ) 9. 8 x y ( 4 x y +3 xy −2 y )
2 2 2 3 2 2 2 3

5. 9 a b ( a b +2 ab −3 a b ) 10. 2 x y ( 5 x y +7 x y −3 xy )
3 2 2 2 3 3 3 3 2 2 3

B. Find the following products.

11. ( 7+ y ) ( 7− y ) 16. ( 9 x 3 y 2 −4 z 2 )( 9 x 3 y 2 +4 z 2 )
12. ( 6 a 3 +5 b2 )( 6 a3 −5 b 2 ) 17. ( 2+a ) ( 2−a ) ( 4+ a2 )
13. ( a 2−b ) ( a2 + b ) ( a 4 + b2 ) 18. ( 3 a−2b )( 3 a+2b ) ( 9 a2 + 4 b 2 )
14. ( 3 xyz+4 a )( 3 xyz−4 a ) 19. a  3a  3a 2  9
15. ( 8 c d +9 ab )( 8 c d−9 ab )
2 2
20. ( 2 x− y )( 2 x+ y ) ( 4 x 2 + y 2 )

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Lesson 6.3 Square of a Binomial

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Define square of a binomial;
2. Apply special products in finding the product of a square of a binomial.

( x+ y )2=x 2 +2 xy + y 2
( x− y )2 =x 2−2 xy + y 2
The square of the sum (or the difference) of two terms is the square of the first
term, plus (or minus) twice the product of the two terms plus the square of the
second term.
The product of the square of a binomial is known as a perfect square
trinomial.

Examples:
2 2 2
1. ( 5 x 4 +7 y 2 ) =( 5 x 4 ) +2 ( 5 x 4 )( 7 y 2) + ( 7 y 2 )
=25 x 8 +70 x 4 y 2 +49 y 4
2 2
2. ( 3m3−8 n ) =( 3 m3 ) −2 ( 3 m3 ) ( 8 n ) + ( 8 n )2
=9 m 6 −48 m3 n+64 n2

3. ( x+5 )2 =( x )2 +2 ( x )( 5 ) + ( 5 )2
2
=x +10 x +25

A trinomial is a perfect square if:

a. Two terms are positive squares
b. The third term is twice the product of the square roots of the squares.

NOTE: the squares are usually the first and third term of the perfect square, and
twice the product of their square roots is the middle term.

Self-Check 6.3
A. Which of the following are perfect squares:
2 2 2
1. a −2 ab+ b 6. 36−12 x−x
2
2 y
2 2 x 2 + xy +
2. 4 a +2 ab+ b 7. 3 9
2 2 2 2
3. 9 x +12 xy−4 y 8. 4 a +20 ab+25 b
2 2 2 4
4. 16 a + 40 ab+ 25 b 9. 1−6 x +9 x
2 4 2
5. 25−10 b+ b 10. 25 x −35 xy+49 y

B. What term is needed to make each of the following a perfect square.

2 2
11. x +4 x+___ 16. 4 x −12 xy+___
2 2
12. 4 a −4 ab+___ 17. x +___ +9
2 2
13. a +___+16 18. b −8 b+___

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 2 2 2
14. 9 x −___ +4 y 19. ___−2ab+b
2 2
15. ___+20 ab+4 b 20. 9 n −___ +49

C. Find the square of the following binomial.

2 2
21. ( x+2 ) 26. ( 4 x −7 )
2 2
22. ( x−3 ) 27. ( 3 x+2 y )
2 2
23. ( 2x−3 ) 28. ( 4 x +3 y )
2 2
24. ( 2 a+3 ) 29. ( 5 a−2b )
2 2
25. ( 3a−5 ) 30. ( 2 y−5 z )

Lesson 6.4 Cube of a Binomial

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Define cube of a binomial;
2. Find the product of a cube of a binomial using special products.

( x+ y )3=x 3 + 3 x 2 y +3 xy 2 + y 3
( x− y )3 =x 3−3 x 2 y +3 xy 2 − y 3
The cube of the sum (or the difference) of two terms is the cube of the first
term plus (or minus) thrice the square of the first times the second plus thrice the
product of the first and the square of the second plus (or minus) the cube of the
second term.

There are always four terms in the cube of a binomial. All terms in the product
are positive in the cube of the sum of two terms. The cube of the difference of a
binomial will have an alternate – and + signs.

Examples:
1. ( 3 x+2 y )3 =( 3 x )3 +3 ( 3 x )2 ( 2 y ) +3 ( 3 x )( 2 y )2 + ( 2 y )3
=27 x 3 +3 ( 9 x 2 ) (2 y ) +3 ( 3 x ) ( 4 y 2 ) +8 y 3
=27 x 3 +54 x2 y+36 xy 2 +8 y3
3 3 2 2 3
2. ( 2a4 −4 b 5 ) =( 2 a 4 ) −3 ( 2 a4 ) ( 4 b5 ) +3 ( 2 a 4 )( 4 b5 ) −( 4 b5 )
=8 a12−3 ( 4 a8 )( 4 b 5 ) +3 ( 2 a 4 )( 16 b 10 )−64 b15
12 8 5 4 10 15
=8 a −48 a b +96 a b −64 b

Note: The exponent of the first term is in descending order while the exponent of
the second term is in ascending order.

Self-Check 6.4
Expand the following:
3 3
1. ( x−5 ) 4. ( m−3 )
3 3
2. ( x+4 ) 5. ( 5 x+ y )
3
3. ( 2 a−b )

Learning Module in Elementary Algebra Page│16

 Problem Set #5
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________
Instructions: 1. Use one whole sheet of paper.
2. Write the answers only.

A. Which of the following are perfect squares:

2 2 2
1. n + 6 n+9 6. m +12 mn+36 n
1 2 1 1
2 x − xy + y 2
2. 9 x +7 x+ 4 7. 9 10 4
2 2 2
3. y −12 y +36 8. a −6 ab +9 b
2 2 4 2 2 4
4. 36 y +13 xy+x 9. 9 b +21 a b + 4 a
2 2 2
5. 64 x −40 x+ x 10. x +4 x+16

B. What term is needed to make each of the following a perfect square.

2 2
11. 4 a +20 a+___ 16. 4 y +4 y+___
2 2
12. ___ + 24 xy+9 y 17. ___ +12 n+n
2 2 2 2
13. c +___ +d 18. 9 x −___+49 y
2 2
14. 36 y −___+81 19. 49 m −28 m+___
2 2 2
15. x −___ +0. 16 y 20. x +___ +25

C. Find the square of the following binomial.

2
21. ( 4 x−3 y )
2
26. ( 2 xy 2−5 z )
2 2
22. ( 7 a−3 b ) 27. ( 1−c )
2
23. ( 3s−4r ) 2 28. ( 6 x− y )
2 2
24. ( 4 p+3 t ) 29. ( n+0 . 3 )

( )
2
1
2 9 x−
25. ( 3 xy+4 z ) 30. 3
Expand the following:
3
31. ( 3 x+1 )
3
34. ( a 2 +b3 )
3
32. ( 2 s−t )
3
35. ( 3 x 2−5 y 3 )
3
33. ( r+4 s )

Learning Module in Elementary Algebra Page│17

Lesson 6.5 Product of Two Binomials

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Find the product of two Binomials using FOIL method .

( ax +by ) ( cx+dy )=acx 2 + ( ad+ bc ) xy +bdy 2

The product of two binomials is obtained by using the FOIL method (F = first O =
outside I = inside L = last):
1. The first term is the product of the first terms of the binomials.
2. The middle term is obtained by adding the product of the outer terms to the
product of the inner terms.
3. The last term is the product of the last terms of the binomials.

Examples:
1. ( 2 x+5 y ) ( 3 x−4 y ) =( 2 x )( 3 x )+ ( 2 x ) (−4 y ) + ( 3 x ) (5 y ) + ( 5 y ) (−4 y )
2 2
=6 x +(−8 xy )+15 xy−20 y
2 2
=6 x +7 xy −20 y
2. ( 3 a3 −2 b5 )( 6 a3 +3 b5 ) =(3 a3 )(6 a3 )+(3 a3 )(3b5 )+(−2 b5 )(6 a3 )+(−2 b5 )(3 b5 )
6 3 5 3 5 10
=18 a +9 a b +(−12 a b )−6 b
6 3 5 10
=18 a −3 a b −6 b
Self-Check 6.5
Expand each of the following.
1. ( x+2 )( x +1 ) 6. ( 3 x+5 )( 2 x−3 )
2. ( x−2 ) ( x−1 ) 7. ( x+7 )( x−4 )
3. ( 3 x+2 ) ( x−2 ) 8. ( 2x−3 y )( 3 x+5 y )
4. ( x+5 )( x−6 ) 9. ( 3 a+b )( 4 a+3 b )
5. ( x+4 )( x−3 ) 10. ( 4 a−b ) (2 a+5 b )

Lesson 6.6 Square of a Trinomial

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Find the product of a square of a trinomial.

( a+b +c )2=a2 + b2 + c2 +2 ab+2 ac+ 2bc

Learning Module in Elementary Algebra Page│18
 The square of a trinomial is equal to the sum of the squares of the terms and
twice the product of each term by every other term, taken separately.

The product of the square of a trinomial consists of six terms. The signs of the
first three terms are positive while the signs of the last three terms depend on the
signs separating two terms in the given trinomial.

