Question

Solve the following linear inequations and graph the solution set on a real number line.
$4\frac{3}{4}\ge x+\frac{5}{6}>\frac{1}{3},x\in R$

Answer 1

Finding the solution set of $x$
Given:
Consider $4\frac{3}{4}\ge x+\frac{5}{6}$
$\frac{19}{4}\ge \frac{(6x+5)}{6}$
$114\ge 24x+20$
By taking the like terms to one side of equation, we get,
$114-20\ge 24x$
$94\ge 24x$
$x\le \frac{94}{24}$
$x\le 3\frac{11}{12}$
Now, consider $x+\frac{5}{6}>\frac{1}{3}$
$\frac{(6x+5)}{6}>\frac{1}{3}$
$18x+15>6$
By taking the like terms to one side of equation,
we get
$18x>6-15$
$18x>-9$
$x>-\frac{9}{18}$
$x>-\frac{1}{2}$
As per the condition given in the equation, $x\in R$
Therefore,solution set $x=\{-\frac{1}{2}<x\le 3\frac{11}{12}\}$