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Question

If the replacement set is set of whole numbers, solve:
(i) $$x+7 \leq 11$$                            (ii) $$3x - 1 > 8$$
(iii) $$ 8 - x > 5$$                              (iv) $$ 7 - 3x \geq -\frac{1}{2}$$


Solution

$$(1) x+7\leq 11$$
$$x\leq 4$$
So, $$x\in [0,4]$$
$$(2) 3x-1>8$$
$$3x>9 \implies x>3$$
So, $$x\in(3,\infty)$$
$$(3) 8-x >5$$
$$x<3$$
$$\implies x\in[0,3)$$
$$(4) 7-3x\geq \cfrac{-1}{2}$$
$$3x\leq 7+\cfrac{1}{2}$$
$$3x\leq \cfrac{15}{2}$$
$$x\leq \cfrac{5}{2}$$
$$x\in [0,\cfrac{5}{2}]$$

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