Question

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

Answer 1

$\left(1\right)x+7\le 11$

$x\le 4$

So, $x\in [0,4]$

$\left(2\right)3x-1>8$

$3x>9⟹x>3$

So, $x\in (3,\infty )$

$\left(3\right)8-x>5$

$x<3$

$⟹x\in [0,3)$

$\left(4\right)7-3x\ge \frac{-1}{2}$

$3x\le 7+\frac{1}{2}$

$3x\le \frac{15}{2}$

$x\le \frac{5}{2}$

$x\in [0,\frac{5}{2}]$