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