Page 160 - ICSE Math 7

(c) 2z – 3 ≥ 5 – 2z
⇒ 2z + 2z ≥ 5 + 3 (Using the rule of transposition)

⇒ 4z ≥ 8
8
⇒ z ≥ (Transposing 4)
4
⇒ z ≥ 2
So, the possible values of z from the replacement set are 2, 3, 4 and 5.

\ Solution set = {2, 3, 4, 5}
–x x
Example 3: Solve – 3 ≥ + 9, x ∈ Z.
2 3
–x x
Solution: – 3 ≥ + 9
2 3
–x x
⇒ – ≥ 9 + 3 (Using the rule of transposition)
2 3
⇒ –3x – 2x ≥ 12
6
–5x
⇒ ≥ 12
6
⇒ –5x ≥ 12 × 6 (Transposing 6)
–5x 72
⇒ ≤ (Dividing both the sides by –5 and reversing the sign)
–5 –5
–72 2
⇒ x ≤ ⇒ x ≤ –14
5 5
So, the possible values of x from the replacement set are …, –17, –16, –15
\ Solution set = {…, –17, –16, –15}

Graphical Representation of Solution Set

We can represent the solution set graphically by plotting the true values of solutions on the number line.
Example 4: Solve the inequation 6x + 2 > 2x + 30, x ∈ N and represent the solution set graphically.

Solution: 6x + 2 > 2x + 30
⇒ 6x – 2x > 30 – 2 (Using the rule of transposition)
⇒ 4x > 28
28
⇒ x > (Transposing 4)
4
⇒ x > 7

So, the possible values of x from the replacement set are 8, 9, 10, …
\ Solution set = {8, 9, 10, …}
To represent the solution set graphically, draw a number line and mark 8, 9, 10, … by
thick dots.
...
or
1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9 10 11

The arrow or three dots above the number line show that the subsequent natural numbers are also
in the solution set.

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