Examples:

2 2 2 2
1. ( 2 x+3 y +5 z ) =(2 x) +(3 y) +(5 z) +2(2 x)(3 y)+2(2 x)(5 z )+2(3 y )(5 z )
=4 x 2 +9 y 2 +25 z 2 +12 xy+20 xz+30 yz
2
( 4 a3−7 b 2−3 c 5 )
2.
3 2 2 2 5 2 3 2 3 5 2 5
=(4 a ) +(−7 b ) +(−3 c ) +2(4 a )(−7 b )+2(4 a )(−3 c )+2(−7 b )(−3 c )
6 4 10 3 2 3 5 2 5
=16 a +49 b +9 c −56 a b −24 a c + 42 b c

Self-Check 6.6
Find the square of the following trinomials.
2 2
1. ( 2 x + y +3 z ) 4. ( 3 x+5 y+2 z )
2 2
2. ( 3 a−2b+d ) 5. ( a+2 b+c )
2
3. ( 4 x +3 y−5 z )

Lesson 6.7 Special Case of a Product of Binomial and Trinomial

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Apply special products in finding the product of a binomial and trinomial.

( x+ y ) ( x 2 −xy + y 2 ) =x 3 + y 3
( x− y ) ( x 2 + xy + y 2 ) =x 3− y 3
When the trinomial factor (based on the binomial factor) is the square of the
first plus (or minus) the product of the first and second plus the square of the
second, the product is the sum (or difference) of two cubes.

This case is so special in the sense that the definite forms of the binomial and
trinomial lead to a very simple product called the sum or difference of two
cubes.

Examples:
1. ( 4 x +2 y ) ( 16 x 2 −8 xy +4 y 2 ) =(4 x)3 +(2 y)3
3 3
=64 x +8 y

2. ( 7 m3 −3 n2 )( 49 m6 +21 m3 n2 + 9 n4 ) =(7 m3 )3 −(3 n 2 )3

9 6
=343 m −27 n

Self-Check 6.7
Find the products of the following.
1. ( 3 a+2 ) ( 9 a −6 a+4 ) 6. ( 3 x−5 ) ( 9 x +15 x+25 )
2 2

2. ( 2 x +3 ) ( 4 x −6 x +9 ) 7. ( 2 x− y ) ( 4 x +2 xy + y )
2 2 2

3. ( 3 x−4 ) ( 9 x +12 x +16 ) 8. ( a+2 b ) ( a −2ab +4 b )

2 2 2

Learning Module in Elementary Algebra Page│19

 4. ( 4 x −3 ) ( 16 x +12 x +9 ) 9. ( 3 a−2b ) ( 9 a +6 ab+4 b )
2 2 2

5. ( 2 a−3 ) ( 4 a +6 a+ 9 ) 10. ( 4 x −3 y ) ( 16 x +12 xy+9 y )

2 2 2

Problem Set #6
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________
Instructions: 1. Use one whole sheet of paper.
2. Show your solution.

Expand each of the following.

1. ( x− y ) ( 2 x+3 y ) 6. ( 5 xy−4 z ) ( 3 xy+5 z )
2. ( 3a−5b )( a+3b ) 7. ( 7 ab−3c )( 3ab+4 c )
3. ( 2 a−3c ) ( 3 a+2 c ) 8. ( 8bc+5 a ) (3bc−4 a )
4. ( 6 a−7 b ) ( 2a−3b ) 9. ( 4 x− y )( 2 x+3 y )
5. ( 2x−3 yz ) ( 3 x+7 yz ) 10. ( 4 a−3 b )( 3a+2b )

Find the square of the following trinomials.

2 2
11. ( 3 a+b−2 c ) 14. ( x+3 y +2 z )
2 2
12. ( 2 a−3 b+4 c ) 15. ( 2 x +5 y−3 z )
2
13. ( 3 x− y +2 c )

Find the products of the following.

16. ( x+4 y ) ( x −4 xy +16 y ) 21. ( 3 a−4 b ) ( 9 a +12 ab+16 b )
2 2 2 2

17. ( a+5 b ) ( a −5 ab+25 b ) 22. ( 4 a−5 b ) ( 16 a +20 ab+ 25 b )

2 2 2 2

18. ( 2 a−5 b ) ( 4 a +10 ab+25 b ) 23. ( 3 a b +7 c )( 9 a b −21 a bc+ 49 c )

2 2 2 4 2 2 2

19. ( 2 x+3 y )( 4 x −6 xy +9 y ) ( x 3 y 2 −3 z 2 )( x6 y 4 +3 x3 y 2 z 2 + 9 z 4 )
2 2 2 4
24.
20. ( 2 x2 y −7 z )( 4 x 4 y 2 +14 x 2 yz +49 z 2 ) 25. ( 2 a+5 bc ) ( 4 a 2−10 abc +25 b2 c 2 )

Learning Module in Elementary Algebra Page│20

 LEARNING MODULE

IN

ELEMENTARY ALGEBRA
(Module 4)

Compiled by:

OMAR U. ENOCK
2021

Learning Module in Elementary Algebra Page│21

 No part of this book may be reproduced or utilized in any form or by any means,
electronic or mechanical, including photocopying, recording, or any information
storage and retrieval system, without permission in writing from the

UNIVERSITY OF SOUTHERN MINDANAO

Kabacan, Cotabato

FACTORS AND FACTORING POLYNOMIALS

In an indicated product, the numbers that are multiplied are called factors.
Thus, since (4)(5) = 20, we say that 4 and 5 are factors of 20. Similarly, since
( x + y )( x− y )=x 2− y 2, we say that ( x + y ) and (x− y ) are both factors of (x ¿ ¿ 2− y 2 )¿ .
In dividing polynomials, we would often think to know the quotient without
doing the actual division process. This may be done if we know the factors of the
numbers. Thus, we say that 8 ¿ 4 is 2 since we know that 2 times 4 is 8. In the
same manner, we say that (x ¿ ¿ 2− y 2 )¿ ¿ (x + y ) is (x− y ) since we know that
( x + y )( x− y )=x 2− y 2.
In this section, we will be dealing with the problem of finding the factors of a
given polynomial. To factor out a polynomial means to express it as a product of
two or more irreducible polynomials with integral coefficients. Irreducible
polynomials are polynomials that cannot be expressed anymore as a product of
two or more polynomials. The process of determining the factors of a polynomial is
called factoring.

Lesson 6.8 Common Monomial Factors

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Define Factoring.
2. Factor completely a polynomial using a common monomial factor.

Definition. The common factor of a given polynomial is the greatest common

number factor and/or common letter factor (with lowest exponent
common to all terms).

To factor a polynomial whose terms, have a common monomial factor:

1. Find the greatest monomial factor of the terms of the polynomial.
2. Divide each term of the polynomial by the greatest monomial factor to get the
other factor.
3. Express the polynomial as the indicated product of the two factors.

Example: Factor each polynomial.

a. 4 x+ 6 y c. 10 a3 b2 c +15 a2 b2 c2 −20 a4 b 3 c
b. 8 x 2 y −12 x 3 y 2 d. 15 x 3+3 x 2 y 2−21 x y 3

Learning Module in Elementary Algebra Page│22

Solutions:
a. 4 x+ 6 y
(1) The greatest common factor is 2.
(2) The other factor is (4 x+ 6 y)÷ 2=2 x+3 y
(3) 4 x+ 6 y=2(2 x+3 y )

b. 8 x 2 y −12 x 3 y 2
(1) The greatest common factor is 4 x2 y .
(2) The other factor is:( 8 x 2 y −12 x 3 y 2) ÷ 4 x 2 y = 2−3 xy
(3) 8 x 2 y −12 x 3 y 2=¿ 4 x2 y (2−3 xy ¿

As a shortcut, we find the greatest monomial factor and write it immediately

to the right of the equal sign after the polynomial. Then, we get the quotient
mentally and do Step 3.

c. 10 a3 b2 c +15 a2 b2 c2 −20 a4 b 3 c=5 a2 b 2 c (2 a+ 3 c−4 a2 b)

d. 15 x 3+3 x 2 y 2−21 x y 3=3 xy (5 x 2+ xy −7 y 2)

Self-check 6.8
Factor each polynomial.
1. x 3−x 2+ x 6. 4 x2 −4 x
2. 15 a+12 b+6 c 7. 15 x 2−50 x −10
3. 8 x 2−18 x 8. 121 a−11 b
4. x 2 y +2 y 9. 64 c3 −56 c 2+ 88 c
5. z 3 +4 z 2 10. 10−10 n+10 n2

Lesson 6.9 Factoring the Difference of Two Squares

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Factor a difference of two perfect squares.

Factoring the difference of two squares like a 2−b2 may be the easiest form
of factoring. It is just the reverse of multiplying the sum of two terms by their
difference. We learned that ( a+ b ) ( a−b )=a2−b 2. So, given the right side, we can
easily get the left side.

