Page 159 - ICSE Math 7

Method of Solving Inequations

Step 1: Simplify both the sides of the inequation.

Step 2: Using the rule of transposition, transpose the terms containing the variable on one side and
the constant terms on the other side.
Step 3: Divide both sides of the inequation by the coefficient of the variable.

Step 4: Form the solution set from the replacement set.

Example 1: Solve 3(x + 1) > 8 where x ∈ W.
Solution: 3(x + 1) > 8

⇒ 3x + 3 > 8
⇒ 3x > 8 – 3

⇒ 3x > 5
5
⇒ x > (Transposing 3)
3
2
⇒ x > 1
3
2
As the replacement set is W, x should be a whole number greater than 1 .
3
\ Solution set = {2, 3, 4, 5, …}

Example 2: Solve the following inequations taking the replacement set as {0, 1, 2, 3, 4, 5}.

(a) 2(x – 2) > –4 (b) 3(x + 2) ≤ 24 – x (c) 2z – 3 ≥ 5 – 2z
Solution: (a) 2(x – 2) > –4

⇒ 2x – 4 > –4
⇒ 2x > –4 + 4 (Transposing –4)

⇒ 2x > 0
⇒ x > 0 (Transposing 2)

So, the possible values of x from the replacement set are 1, 2, 3, 4, and 5.
\ Solution set = {1, 2, 3, 4, 5}

(b) 3(x + 2) ≤ 24 – x
⇒ 3x + 6 ≤ 24 – x

⇒ 3x + x ≤ 24 – 6 (Using the rule of transposition)
⇒ 4x ≤ 18

18
⇒ x ≤ (Transposing 4)
4
1
⇒ x ≤ 4
2
So, the possible values of x from the replacement set are 0, 1, 2, 3 and 4.
\ Solution set = {0, 1, 2, 3, 4}

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