Here we will solve various types of problems on law of inequality.
1. Mark the statement true or false. Justify your answer.
(i) If m + 6 > 15 then m  6 > 3
(ii) If 4k >  24 then  k > 6.
Solution:
(i) m + 6 > 15
⟹ m + 6  12 > 15  12, [Subtracting 12 from both sides]
⟹ m – 6 > 3
Therefore the sentence is true.
(ii) 4k >  24
⟹ \(\frac{4k}{4}\) < \(\frac{24}{4}\), [Dividing both sides by 4]
⟹ k < 6
Therefore the sentence is false.
2. If 3z + 4 < 16 and z ∈ N then find z.
Solution:
3z + 4 < 16
⟹ 3z < 16  4, [Using the Rule of transferring a positive term]
⟹ 3z < 12
⟹ \(\frac{3z}{3}\) < \(\frac{12}{3}\), [using the Rule of division by a positive number]
⟹ z < 4
According to the given question z is natural number.
Therefore, z = 1, 2, and 3.
3. If (m – 1)(6 – m) > 0 and m ∈ N then find m.
Solution:
We know that xy > 0 then x > 0, y > 0 or x < 0, y < 0
Therefore, m – 1 > 0 and 6 – m > 0 ....................... (1)
or, m – 1 < 0 and 6 – m < 0....................... (2)
From (1) we get, m – 1 > 0 ⟹ m > 1,
and 6 – m > 0 ⟹ 6 > m
Therefore form (1), m > 1 as well as m < 6
From (2) we get, m – 1 <0 ⟹ m < 1
and 6 – m < 0 ⟹ 6 < m
Therefore form (2), m < 1 as well as m > 6
This is not possible because is m is less than 1, it cannot be greater than 6.
Thus (1) is possible and it gives 1 < m < 6, i.e., m lies between 1 and 6.
But according to the given question m is natural number. So, m = 2, 3 , 4 and 5.
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