Question

The range of values of $x$ which satisfy the inequality $-2\frac{2}{3}\le x+\frac{1}{3}<3\frac{1}{3}$:$x\in R$ is

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

Answer 1

The correct option is B

$-3\le x<3$

Explanation for the correct option:

Solving $\mathbf{-}\mathbf{2}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{\le }\mathit{x}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{<}\mathbf{3}\frac{\mathbf{1}}{\mathbf{3}}$:

$$\begin{gathered} -2\frac{2}{3}\le x+\frac{1}{3}<3\frac{1}{3} \\ \Rightarrow -\frac{8}{3}\le x+\frac{1}{3}<\frac{10}{3} \\ \Rightarrow -\frac{8}{3}-\frac{1}{3}\le x<\frac{10}{3}-\frac{1}{3} \\ \Rightarrow -\frac{9}{3}\le x<\frac{9}{3} \\ \Rightarrow -3\le x<3 \end{gathered}$$

Therefore, the correct answer is option B.