Question

If $x\in I$, find the solution set of each of the following inequations. Represent it on the number line.
(i) $4-3x\le 13+x$
(ii) $10-x\le 5-4x$

Answer 1

Given $4-3x\le 13+x$,

$=-4x\le 9$

$=-x\le \frac{9}{4}$

$=x\ge -\frac{9}{4}$

$=x\ge -4.5$, Since$x\in I$, $x\ge [-4.5]=x\ge -4$

Given $10-x\le 5-4x$,

$3x\le -5$

$=x\le \frac{-5}{3}$ =$x\le -1.6$

Since $x\in I$, $x\le -1$

The range in which x belongs to is marked in blue colour in the number line.