Graphical representation of the solution set of an inequation:
A number line is used to represent the solution set of an inequation graphically.
● First solve the linear inequation and find the solution set.
● Mark it on the number line by putting a dot.
● In case the solution set is infinite, then put three more dots to indicate infiniteness.
For Example:
1. Solve the inequation 3x - 5 < 4, x ∈ N and represent the solution set graphically.
Solution:
We have 3x - 5 < 4
⇒ 3x - 5 + 5 < 4 + 5 (Add 5 to both sides)
⇒ 3x < 9
⇒ 3x/3 < 9/3 (Divide both sides by 3)
⇒ x < 3
So, the replacement set = {1, 2, 3, 4, 5, ...}
Therefore, the solution set = {1, 2} or S = {x : x ∈ N, x < 3}
Let us mark the solution set graphically.
Solution set is marked on the number line by dots.
2. Solve 2x + 8 ≥ 18
Here x ∈. W represent the inequation graphically
⇒ 2x + 8 - 8 ≥ 18 - 8 (Subtract 8 from both sides)
⇒ 2x ≥ 10
⇒ 2x/2 ≥ 10/2 (Divide both sides by 2)
⇒ x ≥ 5
Replacement set = {0, 1, 2, 3, 4, 5, 6, ...}
Therefore, solution set = {5, 6, 7, 8, 9, ...}
or, S = {x : x ∈ W, x ≥ 5}
Let us mark the solution set graphically.
Solution set is marked on the number line by dots. We put three more dots indicate infiniteness of the solution set.
3. Solve -3 ≤ x ≤ 4, x ∈ I
Solution:
This contains two inequations,
-3 ≤ x and x ≤ 4
Replacement set = {..., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...}
Solution set for the inequation -3 ≤ x is -3, -2, -1, 0, 1, 2, ... i.e., S = {-3, -2, -1, 0, 1, 2, 3, ...} = P
And the solution set for the inequation x ≤ 4 is 4, 3, 2, 1, 0, -1, ... i.e., S = {..., -3, -2, -1, 0, 1, 2, 3, 4} = Q
Therefore, solution set of the given inequation = P ∩ Q
= {-3, -2, -1, 0, 1, 2, 3, 4}
or S = {x : x ∈ I, -3 ≤ x ≤ 4}
Let us represent the solution set graphically.
Solution set is marked on the number line by dots.
A number line is used for representation of the solution set of an inequation.
Now, solution set S = {3, 4, 5, 6, ...} S = (x : x ∈ N, x > 3)
For Example:
4. 2x + 3 ≤ 15
⇒ 2x + 3 - 3 ≤ 15 - 3 (Subtract 3 from both sides)
⇒ 2x ≤ 12
⇒ 2x/2 ≤ 12/2 (Divide both sides by 2)
⇒ x ≤ 6
Now, the solution set S = {1, 2, 3, 4, 5} S' = {x : x ∈ N, x < 6}
Now, S ∩ S’ = {3, 4, 5, 6}
5. 0 < 4x - 9 ≤ 5, x ∈ R
Solution:
Case I: 0 ≤ 4x - 9
0 + 9 ≤ 4x - 9 + 9
⇒ 9 ≤ 4x
⇒ 9/4 ≤ 4x/4
⇒ 2.25 ≤ x
⇒ 2.2 < x
Case II: 4x - 3 ≤ 9
⇒ 4x - 3 + 3 ≤ 9 + 3
⇒ 4x ≤ 12
⇒ x ≤ 3
S ∩ S' = {2.2 < x ≤ 3} x ∈ R
= {x : x ∈ R 3 ≥ x > 2.2}
Arrow on right shows that solution set continues.
● Inequations
Properties of Inequation or Inequalities
Representation of the Solution Set of an Inequation
Practice Test on Linear Inequation
● Inequations - Worksheets
Worksheet on Linear Inequations
8th Grade Math Practice
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