Question

Find the range of values of x which satisfies $-2\frac{2}{3}\le x+\frac{1}{3}<3\frac{1}{3},x\in R$.

• A. $-3\le x<3$
• B. $-3\le x\le 3$
• C. $-3<x\le 3$
• D. $-3<x<3$

Answer 1

The correct option is A $-3\le x<3$

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

$\Rightarrow \frac{-8}{3}\le x+\frac{1}{3}<\frac{10}{3}$

$\Rightarrow \frac{-8}{3}\le x+\frac{1}{3}\text{and}x+\frac{1}{3}<\frac{10}{3}$

$\Rightarrow -\frac{8}{3}-\frac{1}{3}\le x\text{and}x<\frac{10}{3}-\frac{1}{3}$

$\Rightarrow \frac{-8-1}{3}\le x\text{and}x<\frac{10-1}{3}$

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

Rule: If both the sides of an inequation are multiplied or divided by the same negative number, then the sign of the inequality will get reversed.
On multiplying the left hand side inequation by -1, we get;

$\Rightarrow x\ge -3andx<3$

$\therefore$ Range of values is $-3\le x<3.$