Question

Solve the following system of inequalities `(2x + 1)/(7x - 1) > 5, (x + 7)/(x - 8) > 2`

Answer 1

`(2x + 1)/(7x - 1) > 5`

Subtracting 5 both side, we get

`(2x + 1)/(7x - 1) - 5 > 0`

⇒ `(2x + 1- 35x + 5)/(7x - 1) > 0`

⇒ `(6 - 33x)/(7x - 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, `(x + 7)/(x - 8) > 2`

Subtracting 2 both sides, we get

⇒ `(x + 7)/(x - 8) - 2 > 0`

⇒ `(x + 7 - 2x + 16)/(x - 8) > 0`

⇒ `(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.