Question

If $\frac{3x - 4}{2} \ge \frac{x + 1}{4} - 1$, then $x\in$

• A. $[1$, $\infty)$
• B. $(1$, $\infty)$
• C. $(-5$, $5)$
• D. $[-5$, $5]$

Answer 1

Answer: (A)

$[1$, $\infty)$

Answer 2

We have $\frac{3x - 4}{2} \ge \frac{x + 1}{4} - 1$ or, $\frac{3x - 4}{2} \ge \frac{x -3}{4}$ or, $2\left(3x - 4\right) \ge\left(x -3 \right)$ or, $6x - 8 \ge x - 3$ or, $5x \ge 5$ or $x \ge 1$ Thus all real numbers which are greater than or equal to $1$ is the solution set of the given inequality $\therefore\quad x \in [1$, $\infty)$