Answer :
Given
The above inequality can be split into two inequalities.
and
Let us consider the first inequality.
As x > 0, we have x + 1 > 0.
⇒ 4 ≤ 3(x + 1)
⇒ 4 ≤ 3x + 3
⇒ 3x + 3 ≥ 4
⇒ 3x + 3 – 3 ≥ 4 – 3
⇒ 3x ≥ 1
(1)
Now, let us consider the second inequality.
As x > 0, we have x + 1 > 0.
⇒ 3(x + 1) ≤ 6
⇒ 3x + 3 ≤ 6
⇒ 3x + 3 – 3 ≤ 6 – 3
⇒ 3x ≤ 3
⇒ x ≤ 1
∴ x ∈ (–∞, 1] (2)
From (1) and (2), we get
Thus, the solution of the given system of inequations is
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