Example: Factor completely.

a. 4 x2 −9 y 2 c. 25 a 4 b2−36 c 4
b. 81 x 2−100 y 6 d. 2 x5 −2 x y 4

Solutions:
a. 4 x2 −9 y 2 ¿ ( 2 x )2 −¿
¿( 2 x +3 y)(2 x−3 y )

b. 81 x 2−100 y 6 ¿ ( 9 x )2−¿
¿( 9 x +10 y ¿¿ 3)(9 x−10 y 3) ¿

c. 25 a 4 b2−36 c 4 ¿ ¿ ¿
¿(5 a ¿ ¿ 2b +6 c )(5 a b−6 c )¿
2 2 2

d. 2 x5 −2 x y 4 ¿ 2 x ¿ ¿
¿ 2 x(x ¿ ¿ 2+ y )( x − y ) ¿
2 2 2

¿ 2 x( x ¿ ¿ 2+ y )( x− y)(x + y ) ¿
2

Learning Module in Elementary Algebra Page│23

 Observe that in d, a common monomial factor is first removed. Then, after
the factoring as a difference of two squares, one factor is still the difference of two
squares. Hence, there is a need to continue the factoring.

Self-check 6.9
Factor completely.
2 9
1. a 2 b 2−16 6. 64 a − 2
25 b
2. 144 a 2−169 b2 7. 9 a 4 b4 −25 p 4 q 4
3. 1−0.09 a2 8. 50 p2−72 q2
2 1
4. 16 x 2−121 9. m −
169
5. x 4 −256 10. 9 x y 2 −x3

PROBLEM SET #7
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________

Instructions: 1. Use one-half (crosswise) sheet of paper.

2. Write the answers only.

A. Factor each polynomial.

1. 18 k +36 k 2+ 9 k 3 6. 6 q +10 q2 +8 q 3
2. 33 q3 +33 q 2 r + 33 q r 2 7. c 3−5 c
3. a 3 b 2+ a3 b4 + a b4 8. k 4 +2 k 3 +3 k 2 + 4 k
4. 8 kxy + 4 xy +2 k 2 xy 9. 14 dj +14 ej
5. 7 m+7 n+7 p 10. m3 n2−m2 n3

B. Factor completely.
11. 1−121 a2 16. 75 x 3 y 2−108 x y 4
16 2
12. −25 p 17. 81 x 2−49 y 2
49
2 2 25
13. x y − 2 18. x 2 y 2−16
z
14. x y −x 4 y 12
12 4
19. 4 x2 −25
15. 2 x−50 x3 20. a 4−b4

Learning Module in Elementary Algebra Page│24

Lesson 6.10 Factoring the Sum/Difference of Two Cubes

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Factor completely a Sum or Difference of Two Cubes.

In 6.7 we worked with products of the form (x + y )( x ¿ ¿ 2−xy+ y 2)=x 3 + y 3 ¿

and ( x− y )(x ¿ ¿ 2+ xy+ y 2)=x 3− y3 . ¿
Here, we do the opposite procedure, that is, begin with the sum or
difference of two cubes and look for the factors that resulted in the given sum or
difference.

Observe the two factors of the sum or difference of two cubes.

1. One factor is a binomial, the other, a trinomial.
2. The first and third terms of the trinomial are positive squares of the terms of the
binomial.
3. The middle term of the trinomial is the product of the terms of the binomial with
the sign opposite to that of the second term of the binomial.

Example:
Factor:
a. a 3−8 b 3 c. x 3+ 27
b. 27 x 3+ y 3 d. 64 x 4 y−xy 4

Solutions:

a. a 3−8 b 3 ¿ a3 −¿]
¿( a−2 b)(a ¿ ¿ 2+ 2 ab+4 b2) ¿

b. 27 x 3+ y 3 ¿ ( 3 x )3+ y 3=(3 x + y )¿]

¿(3 x + y )(9 x ¿ ¿2−3 xy+ y )¿
2

c. x 3+ 27 ¿( x)3 +¿ [ x 2−x ( 3 )+(3)2 ¿

¿( x +3)( x ¿¿ 2−3 x+ 9)¿

Learning Module in Elementary Algebra Page│25

 d. 64 x 4 y−xy 4=xy ( 64 x 3− y 3 )
¿ xy ¿]
¿ xy ( 4 x− y )[( 4 x)2 +4 x ( y ) +( y )2 ¿
¿ xy ( 4 x− y )(16 x ¿ ¿ 2+ 4 xy + y 2) ¿

Self-check 6.10
Factor each completely.
1. x 3+ 8 6. 27 m3−125
2. a 3+ 64 7. x 3−64
3. a 3+216 8. 432+250 m3
4. 27+ 8 x 3 9. 81 x3 +192
5. a 3−216 10. 500 x 3+256

Lesson 6.11 Factoring a Perfect Square Trinomial

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Factor completely a Perfect Square Trinomial.

Perfect Square Trinomial

2 2
x + 2 xy + y =¿
2 2
x −2 xy+ y =¿

In the formula, we can test if a given trinomial is a perfect square. As can be

seen in the formula, the first term and the third term must be both perfect squares
whose product multiplied by 2 will yield the middle term. Thus, to factor out a
trinomial into a perfect square, we can do the following steps:
1. Write down the square root of the first term.
2. Copy the sign of the second term.
3. Write down the square root of the third term.
4. Square the resulting binomial.

Example:
Factor:
a. x 2+ 4 x +4 c. 25 x 2+ 60 xy +36 y 2
b. 49 x 2−56 x +16 d. 8 x 2−40 xy +50 y 2
Solutions:

a. x 2+ 4 x +4
Write down the square root of the first x
term: x 2
Copy the sign of the second term +
Write down the square root of the third 2
term
Square the resulting binomial (x +2)2

b. 49 x 2−56 x +16
Write down the square root of the first 7x
term: x 2
Copy the sign of the second term -
Write down the square root of the third 4
term
Square the resulting binomial (7 x +2)2

c. 25 x 2+ 60 xy +36 y 2=¿

Learning Module in Elementary Algebra Page│26

d. 8 x 2−40 xy +50 y 2=2(4 m 2−20 mn+25 n2 )
¿ 2(2 m−5 n)2

Observe that in d, a common monomial factor is first removed. Then,

continue factoring the trinomial using steps 1- 4.

Self-check 6.11
Factor each completely. (All are factorable.)
1. 16 x 2+ 40+25 6. 25 x 2+110 x +121
2. 36 v2 −132 v+ 121 7. 25 n2−40 n+16
2
3. 121 m −198 m+81 8. 144 x 2 +264 x+ 121
4. 49 p2−28 p+ 4 9. 36 x 2−60 x+ 25
5. 100 b2−180 b+ 81 10. 4 a2−36 a+81

Problem set #8
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________

Instructions: 1. Use one-half (crosswise) sheet of paper.

2. Write the answers only.

A. Factor each completely.

1. 864−4 u3 6. 216 a 3−125 b3
2. 54 x 3−2 7. 125 x 3+216 y 3
3. 108−4 x 3 8. 32 m3 +500 n3
4. 375−81 a3 9. 64 x −27 y 3
3

5. 500 x 3+256 10. 125 a3 +64 b3

B. Factor each completely. (All are factorable.)

11. 4 n2−28 n+ 49 16. 9 a 2+30 ab +25 b2
2
12. 100 n −60 n+ 9 17. 4 x2 +20 x +25
13. 25 k 2 −80 k +64 18. 121 n2−110 n+25
14. r 2−22r +121 19. x 2+ 24 x+ 144
15. 4−28 y + 49 y 2 20. 8 x 2+ 24 xy +18 y 2

Learning Module in Elementary Algebra Page│27

Lesson 6.12 Factoring Trinomial of the form x 2+ bx+ c

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Factor a trinomial of the form x 2+ bx+ c

The trinomial of the form x 2+ bx+ c can be factored out by mere inspection of
the coefficient of the middle term and the constant term. To get the factors, we find
two numbers whose product is the third term and whose sum is the coefficient of x.

Example:
Factor:
a. x 2+ 5 x−24 d. x 2+ x−12
b. x 2+ 8 x+12 e. x 2−x−12
c. x 2−8 x +12

Solutions:
a. We need two numbers whose product is -24 and whose sum is 5. The
two numbers are -3 and 8. Thus,
x 2+ 5 x−24=(x−3)( x +8)

b. The factors of 12 whose sum is 8 are 6 and 2

x 2+ 8 x+12=( x +6)(x +2)

c. The factors of 12 whose sum is -8 are -6 and -2

2
x −8 x +12=( x−6)( x−2)

d. The two numbers whose product is -12 and whose sum is 1 are 4 and -3.
x 2+ x−12=( x+ 4)(x−3)

e. The two number whose product is -12 and whose sum is -1 are -4 and 3.
2
x −x−12=( x−4)(x+ 3)

Learning Module in Elementary Algebra Page│28

Self-check 6.12
Factor:
1. x 2+ 7 x +12 6. x 2−2 x−15
2. m2 +10 m+21 7. x 2−6 x +8
3. y 2−7 y −8 8. y 2−7 xy +10 x 2
4. x 2−6 x +5 9. a 2−11 ab−60 b 2
5. x 2+ 4 x−32 10. x 2+ 8 xy+12 y 2

Lesson 6.13 Factoring Trinomial of the form ax 2 +bx +c

Intended Learning Outcome:
At the end of this lesson, you are expected to:
1. Factor a trinomial of the form ax 2 +bx +c

Factoring Trinomials using Trial and Error.

What happens when the leading coefficient is not 1 and there is no GCF?
There are several methods that can be used to factor these trinomials. First we will
use the Trial and Error method.

