Fill in the blanks with correct inequality sign
Question:

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) $u-v=2 \Rightarrow u \ldots \ldots \ldots . v$

Solution:

(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) $a<b$ and $c>0$ $\Rightarrow \frac{a}{c}$ ……… $\frac{\mathrm{b}}{\mathrm{c}}$

As

$a<b \ldots(1)$

$c>0$

Dividing both sides by c in equation (1)

Then,

$\frac{a}{c}<\frac{b}{c}$

Therefore

$a<b$ and $c>0 \Rightarrow \frac{a}{c}<\frac{b}{c}$

(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,

$p-q=-3 \Rightarrow p<q$

(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,

$u-v=2 \Rightarrow u>v$