Fill in the blanks with correct inequality sign (>, <, ≥, ≤)
(i) $5 x<20 \Rightarrow x$ 4
(ii) $-3 x>9 \Rightarrow x \ldots \ldots \ldots .3$
(iii) $4 x>-16 \Rightarrow x$ $-4$
(iv) $-6 x \leq-18 \Rightarrow x \ldots \ldots \ldots .3$
(v) $x>-3 \Rightarrow-2 x \ldots \ldots \ldots .6$
(vi) $\mathrm{a}<\mathrm{b}$ and $\mathrm{c}>0 \Rightarrow \frac{\mathrm{a}}{\mathrm{c}} \ldots \ldots \ldots \frac{\mathrm{b}}{\mathrm{c}}$
(vii) $p-q=-3 \Rightarrow p \ldots \ldots \ldots . q$
(viii) $u-v=2 \Rightarrow u \ldots \ldots \ldots . v$
(i) $5 x<20 \Rightarrow x \ldots \ldots \ldots 4$
As, $5 x<20$
Then,
Dividing both the sides by 5
$\frac{x}{5}<\frac{20}{5}$
$x<4$
Therefore,
$5 x<20 \Rightarrow x<4$
(ii) $-3 x>9 \Rightarrow x \ldots \ldots \ldots-3$
As, $-3 x>9$
Then, Dividing both the sides by 3
$\frac{x}{3}>-\left(\frac{9}{3}\right)$
$x>-3$
Therefore
$-3 x>9 \Rightarrow x>-3$
(iii) $4 x>-16 \Rightarrow x$ $-4$
As, $4 x>-16$
Then, Dividing both the sides by 4
$\frac{x}{4}>-\left(\frac{16}{4}\right)$
$x>-4$
Therefore,
$4 x>-16 \Rightarrow x>-4$
(iv) $-6 x \leq-18 \Rightarrow x \ldots \ldots \ldots 3$
As $-6 x \leq-18$
Then, Dividing both the sides by 6
$\frac{-x}{6} \leq\left(\frac{-18}{6}\right)$
-x \leq-3
Now multiplying by -1 on both sides
$-x(-1) \leq-3(-1)$
x ≥ 3 (inequality sign reversed)
Therefore,
$-6 x \leq-18 \Rightarrow x \geq 3$
(v) $x>-3 \Rightarrow-2 x \ldots \ldots \ldots 6$
As, $x>-3$
Multiplying both sides by 2
Then
2x > -6
Now multiplying both the sides by -1
$2 x(-1)<6(-1)$
$-2 x>6$
Therefore,
$x>-3 \Rightarrow-2 x>6$
(vi) $a
As
$a
$c>0$
Dividing both sides by c in equation (1)
Then,
$\frac{a}{c}<\frac{b}{c}$
Therefore
$a
(vii) $p-q=-3 \Rightarrow p \ldots \ldots \ldots q$
As,
$p-q=-3$
$p=q-3$
From the above equation it is clear that p would always be less than q
Therefore,
$p-q=-3 \Rightarrow p (viii) $u-v=2 \Rightarrow u \ldots \ldots . . v$ As, $u-v=2$ $u=v+2$ From the above equation it is clear that u would always be greater than v Therefore, $u-v=2 \Rightarrow u>v$