How to factor trinomials of the form ax 2 +bx +c using trial and error?
Step 1. Write the trinomial in descending order of degrees.
Step 2. Find all the factor pairs of the first term.
Step 3. Find all the factor pairs of the third term.
Step 4. Test all the possible combination of the factors until the correct product is
found.
Step 5. Check by multiplying.
Example 1. Let us factor the trinomial 3 x 2+22 x +7

Solution:
2
Step 1. Write the The trinomial is already 3 x +22 x +7
trinomial in descending in descending order
order
Step 2. Find all the The only factors of 3 x 2 2
3 x +22 x +7
factor pairs of the first are 1 x , 3 x 1 x∙3x
term. Since there is only one
pair, we can put them in (x)(3 x )
the parentheses.
Step 3. Find all the The only factors of 7 are 2
3 x +22 x +7
factor pairs of the third 1, 7. 1 x∙3x 1 ∙7
term. (x)(3 x )

Step 4. Test all the Possible factors Product

possible combination of (x +1)(3 x +7) 2
3 x +10 x +7
the factors until the (x +7)(3 x+1) 3 x 2+22 x +7
correct product is found.
Step 5. Check by ( x +7)(3 x+1)
multiplying. 2
3 x +22 x +7 ✓

Example 2. Factor completely: 6 x 2−13 x+ 5

Solution:
The trinomial is already in descending 6 x 2−13 x+ 5
order
2
Find all the factor pairs of the first term. 6 x −13 x+ 5
1 x∙6 x
2 x∙3x
Find all the factor pairs of the third 2
6 x −13 x+ 5
term. Consider the signs. Since the 1 x∙6 x −1 ∙−5

Learning Module in Elementary Algebra Page│29

last term, 5 is positive its factors must 2 x∙3x
both be positive or both be negative.
The coefficient of the middle term is
negative so we use the negative
factors.

Consider all the combination of factors.

2
6 x −13 x+ 5
Possible factors Product
( x−1)(6 x−5) 2
6 x −11 x+5
( x−5)(6 x−1) 6 x 2−31 x+5
(2 x−1)(3 x−5) 2
6 x −13 x+ 5
(2 x−5)(3 x−1) 6 x 2−17 x +5
The correct factors are those whose (2 x−1)(3 x−5)
product is the original trinomial.
Check by multiplying. ( 2 x−1 )( 3 x−5 )=6 x 2−10 x−3 x+5
¿ 6 x 2−13 x+5 ✓

Factor Trinomials using the “ac” Method

Another way to factor trinomials of the form ax 2 +bx +c is the “ac” method.
(The “ac” method is sometimes called the grouping method.) The “ac” method is
actually an extension of the method you used in the last section to factor trinomials
with leading coefficient one. This method is very structured (that is step by step),
and it always work!

How to Factor Trinomials of the form ax 2 +bx +c using the “ac” Method.
Step 1. Factor any GCF.
Step 2. Find the product ac.
Step 3. Find two numbers m and n that;
Multiply to ac m∙ n=a ∙ c
Add to b m+n=b
Step 4. Split the middle term using m and n
2
ax +bx +c
ax 2 +mx+ nx+ c
Step 5. Factor by grouping
Step 6. Check by multiplying the factors.

When the third term of the trinomial is negative, the factors of the third term
will have opposite signs.

Example 1. Factor 6 x 2+ 7 x+2

Solution:
Step 1. Factor any GCF. Is there a greatest 6 x 2+ 7 x+2
common factor? No.
Step 2. Find the product a∙c ax 2 +bx +c
ac. 6∙2 2
6 x + 7 x+2
12
Step 3. Find two Find two numbers that
numbers m and n that; multiply to 12 and add to
Multiply to ac 7. Both factors must be
m∙ n=a ∙ c positive.
Add to b 3 ∙ 4=12 3+ 4=7
m+n=b
Step 4. Split the middle Rewrite 7x as 3 x+ 4 x . 6 x 2+ 7 x+2
term using m and n Notice that

Learning Module in Elementary Algebra Page│30

 ax 2 +bx +c 6 x 2+ 3 x + 4 x+ 2 is equal to 6 x 2+ 3 x + 4 x+ 2
2 2
ax +mx+nx+ c 6 x + 7 x+2 . We just split
the middle term to get a
more useful form.
Step 5. Factor by 3 x ( 2 x+ 1 )+2( 2 x +1)
grouping (2 x+1)(3 x+ 2)

Step 6. Check by (2 x+1)(3 x+ 2)

multiplying the factors. 2
6 x + 4 x +3 x+ 2
2
6 x + 7 x+2 ✓

Example 2. 8 x 2−17 x−21

Solution:
Is there a greatest 2
ax +bx +c
common factor? No. 8 x 2−17 x−21
Find a ∙ c a∙c
8(−21)
−168

Find two numbers that multiply to -168 and add to -17. The larger factor must be
negative.
Factors of -168 Sum of factors
1, -168 1 + (-168) = - 167
2, -84 2 + (-84) = - 82
3, -56 3 + (-56) = - 53
4, -42 4 + (-42) = - 38
7, -24 7 + (-24) = - 17
8, -21 8 + (-21) = - 13
Split the middle term using 7x and – 2
8 x −17 x−21
24x 2
8 x + 7 x −24 x−21
Factor by grouping (8 x ¿¿ 2+7 x)−(24 x +21)¿
x (8 x+ 7)−3(8 x +7)
(8 x +7)( x−3)
Check by multiplying
(8 x +7)( x−3)
2
8 x −24 x +7 x−21
8 x 2−17 x−21✓

Self-check 6.13
Factor:
1. 2 x2 −5 x −3 6. 4 y 2+ 8 y+ 3
2. 3 x 2+10 x−8 7. 3 x 2+ 4 x−7
3. 2 y 2+15 y +7 8. 2 x2 +13 x +15
4. 7 a 2−11 a+4 9. 9 y 2 +6 y−8
5. 5 n2 +17 n+6 10. 6 x 2−7 x−20

Learning Module in Elementary Algebra Page│31

 Problem set #9
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________

Instructions: 1. Use one-half (crosswise) sheet of paper.

2. Write the answers only.

A. Factor:
1. 51−20 k+ k 2 6. x 2−3 x−40
2. y 2 +6 y−72 7. 45+ 14 y + y 2
3. x 2−x−6 8. x 2−13 x+ 36
4. y 2 +3 y−18 9. a 2−14 ab+ 24 b2
5. b 2+7 b−18 10. v 2+12 v +2

B. Factor:
2
11. 2 n −3 n−14 16. 5 z 2+7 z +2
2 2
12. 10 x +13 x−30 17. 2 n −11 n+5
2
13. 12 y +7 y +1 18. 3 z 2+ z −2
14. 2 n2 +9 n−5 19. 5 h2−2 h−7
2
15. 2 x +7 x +6 20. 8 s2−10 st+3 t 2

Learning Module in Elementary Algebra Page│32

 CHAPTER 7
LINEAR EQUATIONS AND INEQUALITIES
The coordinate Plane

Introduction
 
The coordinate plane was developed centuries ago and refined by the
French mathematician René Descartes. In his honor, the system is sometimes
called the Cartesian coordinate system. The coordinate plane can be used to plot
points and graph lines. This system allows us to describe algebraic relationships in
a visual sense, and also helps us create and interpret algebraic concepts.

The coordinate plane is consist of a horizontal axis and a vertical axis,

number lines that intersect at right angles. (They are perpendicular to each other.)
 

Learning Module in Elementary Algebra Page│33

 The horizontal axis in the coordinate plane is called the x-axis. The vertical
axis is called the y-axis. The point at which the two axes intersect is called
the origin. The origin is at 0 on the x-axis and 0 on the y-axis.
 
The intersecting x- and y-axes divide the coordinate plane into four
sections. These four sections are called quadrants. Quadrants are named using
the Roman numerals I, II, III, and IV beginning with the top right quadrant and
moving counter clockwise.
 
Locations on the coordinate plane are described as ordered pairs. An
ordered pair tells you the location of a point by relating the point’s location along
the x-axis (the first value of the ordered pair) and along the y-axis (the second
value of the ordered pair).
 
In an ordered pair, such as (x, y), the first value is called the x-
coordinate and the second value is the y-coordinate. Note that the x-coordinate
is listed before the y-coordinate. Since the origin has an x-coordinate of 0 and a y-
coordinate of 0, its ordered pair is written (0, 0).

Consider the point below. 

 

To identify the location of this point, start at the origin (0, 0) and move right
along the x-axis until you are under the point. Look at the label on the x-axis. The
4 indicates that, from the origin, you have traveled four units to the right along
the x-axis. This is the x-coordinate, the first number in the ordered pair.
 
From 4 on the x-axis move up to the point and notice the number with which
it aligns on the y-axis. The 3 indicates that, after leaving the x-axis, you traveled 3
units up in the vertical direction, the direction of the y-axis. This number is the y-
coordinate, the second number in the ordered pair. With an x-coordinate of 4 and
a y-coordinate of 3, you have the ordered pair (4, 3).
Let’s look at another example.

Example
Problem Describe the point shown as an ordered pair.

Learning Module in Elementary Algebra Page│34

 (5, y) Begin at the origin and move along the x-
axis. This is the x-coordinate and is
written first in the ordered pair.
(5, 2) Move from 5 up to the ordered pair and
read the number on the y-axis. This is
the y-coordinate and is written second in
the ordered pair.
Answer The point shown as an ordered pair is (5, 2).
 
