Question

The solution set for the following inequation is:
$-\frac{x}{3}\le \frac{x}{2}-1\frac{1}{3}<\frac{1}{6},xϵR$

• A. $\{\frac{8}{5}\le x<3,xϵR\}$
• B. $\{\frac{8}{5}<x<3,xϵR\}$
• C. $\{\frac{8}{5}<x\le 3,xϵR\}.$
• D. $\{\frac{8}{5}\le x\le 3,xϵR\}$

Answer 1

The correct option is A $\{\frac{8}{5}\le x<3,xϵR\}$

Given: $-\frac{x}{3}\le \frac{x}{2}-1\frac{1}{3}<\frac{1}{6}$

$\Rightarrow -\frac{x}{3}\le \frac{3x-8}{6}\text{and}\frac{3x-8}{6}<\frac{1}{6}$


Rule: If both the sides of an inequation are multiplied or divided by the same positive number, then the sign of the inequality will remain the same.
On multiplying left hand side inequation by 6 and right hand side inequation by 6 in both the inequations, we get;

$\Rightarrow -2x\le 3x-8\text{and}3x-8<1$

$\Rightarrow 8\le 3x+2x\text{and}3x<1+8$

$\Rightarrow 8\le 5x\text{and}3x<9$

$\Rightarrow \{\frac{8}{5}\le x<3,xϵR\}.$