Given,
(7x-1)/2 < -3, (3x+8)/5 + 11 < 0
Let us consider the first inequality.
\(\frac{7x-1}{2}\) < -3
⇒ \((\frac{7x-1}{2})\) \(\times\)2 < -3 \(\times\) 2
⇒ 7x – 1 < –6
⇒ 7x – 1 + 1 < –6 + 1
⇒ 7x < –5
⇒ \(\frac{7x}{7}\) < \(\frac{-5}{7}\)
⇒ x < \(-\frac{5}{7}\)
∴ x ∈ (-∞,\(-\frac{5}{7}\)) ...(1)
Now,
Let us consider the second inequality.
⇒ 3x + 63 < 0
⇒ 3x + 63 – 63 < 0 – 63
⇒ 3x < –63
⇒ \(\frac{3x}{3}\) < \(\frac{-63}{3}\)
⇒ x < –21
∴ x ∈ (–∞, –21) ...(2)
From (1) and (2), we get
x ∈ (-∞,\(-\frac{5}{7}\)) ∩ (–∞, –21)
∴ x ∈ (–∞, –21)
Thus,
The solution of the given system of inequations is (–∞, –21).
