Question

Solve the following inequation and graph the solution on the number line.
$-2\frac{2}{3}\le x+\frac{1}{3}<3\frac{1}{3};xϵR$

Answer 1

$Given-2\frac{2}{3}\le x+\frac{1}{3}<3\frac{1}{3};xϵR$
First we take $-2\frac{2}{3}\le x+\frac{1}{3}$
$\Rightarrow -\frac{8}{3}\le x+\frac{1}{3}\Rightarrow -\frac{8}{3}-\frac{1}{3}\le x\Rightarrow -\frac{9}{3}\le x\Rightarrow -3\le x...\left(i\right)$
Now we take $x+\frac{1}{3}<3\frac{1}{3}$
$\Rightarrow x<\frac{10}{3}-\frac{1}{3}\Rightarrow x<\frac{9}{3}\Rightarrow x<3$ ...(ii)
From equations (i) and (ii), we get:
$-3\le x<3.$
Hence, the solution set $=\{x:xϵR,-3\le x<3\}.$
The graph of the solution set is shown by the thick portion of the number line. The hollow circle at 3 indicates that the number 3 is not included among the solutions.