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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