Given,
(2x+1)/(7x-1)>5, and (x+7)/(x-8)>2
Let us consider the first inequality.
\(\frac{2x+1}{7x-1}\) > 5
For this inequation to be true,
There are two possible cases.
i. 11x – 2 > 0 and 7x – 1 < 0
⇒ 11x – 2 + 2 > 0 + 2 and
7x – 1 + 1 < 0 + 1
⇒ 11x > 2 and 7x < 1
Hence,
This case is not possible.
ii. 11x – 2 < 0 and 7x – 1 > 0
⇒ 11x – 2 + 2 < 0 + 2 and
7x – 1 + 1 > 0 + 1
⇒ 11x < 2 and 7x > 1
Hence,
x ∈ (\(\frac{1}{7},\frac{2}{11}\)) ....(1)
Now,
Let us consider the second inequality.
For this inequation to be true,
There are two possible cases.
i. x – 23 > 0 and x – 8 < 0
⇒ x – 23 + 23 > 0 + 23 and
x – 8 + 8 < 0 + 8
⇒ x > 23 and x < 8
∴ x ∈ (23, ∞) ∩ (–∞, 5)
However,
(23, ∞) ∩ (–∞, 5) = ∅
Hence,
This case is not possible.
ii. x – 23 < 0 and x – 8 > 0
⇒ x – 23 + 23 < 0 + 23 and
x – 8 + 8 > 0 + 8
⇒ x < 23 and x > 8
∴ x ∈ (–∞, 23) ∩ (8, ∞)
However,
(–∞, 23) ∩ (8, ∞) = (8, 23)
Hence,
x ∈ (8, 23) ....(2)
From (1) and (2), we get
x ∈ (\(\frac{1}{7}\),\(\frac{2}{11}\)) ∩ (8,23)
∴ x ∈ ∅
Thus,
There is no solution of the given system of inequations.