 

Plotting Points in the Coordinate Plane

 
Now that you know how to use the x- and y-axes, you can plot an ordered
pair as well. Just remember, both processes start at the origin—the beginning! The
example that follows shows how to graph the ordered pair (1, 3).
 
Example
Problem Plot the point (1, 3).

The x-coordinate is 1 The y-coordinate is 3
because it comes first in the because it comes second in
ordered pair. Start at the the ordered pair. From here
origin and move a distance move directly 3 units in a
of 1 unit in a positive positive direction (up). If you
direction (to the right) from look over to the y-axis, you
the origin along the x-axis. should be lined up with 3 on
that axis.
Answer Draw a point at this location and label the point (1, 3).

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In the previous example, both the x- and y-coordinates were positive. When
one (or both) of the coordinates of an ordered pair is negative, you will need to
move in the negative direction along one or both axes. Consider the example
below in which both coordinates are negative.
 
Example
Problem Plot the point (−4, −2).

The x-coordinate is −4 The y-coordinate is −2

because it comes first in because it comes second in
the ordered pair. Start at the ordered pair. Now move 2
the origin and move 4 units units in a negative direction
in a negative direction (left) (down). If you look over to
along the x-axis. the y-axis, you should be lined
up with −2 on that axis.
Answer Draw a point at this location and label the point (−4, −2).
The steps for plotting a point are summarized below.
 
Steps for Plotting an Ordered Pair (x, y) in the Coordinate Plane
o       Determine the x-coordinate. Beginning at the origin, move horizontally,
the direction of the x-axis, the distance given by the x-coordinate. If the x-
coordinate is positive, move to the right; if the x-coordinate is negative,
move to the left.
o       Determine the y-coordinate. Beginning at the x-coordinate, move
vertically, the direction of the y-axis, the distance given by the y-
coordinate. If the y-coordinate is positive, move up; if the y-coordinate is
negative, move down.
o       Draw a point at the ending location. Label the point with the ordered pair.

 
 
The Four Quadrants
 
Ordered pairs within any particular quadrant share certain characteristics.
Look at each quadrant in the graph below. What do you notice about the signs of
the x- and y-coordinates of the points within each quadrant?
 

Learning Module in Elementary Algebra Page│36

 
Within each quadrant, the signs of the x-coordinates and y-coordinates of
each ordered pair are the same. They also follow a pattern, which is outlined in the
table below.
 
General Form
Quadrant of Point in this Example Description
Quadrant
Starting from the origin, go along
the x-axis in a positive direction
I    (+, +)   (5, 4)
(right) and along the y-axis in a
positive direction (up).
Starting from the origin, go along
the x-axis in a negative direction
II    (−, +)   (−5, 4)
(left) and along the y-axis in a
positive direction (up).
Starting from the origin, go along
the x-axis in a negative direction
III    (−, −)   (−5, −4)
(left) and along the y-axis in a
negative direction (down).
Starting from the origin, go along
the x-axis in a positive direction
IV    (+, −)   (5, −4)
(right) and along the y-axis in a
negative direction (down).
 
Once you know about the quadrants in the coordinate plane, you can
determine the quadrant of an ordered pair without even graphing it by looking at
the chart above. Here’s another way to think about it.
 

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The example below details how to determine the quadrant location of a
point just by thinking about the signs of its coordinates. Thinking about the
quadrant location before plotting a point can help you prevent a mistake. It is also
useful knowledge for checking that you have plotted a point correctly.
 
Example
Problem In which quadrant is the point (−7, 10) located?
(−7, 10) Look at the signs of
the x- and y-coordinates. For
this ordered pair, the signs
are (−, +).
Points with the pattern Using the table or grid above,
(−, +) are in Quadrant II. locate the pattern (−, +).
Answer The point (−7, 10) is in Quadrant II.
 
 
Example
Problem In which quadrant is the point (−10, −5) located?
(−10, −5) Look at the signs of
the x- and y-coordinates. For
this ordered pair, the signs
are (−, −).
Points with the pattern Using the table or grid above,
(−, −) are in Quadrant III. locate the pattern (−, −).
Answer The point (−10, −5) is in Quadrant III.
 
What happens if an ordered pair has an x- or y-coordinate of zero? The
example below shows the graph of the ordered pair (0, 4).
 

 
A point located on one of the axes is not considered to be in a quadrant. It
is simply on one of the axes. Whenever the x-coordinate is 0, the point is located
on the y-axis. Similarly, any point that has a y-coordinate of 0 will be located on
the x-axis.
 
  
Lesson 7.1 Slope of a Line

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Define slope,

Learning Module in Elementary Algebra Page│38

 2. Find the slope given the two points.

 Introduction
The idea of slope is something you encounter often in everyday life. Think
about rolling a cart down a ramp or climbing a set of stairs. Both the ramp and the
stairs have a slope. You can describe the slope, or steepness, of the ramp and
stairs by considering horizontal and vertical movement along them. In
conversation, you use words like “gradual” or “steep” to describe slope. Along a
gradual slope, most of the movement is horizontal. Along a steep slope, the
vertical movement is greater.
 
Defining Slope
The mathematical definition of slope is very similar to our everyday one. In
math, slope is used to describe the steepness and direction of lines. By just
looking at the graph of a line, you can learn some things about its slope, especially
relative to other lines graphed on the same coordinate plane. Consider the graphs
of the three lines shown below:
 

 
First, let’s look at lines A and B. If you imagined these lines to be hills, you
would say that line B is steeper than line A. Line B has a greater slope than line A.
 
Next, notice that lines A and B slant up as you move from left to right. We
say these two lines have a positive slope. Line C slants down from left to right.
Line C has a negative slope. Using two of the points on the line, you can find the
slope of the line by finding the rise and the run. The vertical change between two
points is called the rise, and the horizontal change is called the run. The slope
equals the rise divided by the run 

 
 
Finding the Slope of a Line from a Graph
You can determine the slope of a line from its graph by looking at the rise
and run. One characteristic of a line is that its slope is constant all the way along it.
So, you can choose any 2 points along the graph of the line to figure out the slope.
Let’s look at an example.
 
 

Learning Module in Elementary Algebra Page│39

 Example
Problem Use the graph to find the slope of the line.

rise = 2 Start from a point on the line,

such as (2, 1) and move
vertically until in line with
another point on the line, such
as (6, 3). The rise is 2 units. It is
positive as you moved up.
run = 4 Next, move horizontally to the
point (6, 3). Count the number
of units. The run is 4 units. It is
positive as you moved to the
right.

rise
Slope =
run

2 1
Slope =  =
4 2
Answer 1
The slope is  .
2
 
1
This line will have a slope of   no matter which two points you pick on the
2
line. Try measuring the slope from the origin, (0, 0), to the point (6, 3). You will find
rise 3 1
that the rise = 3 and the run = 6. The slope is  = = . It is the same!
run 6 2
 
Let’s look at another example.
 
Example
Proble Use the graph to find the slope of the two lines.
m

Notice that both of these lines have

Learning Module in Elementary Algebra Page│40

 positive slopes, so you expect your
answers to be positive.
Blue line

rise = 4 Start with the blue line, going from

point (-2, 1) to point (-1, 5). This line
has a rise of 4 units up, so it is
positive.
run = 1 Run is 1 unit to the right, so it is
positive.
Substitute the values for the rise and
run in the formula
rise
Slope =
run

4
Slope = =4
1
Red line

rise = 1 The red line, going from point (-1, -2)

to point (3, -1) has a rise of 1 unit.
run = 4 The red line has a run of 4 units.
Substitute the values for the rise and
run into the formula
rise
Slope =
run

1
Slope =
4
Answer 1
The slope of the blue line is 4 and the slope of the red line is  .
4
 

When you look at the two lines, you can see that the blue line is steeper
than the red line. It makes sense the value of the slope of the blue line, 4, is
1
greater than the value of the slope of the red line,  . The greater the slope, the
4
steeper the line.
 
The next example shows a line with a negative slope.
 
Example
Proble Find the slope of the line graphed below.
m

rise = −3 Start at Point A, (0, 4) and

Learning Module in Elementary Algebra Page│41

 rise −3.This means moving 3
units in a negative direction.
run = 2 From there, run 2 units in a
positive direction to Point B (2,
1).
rise
Slope = 
run

−3
Slope = 
2
Answer −3
The slope of the line is  .
2
 
 
Direction is important when it comes to determining slope. It’s important to
pay attention to whether you are moving up, down, left, or right; that is, if you are
moving in a positive or negative direction. If you go up to get to your second point,
the rise is positive. If you go down to get to your second point, the rise is negative.
If you go right to get to your second point, the run is positive. If you go left to get to
your second point, the run is negative. In the example above, you could have
found the slope by starting at point B, running −2, and then rising +3 to arrive at
rise +3 −3
point A. The result is still a slope of  = =
run −2 2
  
 Finding the Slope of a Line Given Two Points
You’ve seen that you can find the slope of a line on a graph by measuring
the rise and the run. You can also find the slope of a straight line without its graph
if you know the coordinates of any two points on that line. Every point has a set of
coordinates: an x-value and a y-value, written as an ordered pair (x, y).
The x value tells you where a point is horizontally. The y value tells you where the
point is vertically.
 
