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1. LINEAR
EQUATIONS and
INEQUALITIES in
ONE VARIABLE

2. Linear equations and
Inequalities in One
Variable
Equation and Inequalities are relations
between two quantities.

3. Equation is a mathematical sentence indicating that
two expressions are equal. The symbol “=“ is used
to indicate equality.
Ex.
2x + 5 = 9 is a conditional equation
since its truth or falsity depends on
the value of x
2 + 9 = 11 is identity equation since both of its
sides are identical to the same
number 11.

4. Inequality is a mathematical sentence indicating
that two expressions are not equal. The symbols
<, >, are used to denote inequality.
Ex.
3 + 2 ≠ 4 is an inequality
If two expressions are unequal, then their
relationship can be any of the following, >, ≥, < or
≤.

5. Linear equation in one variable is an
equation which can be written in the form
of ax + b = 0, where a and b are real-
number constants and a ≠ 0.
Ex.
x + 7 = 12

6. Solution Set of a Linear Equation
Example
C. 4x + 2 = 10 this statement is either true of
false
is false because 4(1) + 2 is ≠ 10
is true because 4(2) + 2 = 10
If x = 1, then 4x + 2 = 10
If x = 2, then 4x + 2 = 10

7. B. x – 4 < 3 this statement is either true or false
If x =6, then x – 4 is true because 6 – 4 < 3
If x = 10 , then x – 4 is false because 6 – 4 is not < 3
When a number replaces a variable in an equation (or
inequality) to result in a true statement, that number is a
solution of the equation (or inequality). The set of all
solutions for a given equation (or inequality) as called the
solution set of the equation (or inequality).

8. Solution Set of Simple Equations and
Inequalities in One Variable by
Inspection
To solve an equation of inequality means to
find its solution set. There are three(3)
ways to solve an equation or inequality by
inspection

9. A. Guess-and-Check
In this method, one guesses and substitutes
values into an equation of inequality to see
if a true statement will result.

10. Consider the inequality x – 12 < 4
If x = 18, then 18 – 12 is not < 4
If x = 17, then 17 – 12 is not < 4
If x = 16, then 16 – 12 is not < 4
If x = 15, then 15 – 12 < 4
If x = 14, then 14 – 12 < 4
Inequality x - 12 < 4 is true for all values of x which are less
than 16. Therefore, solution set of the given inequality is x <
16.

11. Another example
X + 3 = 7
If x = 6, then 6 + 3 ≠ 7
If x = 5, then 5 + 3 ≠ 7
If x = 4, then 4 + 3 = 7
Therefore x = 4

12. B. Cover-up
In this method , one covers up the term
with the variable.

13. Example
Consider equation x + 9 = 15
x + 9 = 15
+ 9 = 15
To result in a true statement, the
Therefore x = 6
must be 6.

14. Another example
X – 1 = 3
– 1 = 3
x = 4

15. C. Working Backwards
In this method, the reverse procedure is
used

16. Consider the equation 2x + 6 = 4
times equals plus equals
2 2x 6
Start
14
End
2 8 6
equals divided equals minus
x

17. Example: 4y = 12
times equals
4
Start 12 End
4
equals divided Therefore y = 3
y

18. Properties of
Equality and
Inequality

19. Properties of Equality
Let a, b, and c be real numbers.
C. Reflexive Property
a = a
Ex. 3 = 3, 7 = 7 or 10.5 = 10.5

20. B. Symmetric Property
If a = b, then b = a
Ex. If 3 + 5 = 8, then 8 = 3 + 5
If 15 = 6 + 9, then 6 + 9 = 15
If 20 = (4)(5), then (4)(5) = 20

21. C. Transitive Property
If a = b and b = c, then a = c
Ex. 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)

22. D. Addition Property
If a = b, then a + c = b + c
Ex. If 3 + 5 = 8, then (3 + 5) = 3 = 8 +3

23. E. Subtraction Property
If a = b, then a – c = b – c
Ex. 3 + 5 = 8, then (3 + 5) – 3 = 8 - 3

24. F. Multiplication Property
If a = b, then ac = bc
Ex. (4)(6) = 24, then (4)(6)(3) = (24)(3)

25. G. Division Property
If a = b, and c ≠ 0, then a/c = b/c
Ex. If (4)(6) = 24, then (4)(6)/3 = 24/3

26. Properties of Inequality
Let a, b and c be real numbers.
Note: The properties of inequalities will still hold
true using the relation symbol ≤ and ≥.

27. A. Addition Property
If a < b, then a + c < b + c
Ex. If 2 < 3, then 2 + 1 < 3 + 1

28. B. Subtraction Property
If a < b, then a – c < b – c
Ex. If 2 < 3, then 2 – 1 < 3 – 1

29. C. Multiplication Property
If a < b and c > 0, then ac < bc
IF a < b and c < 0, then ac > bc
Ex. If 2 < 3, then (2)(2) < (3)(2)
If 2 < 3, then (2)(-2) > (3)(-2)

30. D. Division Property
If a < b and c > 0, then a/c < b/c
If a < b and c < 0, then a/c > b/c
Ex. If 2 < 3, then 2/3 < 3/3
If 2 < 3, then 2/-3 > 3/-3

31. Solving Linear
Equations in One
Variable

32. Example:
Solve the following equations:
3. x – 5 = 8
x – 5 + 5 = 8 + 5
x + 0 = 13
x = 13
add 5 to both sides
of the equation
Recall that if the same number is added to
both sides of the equation, the resulting
sums are equal.

33. 2. x – 12 = -18
x – 12 + 12 = -18 + 12 add 12 to both
sides
x + 0 = -6
x = -6
This problem also uses the addition property
of equalities.

