Solve the following system

According to the question,

$\frac{2 x+1}{7 x-1}>5$

Subtracting 5 both side, we get

$\frac{2 x+1}{7 x-1}-5>0$

$\Rightarrow \frac{2 x+1-35 x+5}{7 x-1}>0$

$\Rightarrow \frac{6-33 x}{7 x-1}>0$

For above fraction be greater than 0, either both denominator and numerator should be greater than 0 or both should be less than 0.

⇒ 6 – 33x > 0 and 7x – 1 > 0

⇒ 33x < 6 and 7x > 1

⇒ x < 2/11 and x > 1/7

⇒ 1/7 < x < 2/11 …(i)

Or

⇒ 6 – 33x < 0 and 7x – 1 < 0

⇒ 33x > 6 and 7x < 1

⇒ x > 2/11 and x < 1/7

⇒ 2/11< x < 1/7 …(which is not possible since 1/7 > 2/11)

Also,

$\frac{x+7}{x-8}>2$

Subtracting 2 both sides, we get

$\Rightarrow \frac{x+7}{x-8}-2>0$

$\Rightarrow \frac{x+7-2 x+16}{x-8}>0$

$\Rightarrow \frac{23-x}{x-8}>0$

For above fraction be greater than 0, either both denominator and numerator should be greater than 0 or both should be less than 0.

⇒ 23 – x > 0 and x – 8 > 0

⇒ x < 23 and x > 8

⇒ 8 < x < 23 …(ii)

Or

23 – x < 0 and x – 8 < 0

⇒ x > 23 and x < 8

⇒ 23 < x < 8 …(which is not possible, as 23 > 8]

Therefore, from equations (i) and (ii), we infer that there is no solution satisfying both inequalities.

Hence, the given system has no solution.