Consider two points on a line Point 1 and Point 2. Point 1 has coordinates
(x1, y1) and Point 2 has coordinates (x2, y2).
 

 
The rise is the vertical distance between the two points, which is the
difference between their y-coordinates. That makes the rise y2 − y1. The run
between these two points is the difference in the x-coordinates, or x2 − x1.
 
rise y 2− y 1
So,  Slope=  or m=  
run x 2−x 1
 

Learning Module in Elementary Algebra Page│42

 In the example below, you’ll see that the line has two points each indicated
as an ordered pair. The point (0, 2) is indicated as Point 1, and (−2, 6) as Point 2.
So you are going to move from Point 1 to Point 2. A triangle is drawn in above the
line to help illustrate the rise and run.
 

 
You can see from the graph that the rise going from Point 1 to Point 2 is 4,
because you are moving 4 units in a positive direction (up). The run is −2, because
you are then moving in a negative direction (left) 2 units. Using the slope formula,
rise 4
ope= = =−2 .
run −2
 
You do not need the graph to find the slope. You can just use the
coordinates, keeping careful track of which is Point 1 and which is Point 2. Let’s
organize the information about the two points:
 
Name Ordered Pair Coordinates
x1 = 0
Point 1 (0, 2)
y1 = 2
x2 = -2
Point 2 (−2, 6)
y2 = 6
 
y 2− y 1 6−2 4
The slope, m= = = =−2   . The slope of the line, m, is −2.
x 2−x 1 −2−0 −2
 
It doesn’t matter which point is designated as Point 1 and which is Point 2.
You could have called (−2, 6) Point 1, and (0, 2) Point 2. In that case, putting the
2−6 −4
coordinates into the slope formula produces the equation  m= = =−2.
0−(−2) 2
Once again, the slope m = −2. That’s the same slope as before. The
important thing is to be consistent when you subtract: you must always subtract in
the same order y2 − y1 and x2 − x1.
 
Example
Problem What is the slope of the line that contains the
points (5, 5) and (4, 2)?
x1 = 4 (4, 2) = Point 1, (x1, y1)
y1 = 2
x2 = 5 (5, 5) = Point 2, (x2, y2)
y2 = 5
Substitute the values into the slope
formula and simplify.

Learning Module in Elementary Algebra Page│43

 m = 3
Answer The slope is 3.
 
The example below shows the solution when you reverse the order of the
points, calling (5, 5) Point 1 and (4, 2) Point 2.
 
Example
Problem What is the slope of the line that contains the
points (5, 5) and (4, 2)?
x1 = 5 (5, 5) = Point 1, (x1, y1)
y1 = 5
x2 = 4 (4, 2) = Point 2, (x2, y2)
y2 = 2
Substitute the values into the slope
formula and simplify.

m = 3
Answer The slope is 3.
 
Notice that regardless of which ordered pair is named Point 1 and which is
named Point 2, the slope is still 3. 
 
Finding the Slopes of Horizontal and Vertical Lines
 
So far you’ve considered lines that run “uphill” or “downhill.” Their slopes
may be steep or gradual, but they are always positive or negative numbers. But
there are two other kinds of lines, horizontal and vertical. What is the slope of a flat
line or level ground? Of a wall or a vertical line?
 
Let’s consider a horizontal line on a graph. No matter which two points you
choose on the line, they will always have the same y-coordinate. The equation for
this line is y = 3. The equation can also be written as y = (0)x + 3.
 

 
Using (−3, 3) as Point 1 and (2, 3) as Point 2, you get:
 

Learning Module in Elementary Algebra Page│44

  
The slope of this horizontal line is 0.
 
Let’s consider any horizontal line. No matter which two points you choose
on the line, they will always have the same y-coordinate. So, when you apply the
slope formula, the numerator will always be 0. Zero divided by any non-zero
number is 0, so the slope of any horizontal line is always 0.
  
How about vertical lines? In their case, no matter which two points you choose,
they will always have the same x-coordinate.
 

 
 
So, what happens when you use the slope formula with two points on this
vertical line to calculate the slope? Using (2, 1) as Point 1 and (2, 3) as Point 2,
you get:
 

 
But division by zero has no meaning for the set of real numbers. Because of
this fact, it is said that the slope of this vertical line is undefined. This is true for all
vertical lines— they all have a slope that is undefined.
 
Example
Problem What is the slope of the line that contains the
points (3, 2) and (−8, 2)?

(3, 2) = Point 1, 

(−8, 2) = Point 2, 

Substitute the values into the slope

formula and simplify.

m = 0
Answer The slope is 0, so the line is horizontal.
  
Summary
Slope describes the steepness of a line. The slope of any line remains
constant along the line. The slope can also tell you information about the direction

Learning Module in Elementary Algebra Page│45

of the line on the coordinate plane. Slope can be calculated either by looking at the
graph of a line or by using the coordinates of any two points on a line. There are
rise y 2− y 1
two common formulas for slope:  Slope=  and m=  where m = slope
run x 2−x 1
and ( x 1 , y 1 ) and (x 2 , y 2 ) are two points on the line.
 
The images below summarize the slopes of different types of lines.
 

Self-check 7.1
Find the slope, if it exists, of the line passing through each pair of points. Tell
whether the line involved slopes upwards or downwards, is horizontal or is vertical.

1. (- 2, 3) ; (2, 8) 6. (3, -4) ; (5, 1)

2. (4, 3) ; (0, 9) 7. (4, - 5) ; (4, 7)
3. (4, 5) ; (7, 5) 8. (- 4, 3) ; (2, 3)
4. (3, - 2) ; (3, 4) 9. ( - 2, - 3) ; (- 2, 5)
5. (- 2, - 3) ; (2, 2) 10. (- 5, - 3) ; (3 , 5)

THE EQUATIONS OF A LINE

The standard form or general equation of a line is ax +by =c where a, b and

c are real numbers and not both a and b are zero.

Lesson 7.2 The Slope-Intercept Form

Intended Learning Outcome:
At the end of this lesson, you are expected to:
1. Write the equation of a line given the slope, m, and the y-intercept, b.

The slope intercept is the most “popular” form of a straight line. Many
students find this useful because of its simplicity. One can easily describe the
characteristics of the straight line even without seeing its graph because the slope
and y-intercept can easily be identified or read off from this form.
The equation of the line with a slope of m and a y-intercept of b is:
y=mx +b ; this is called the slope-intercept form of the equation of the line.
The slope m measures how steep the line is with respect to the horizontal.
The y-intercept is the point where the line crosses the y-axis.

Example: Given the slope m and the y-intercept b, write the equation of the line.
Then re-write it in standard form.
−3 −5 3
a. m = -5, b = 3 b. m = 5 , b = -2 c. m = 6 ; b = 4
Solutions:
a. The needed information to write the equation of the line in the form
ax +by =c are clearly given the problem since

Learning Module in Elementary Algebra Page│46

 m = -5 (slope)
b= 3 (y-intercept)
y=mx+b slope-intercept formula
y=(−5)x +(3) by substitution
y=−5 x +3 slope-intercept form
5 x+ y=3 standard form

−3
b. m = , b = -2
5
y=mx+b slope-intercept formula
−3
y= x−2 by substitution
5
5 y=−3 x−10 Multiplying both sides by 5
3 x+ 5 y =10 standard form

−5 3
c. m = 6 ; b = 4
y=mx+b slope-intercept formula
−5 3
y= x+ by substitution
6 4
12 y=−10 x+ 9 Multiplying both sides by 12(LCD)
10 x+12 y=9 standard form

Self-check 7.2
Given the slope and the intercept, write the standard form of each line.
1. m = 3 ; b = 4 6. m = - 4 ; b = 5
3 2
2. m = 5; b = 4 7. m = 5 ; b = - 4
1 3
3. m= 2 ;b=3 8. m = - 2 ; b = 5
−3 1 1 −2
4. m= 2 ;b= 3 9. m = 2 ; b = 3
4 5
5. m=2;b= 3 10. m = 2 ; b = 3
Problem set #10
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________

Instructions: 1. Use one-half (crosswise) sheet of paper.

2. Write the answers only.

A. Find the slope, if it exists, of the line passing through each pair of points. Tell
whether the line involved slopes upwards or downwards, is horizontal or is
vertical.
1. (2, - 3), (6, - 3)
2. (- 6, 2), (1, 8)
3. (2, - 3), (- 3, 2)
4. (5, - 2), (- 3, 6)
5. (1, - 2), (3, 4)

B. Given the slope and the intercept, write the standard form of each line.
1. m = 1, b=1
1
2. m = , b = -2
4
3. m = -3, b = -3
2
4. m = , b=3
3
−3
5. m = , b=0
2

Learning Module in Elementary Algebra Page│47

Lesson 7.3 The Two-Point Form

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Write the equation of a line in standard form given the two points.

The two-point form equation of a line containing two points ( x 1 , y 1 ) and

( x 2 , y 2 ) is:
y 2− y1
y− y 1 = ( x−x 1 )
x 2−x 1
Example1. Find the equation of the straight line passing through the points (2,3)
and (5 ,−6).