34. 2. x + 4 = 6
x + 4 – 4 = 6 – 4
sides of
x + 0 = 2
x = 2
subtract 4 to both
the equation
Recall that if the same number is subtracted
to both sides of the equation, the
differences are equal.

35. 2. x + 12 = 25
x + 12 – 12 = 25 – 12
x + 0 = 25 – 12
subtract 12 to both
sides
This problem also uses the subtraction
property of equalities.

36. 2. x/2 = 3
x/2 . 2 = 3 . 2
x = 6
multiply both sides by 2
Recall that if the same number is multiplied
to both sides of the equation, the
products are equal.

37. 6. x/7 = -5
x/7 . 7 = -5 .7
x = -35
multiply both sides by 7
This problem also uses multiplication
property of equalities.

38. 7. 5 x = 35
5x/5 = 35/5
is
X = 7
both sides of the equation
divided by the numerical
coefficient of x to make
the coefficient of x
equals to 1
Recall the if both sides of the equation is
divided by a non-zero number, the
quotients are equal.

39. 8. 12y = -72
12y/12 = -72/12
y = -6
divide both sides by 12
This problem also uses the division property
of equalities.

40. Other equations in one variable are solved
using more than on property of equalities.
9. 2x + 3 = 9
2x+ 3 – 3 = 9 – 3 subtraction property
2x = 6
2x/2 = 6/2 division property
x = 3

41. 10. 5y – 4 = 12 – y
5y – 4 + 4 = 12 – y + 4
5y = 16 – y
5y + y = 16 – y + y
6y = 16
6y/5 = 16/5
y = 2 4/6
addition property
addition property
division property

42. Solving Linear
Inequalities in
One Variable

43. The solution set o inequalities maybe
represented on a number line.
Recall that a solution of a linear
inequality in one variable is a real number
which makes the inequality true.
Example:
1. Graph x > 6 on a number line
O
0 1 2 3 4 5 6 7 8 9 10 11
x>6
The ray indicates the solution set of x > 6

44. The ray indicates the that he solution set, x
> 6 consist of all numbers greater than 6.
The open circle of 6 indicates that 6 is not
included.

45. 2. Graph the solution set x ≤ -1 on a number
line.
x ≤ -1
-2 -1 0 1
The ray indicates that the solution set of x ≤
-1 consist of all the numbers less than or
equal to -1. The solid circle of -1 indicates
that -1 is included in the solution set.

46. Applying the Properties of Inequalities in
Solving Linear Inequalities:
1. Solve x – 2 > 6 and graph the solution set.
x – 2 > 6
x – 2 + 2 > 6 + 2
the
x + 0 > 8
x > 8
add 2 to both sides of
inequality
O
8
x > 8

47. 2. x + 15 < -7
x + 15 – 15 < -7 – 15 subtract 15 from both
sides of the
x +0 < - 22 inequalities.
x < -22
x < -22 o
-22

48. Solving Word
Problems
Involving Linear
Equations

49. Steps in solving word problems:
2. Read and understand the problem. Identify what
is given and what is unknown. Choose a variable to
represent the unknown number.
3. Express the other unknown, if there are any., in
terms of the variable chosen in step 1.
4. Write a equation to represent the relationship
among the given and unknown/s.
5. Solve the equation for the unknown and use the
solution to find for the quantities being asked.
6. Check by going back to the original statement.

50. Example:
2. 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: Let x – 3 be the second number.
Step 3: x + ( x – 3) = 49
Step 4: x + x – 3 = 49
2x – 3 = 49
2x = 49 + 3
2x = 52
x = 26
x – 3 = 23
the first number
the second number
Step 5: Check: The sum of 26 and 23 is 49,
and 23 is 3 less than 26.

51. 2. 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?
Solution:
Step 1: Let x be Leila’s age now
3x is Mrs. dela Rosa’s age now
Step 2: x – 6 is Leila’s age 6 years ago
3x – 6 is Mrs. dela Rosa’s age 6 years ago
Step 3: 5(x – 6) = 3x – 6
Step 4: 5(x – 6) = 3x – 6
5x – 30 = 3x – 6
5x – 30 + 30 = 3x – 6 + 30
5x = 3x + 24
5x – 3x = 3x +24 – 3x
2x = 24
2x/2 = 24/ 2
X = 12
3x = 36
Leila’s age now
Mrs. dela Rosa’s age now
Step 5: Check: Thrice of Leila’s present age, 12, is Mrs. dela Rosa’s presnt age, 36. Six
years ago, Mrs. dela Rosa was 36 – 6 = 30years old which was five times Leila’s age, 12
– 6 = 6.

52. Activity: Tell whether the ff. are algebraic expression, linear
equation and linear inequality
1.2x = 7 8. 74x – 5
2.6x – 3 9. 43y -2 ≤ 3
3.8x < 7 10. 654 – x = 15
4.-56x + 7 11. 16 + 2
5. 98a > 6 12. 6x + 4 > 2
6. 9b ≠ 7 13. 8𝑥2
− 3
7. -11 ≥ 8x 14. 76 + 6
For 15-20. give 2 examples of each

53. Activity: Tell whether the ff. are algebraic expression, linear
equation and linear inequality
1.2x = 7 l.e 8. 74x – 5 ae
2.6x – 3 a.e 9. 43y -2 ≤ 3 li
3.8x < 7 l.i 10. 654 – x = 15 le
4.-56x + 7 ae 11. 16 + 2 ae
5. 98a > 6 li 12. 6x + 4 > 2 li
6. 9b ≠ 7 li 13. 8𝑥2
− 3 𝑎𝑒
7. -11 ≥ 8x li 14. 76 + 6 = 3 le

54. GRAPH THE FOLLOWING LINEAR INEQUALITIES:
1.x> -2
2.x< 8
3.x≥ -2
4.x≤ 0
5.x> 2