The document explains the concepts of linear equations and inequalities in one variable, including their definitions and forms. It provides several examples and detailed steps for solving them, including techniques like the distributive property and checking solutions through substitution. The material is prepared by Mr. Jayson B. Gulla.

1.  Differentiates between equations and inequalities
 Find the solution of linear equation or inequality in one variable
 Solves linear equation or inequality in one variable

2.  An Equation is a Mathematical Statement in
which two expressions represent the same
quantity.
 An Equation has three parts:
Example: 7x = 3x + 8
Left Side Right Side
Equal Sign

3.  A Linear Equation in one variable
is any equation that can be
written in the form
ax + b = 0
in which a and b are real numbers
and a ≠ 0.

4.  A Linear Inequality in one
variable is any inequality that can
be written in the form
ax + b > 0 ax + b < 0
ax + b ≥ 0 ax + b ≤ 0
in which a and b are real numbers
and a ≠ 0.

5.  1. x – 6 = 5 2. 2x/3 – 1 ≥ 6
 3. x + 3 = 4 4. 3x ≤ 9
 5. 2x > 6 6. 2(x-5) = 4
 7. 3x -2 < 4 8. x > 5
 9. x + 1 = 0 10. -2x = 8

6. Step 1: If parenthesis are present,
use Distributive Property of
Equality.
Step 2: Combine any like terms on
each side of the equation.
Step 3: Use APE to rewrite the
equation so that variable terms are
on one side and constant term are
on the other side.

7. Step 4: Use the MPE to multiply
both sides by the reciprocal of the
numerical coefficient of the
variable and solve, then simplify.
Step 5: Check the solution from the
original equation by substitution.

8. Step 1: Start by writing the original x – 6 = 5
equation.
Step 2: Undo subtraction x – 6 + 6 = 5 + 6
Step 3: Simplify each side x = 11
Step 4: Check: Substitute 11 to x. x – 6 = 5
11 – 6 = 5
5 = 5
True statement

9. Solution:
k- (-15) > 31
k+15 > 31
k+15 – 15 > 31 – 15
k > 16
To Check: Substitute 16 to k.
k – (-15) > 31
17 – (-15) > 31
32 > 31 True Statement

10. Solution:
x/5 = -2
(x/5)(5) = -2(5)
x = -10
To Check:
x/5 = -2
(-10)/5 = -2
-2 = -2

11. Solution:
5x = -25 Write the original equation.
5x/5 = -25/5 Divide each side by 5.
x = -5 Simplify.
Checking:
5x = -25
5(-5) = 25 Replace x with -5
-25 = -25 True Statement

12. Solution:
6x + 2(2x + 3) = 16 Apply the DPE.
6x + 4x + 6 = 16 Combine like terms.
10x + 6 = 16 Apply APE add -6 both sides
10x + 6 – 6 = 16 – 6
10x = 10 Divide both sides by 10.
10x/10 = 10/10 Simplify.
X = 1
Checking: Substitute 1 to x.

13. 1.x – 5 = 10
2.x + 2 = 5
3.x/2 = -3
4.4x = -20
5.2(x + 1) = 2

14. PREPARED BY:
MR. JAYSON B. GULLA
THANK YOU FOR LISTENING