Solution:
The equation of the straight line passing through the points ( x 1 , y 1 ) =¿ (2 , 3)
and ( x 2 , y 2 ) =¿ (6 ,−5) or
x 1=2 , y1 =3
x 2=5 , y 2=−6

y 2− y 1
y− y1 = ( x−x 1) two-point form formula
x2− x1

Learning Module in Elementary Algebra Page│48

 −5−3
y−3= ( x−2) by substitution
6−2
−8
y−3= (x−2) simplify
4
y−3=−2( x−2)
y−3=−2 x+ 4
2 x+ y =7 the required equation.

Example 2. Find the equation of the straight line joining the points (−3 , 4 )
and (5 ,−2).

Solution:
The equation of the straight line joining the points ( x 1 , y 1 ) =¿ (−3,4)
and ( x 2 , y 2 ) =¿ (5 ,−2) or
x 1=−3 , y 1=4
x 2=5 , y 2=−2

y 2− y 1
y− y1 = ( x−x 1) two-point form formula
x2− x1
−2−4
y−4= [ x−(−3)] by substitution
5−(−3 )
−6
y−4= (x+ 3) simplify
8
−3
y−4= ( x+3)
4
4 ( y −4)=−3( x +3)
4 y−16=−3 x−9
3 x+ 4 y =7 the required equation

Self-check 7.3
Given two of its points, find the equation of its line. Write each equation in
standard form.
1. (5, - 2) , (- 2, 5) 6. (- 3, -5), (2, - 15)
2. (- 6, - 3), (- 7, - 1) 7. (-1, 1), (2, -3)
3. (- 7, 4), (-2, 19) 8. (2, 3), (-3, -3)
4. (4, 5). (0, 3) 9. (-3, 4), (2, 2)
5. (2, 10), (5, 1) 10. (2, 2), -3, -3)

Lesson 7.4 The Point-slope Form

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Find the equation of a line in standard form given a point, P, and a slope, m.

y 2− y 1
In the two-point form equation of a line, if we replace the slope, by
x 2−x 1
m, we get the point-slope form of the equation.
The point-slope form of the equation of a line with slope m and containing
the point ( x1 , y 1 ) is:
y− y 1 =m( x−x 1 )

Example 1. Find the equation of the line with a slope of 2 and containing the point
(3, -1). Write the equation in standard form.

Learning Module in Elementary Algebra Page│49

Solution:
m = 2, the given point is (3, -1)
y− y1 =m( x−x 1) point-slope form equation
y−(−1)=2( x−3) substitution
y +1=2 x−6 simplify
−2 x+ y=−7 The required equation

3
Example 2. Find the equation of the line with a slope of and containing the point
2
(-2, 4). Write the equation in standard form.

Solution:
3
m = , the given point is (-2, 4)
2
y− y1 =m( x−x 1) point-slope form equation
3
y−4= [x−(−2)] substitution
2
3
y−4= ( x +2) simplify
2
2( y −4)=3(x +2)
2 y−8=3 x+6
2 y−3 x=14 the required equation

−4
Example 3. Find the equation of the line with a slope of and containing the
3
point (-2, -5). Write the equation in standard form.

Solution:
−4
m= , the given point is (-2, -5)
3
y− y1 =m( x−x 1) point-slope form equation
−4
y−(−5)= [x−(−2)] substitution
3
−4
y +5= (x +2) simplify
3
3 ( y +5 )=−4 ( x+2)
3 y +15=−4 x−8
4 x+3 y =−23 the required equation

Self-check 7.4
Find the equation of each line given each slope and one point on it. Write each
equation in standard form.
2
1. m = 2; (- 2, 5) 6. m = 3 ; (2, - 3)
−3
2. m = - 3; (- 1, 3) 7. m = 2 ; (- 3, 2)
3
3. m = 4 ; (4, 1) 8. m = 4; (- 3, - 5)

Learning Module in Elementary Algebra Page│50

 −2
4. m = 3 ; (3, - 1) 9. m = - 5 ; (2, -1)
1 3
5. m = 2 ; (3, -5) 10. m = 2 ; (-2, -3)

Lesson 7.5 The Intercepts Form

Intended Learning Outcome:

At the end of this lesson, you are expected to:
1. Find the equation of a line given the x-intercept, a, and y-intercept, b, in
standard form
5
2 x−3 y=5 x=
In the equation, setting y to 0, gives 2 , the x-intercept. Setting x
−5
y=
to 0, gives 3 , the y-intercept.
If we divide the original equation by 5 on both sides, we get,
x y
+ =1
2x 3 y 5 −5
− =1
5 5 , which can we re-write as 2 3
A study of the last equation shows that the denominators are the intercepts of
the line. This last equation is called the intercepts form of the equation of a line.

In general formula, if we begin with ax+by=c , the intercepts are:

c c
If x=0 , by=c or y= , If y=0 , ax=c or x= .
b a

If we divide both sides of the equation ax +by =c by c, we get,

x y
ax by + =1.
+ =1, which we can re-write as c c
c c
a b
Again, following the convention that (a, 0) and (0, b) are the points where a line
intersects the x-axis and y-axis respectively we have the following:

x y
+ =1
The equation of a line in intercepts form is a b , where a and b are
the x-intercept and y-intercept respectively.

Examples:
1. Find the equation of a line in standard form if the x and y intercepts are 2
and -3 respectively.

2. Find the equation of a line in standard form if the x and y intercepts are
−3 5
and respectively.
2 3

Solutions:

1. x-intercept or a = 2
y-intercept or b= -3

Learning Module in Elementary Algebra Page│51

 x y
The equation is + =1 Intercepts form
a b

x y
+ =1 by substitution
2 −3

3 x−2 y=6 Multiplying both sides by 6

Standard form

−3
2. x-intercept or a =
2
5
y-intercept or b =
3

x y
The equation is + =1 Intercepts form
a b

x y
+ =1
−3 5 by substitution
2 3

2x 3 y
− =1 simplifying
−3 5
−10 x+ 9 y=15 Multiplying both sides by 15
Standard form

Self-check 7.5
Given the intercepts of a line, find its equation in standard form.
5 1
1. a = 2; b = 3 6. a = 6 ; b = 3
3 1
2. a = - 4; b = - 3 7. a = 2 ; b = 4
2 −1
3. a = 3 ; b = - 2 8. a = - 7 ; b = 2
3
4. a = - 5; b = 6 9. a = 5 ; b = 4
−5 7 −3
5. a = 3; b = 2 10. a = 3 ; b = 4

Problem set #11

Learning Module in Elementary Algebra Page│52
Name: _________________________ Score: _______
Grade/Section: ___________________ Date: ________

Instructions: 1. Use one-half (crosswise) sheet of paper.

2. Write the answers only.

A. Given two of its points, find the equation of its line. Write each equation in
standard form.
1. (2, -1), (-1, 2)
2. (3, -4), (0, 0)
3. (3, 0), (0, -3)
4. (2, 3), (3, 2)
5. (6, 2), (3, 5)

B. Find the equation of each line given each slope and one point on it. Write each
equation in standard form.
−3
6. m= , (- 1, - 1)
2
3
7. m= , (5, -2)
5
8. m=−8 , (- 4, -1)
9. m=−3 , (- 1, 15)
10. m=9 , (0, - 2)

C. Given the intercepts of a line, find its equation in standard form.

11. a = - 3; b = 2
−3
12. a = 2 ; b=4
3 5
13. a = 4, b= 2
5 −2
14. a = 3; b= 3
7 −4
15. a = 2 ; b= 3

Learning Module in Elementary Algebra Page│53

References

De Leon, C., Bernabe, J., Elementary Algebra, JTW Corporation, Quezon City,
2002

Department of Education. Grade 7 Math Learning Guide

Dictionary.com http://www.dictionary.com

https://openstax.org/books/elementary-algebra-2e/pages/7-3-factor-trinomials-of-1
the-form-ax2-bx-c

https://courses.lumenlearning.com/suny-begin/algebra/chapter/read-or-watch-find-
slope-from-a-graph/

Malaborbor, P., Carreon, E., Lorenzo, JR., Sabangan, L., Elementary Algebra,
Diamond Offset Press, Inc., March 2002

Marasigan, J., Coronel, A., Coronel I., Elementary Algebra, Union Publishing, Phil,
2003

Morala, N., Sandoval, C., Crisostomo, E., Dela Cruz, L., Bernardo, A., Piao, L.
College Algebra (Worktext), Trinitas Publishing, Inc., 2005

Properties of Equality (n.d). Retrieved from

www.basic-mathematics.com/properties-of-equality.html on January 26,
2021.

Properties of Real Numbers (n.d.). Retrieved from

www.chilimath.com/lessons/introductory-algebra/properties-of-real-numbers
on January 26, 2021.

Properties of Real Numbers (n.d.). Retrieved from

www.mathsisfun.com/sets/real-numbers-properties.html on January 26,
2021.

Special Products (n.d). Retrieved from

www.Intmath.com/factoring-fraction/1-special-products.php on January 26,
2021.

Special Products (n.d). Retrieved from

www.mathsisfun.com/algebra/special-binomial-products.html on January
26, 2021.

Learning Module in Elementary Algebra Page│54

ANSWER KEY

Self-Check 5.1
A. B. C.
1. Closure Property 6. 15+(5+8), Associative Property 11. 3(4) + 3(5)
2. Commutative Property 7. 23, Commutative Property 12. 7(6) + 7(5)
3. Closure Property 8. 90, Closure Property 13. 5a + 5b
4. Distributive Property 9. 1, Identity Property 14. m(8 – 15)
5. Commutative Property 10. 4, Distributive Property 15. 1.6(3) + 1.6(4)

Self-Check 5.2
A. B.
1. Reflexive Property 7. 12
2. Symmetric Property 8. a - 4
3. Symmetric Property 9. 6
4. Reflexive Property 10. 38
5. Transitive Property
6. Multiplicative Property of Zero

Self-Check 5.3
9
1. x = 50 6. x = 11. x = -4
4
2. x = 53 7. x = -15 12. x=2
3. x = -9 8. x = 30 13. x=3
4. x = -59 9. y = 4 14. x = -9
5. x = -40 10. x = 3 15. a=4

Self-Check 5.4
1. x < -6 6. x > -70
2. x > 25 7. x ≥ -8.4
1
3. x > 8. x > -3
10
4. x > 53 9. x > 4.48
5. x ≤ 3.2 10. x > 0.9

Self-Check 5.5
1. 24, 96
2. 75
3. 9, 11, 13
4. 4, 6, 8, 10
5. 20 – number of 25-centavo coins
30 – number of 10-centavo coins
6. 42 – number of 5-peso coins
53 – number of 1- peso coins
7. 5 – Fe’s age
3 – Anita’s age
8. 11 – Ben’s age
44 – Ariel’s age

Self-Check 6.1
3 2 4 3
14 x+21 12 a b −18 a b
1. 6.
2
12 a−20 3 x −2 xy
2. 7.
21 x+3 xy 2
15 a b−12 ab
2
3. 8.
2 2 3
18 x y−27 xy 28 x −12 xy
4. 9.
2
12xy−3x 40 x 2 y 2 −35 x 3 y 3
5. 10.

Self-Check 6.2
2 2 2
x −16 9 a −b
1. 6.
2
a −25 25 a 4 −4 b 2
2. 7.

Learning Module in Elementary Algebra Page│55

 2 2 2
4 a −49 16 x −25 y
3. 8.
c 2 −d 2 4 a4 b4 −c 6
4. 9.
4 x2− y2 16 x 4 y 2 −9 z 2
5. 10.

Self-Check 6.3
A.
1. Perfect Square 6. Not a Perfect Square
2. Not a Perfect Square 7. Perfect Square
3. Not a Perfect Square 8. Perfect Square
4. Perfect Square 9. Perfect Square
5. Perfect Square 10. Not a Perfect Square

B.
2
9y
11. 4 16.
b2 6x
12. 17.
8a 16
13. 18.
12xy a2
14. 19.
2
25 a 42n
15. 20.

C.
2 2
x +4 x+ 4 16 x −56 x +49
21. 26.
2
x −6 x +9 9 x 2 +12 xy+4 y 2
22. 27.
2 2 2
4 x −12 x+ 9 16 x +24 xy+9 y
23. 28.
4 a2 +12 a+ 9 25 a2 −20 ab+4 b 2
24. 29.
2
9 a −30 a+25 4 y 2−20 yz+25 z 2
25. 30.

Self-Check 6.4
3 2
x −15 x +75 x−125
1.
3 2
x +12 x +48 x +64
2.
8 a3 −12 a2 b +6 ab 2 −b 3
3.
3 2
m −9 m +27 m−27
4.
3 2 2 3
125 x +75 x y+15 xy + y
5.

Self-Check 6.5
2 2
x +3 x+ 2 6 x +x−15
1. 6.
2 2
x −3 x +2 x +3 x−28
2. 7.
2
3 x −4 x−4 6 x 2+xy−15 y 2
3. 8.
2 2 2
x −x−30 12 a +13 ab+ 3 b
4. 9.
2 2 2
x + x−12 8 a +18 ab−5 b
5. 10.

Learning Module in Elementary Algebra Page│56

Self-Check 6.6
2 2 2
4 x + y +9 z +4 xy+12 xz+6 yz
1.
9 a2 +4 b2 + d2 −12 ab+6 ad−4 bd
2.
2 2 2
16 x +9 y +25 z +24 xy −40 xz−30 yz
3.
9 x 2 +25 y 2 +4 z 2 +30 xy +12 xz+20 yz
4.
2 2 2
a +4 b + c + 4 ab+2 ac+ 4 bc
5.

Self-Check 6.7
3 3
27 a +8 27 x −125
1. 6.
3 3 3
8 x +27 8x +y
2. 7.
3 3 3
27 x −64 a +8 b
3. 8.
3 3 3
64 x −27 27 a +8 b
4. 9.
3 3 3
8 a −27 64 x −27 y
5. 10.

Self-check 6.8
1. x ( x ¿¿ 2−x+ 1)¿ 6. 4 x( x−1)
2. 3(5 a+ 4 b +2 c) 7. 5(3 x ¿¿ 2−10 x−2)¿
3. 2 x(4 x−9) 8. 11(11 a−b)
4. y ( x ¿¿ 2+2)¿ 9. 8 c (8 c ¿¿ 2−7 c+11)¿
5. z 2 ( z+ 4) 10. 10(1−n+n¿¿ 2)¿

Self-check 6.9
3 3
1. (ab +4)( ab−4) 6. (8 a+
)(8 a− )
5b 5b
2. (12 a+13 b)(12 a−13b) 7. (3 a ¿ ¿ 2 b2 +5 p2 q2 )(3 a ¿ ¿ 2 b2−5 p2 q2 )¿ ¿
3. (1+0.3 a)(1−0.3 a) 8. 2(5 p+6 q)(5 p−6 q)
1 1
4. (4 x+11)( 4 x−11) 9. (m+ )(m− )
13 13
5. (x ¿¿ 2+16)( x+ 4)(x−4) ¿ 10. x (3 y + x)(3 y−x )

Self-check 6.10
1. (x +2)( x¿¿ 2−2 x +4 )¿ 6. (3 m−5)(9 m¿¿ 2+15 m+ 25)¿
2. (a+ 4)(a¿¿ 2−4 a+16)¿ 7. (x−4)(x ¿¿ 2+ 4 x +16)¿
3. (a+ 6)(a¿¿ 2−6 a+36)¿ 8. 2(6+ 5 m)(36−30 m+25 m¿¿ 2)¿
4. (3+2 x)(9−6 x +4 x¿¿ 2) ¿ 9. 3(3 x +4 )(9 x¿¿ 2−12 x +16) ¿
5. (a−6)(a ¿¿ 2+ 6 a+36)¿ 10. 4 (5 x +4)( 25 x ¿¿ 2−20 x +16)¿

Self-check 6.11
1. (4 x+5)2 6. (5 x+ 11)2
2. (6 v−11)2 7. (5 n−4)2
3. (11 m−9)2 8. (12 x+11)2
4. (7 p−2)2 9. (6 x−5)2
5. (10 b−9)2 10. (2 a−9)2

Self-check 6.12
1. (x +3)(x+ 4) 6. (x +3)( x−5)
2. (m+3)(m+7) 7. (x−2)(x−4)
3. ( y−8)( y+ 1) 8. ( y−5 x )( y−2 x)
4. (x−1)(x−5) 9. (a+ 4 b)( a−15 b)
5. ( x−4)(x +8) 10. ( x +2 y)( x +6 y )

Self-check 6.13

Learning Module in Elementary Algebra Page│57

1. (2 x+1)( x−3) 6. (2 y +3)(2 y +1)
2. (3 x−2)( x+ 4) 7. (3 x+ 7)( x−1)
3. (2 y +1)( y +7) 8. (2 x+3)( x +5)
4. (7 a−4 )(a−1) 9. (3 y +4 )(3 y−2)
5. (5 n+2)(n+3) 10. (3 x+ 4)(2 x−5)

Self-check 7.1
5 5
1. m= , upward 6. m= , upward
4 2
3
2. m= , downward 7. m=¿ undefined, vertical
−2
3. m=0 , horizontal 8. m=0 , horizontal
4. m=¿ undefined, vertical 9. m=¿ undefined, vertical
5
5. m= , upward 10. m=1 , upward
4

Self-check 7.2
1. −3 x+ y=4 6. 4 x+ y=5
2. −20 x+ 4 y =3 7. −2 x+5 y =−20
3. −x +2 y=6 8. 10 x+5 y =3
4. 9 x +6 y=2 9. −3 x+ 6 y=−4
5. −6 x +3 y=4 10. −6 x−3 y=5

Self-check 7.3
1. x + y=3 6. 2 x+ y =−11
2. 2 x+ y =−15 7. 4 x+3 y =−1
3. −3 x+ y=25 8. 6 x−5 y=−3
4. −x +2 y=6 9. 2 x+5 y =14
5. 3 x+ y=16 10. x−4 y=10

Self-check 7.4
1. −2 x+ y=9 6. −2 x+3 y =−13
2. 3 x+ y=0 7. 3 x+ 2 y =−5
3. −3 x+ 4 y =−8 8. −4 x+ y=7
4. 2 x+3 y =3 9. 5 x+ y=9
5. −x +2 y=−13 10. −3 x+ 2 y =0

Self-check 7.5
1. 3 x+ 2 y =6 6. 6 x +15 y=5
2. 3 x+ 4 y =−12 7. 2 x+12 y=3
3. 3 x− y=2 8. x +14 y=−7
4. 6 x−5 y=−30 9. 20 x+ 3 y=12
5. 5 x−6 y=15 10. 9 x−28 y=21

Learning Module in Elementary Algebra Page│58