Question

If $\frac{21x-6}{4}-9\le0$ and $\frac{x-1}{3}+1\ge0, x\in\mathbb{R}$ then $x$ lies in the interval

• A. [-2,1]
• B. [-2,2]
• C. [-1,2]
• D. [2,4]
• E. [-2,4]

Hint:

Logic Tip: Always express compound inequalities on a mental (or sketched) number line. "Less than or equal to" points left, and "greater than or equal to" points right. The region where the two lines overlap is your final answer.

Answer 1

The Correct Option is B

Concept:
To find the interval in which $x$ lies, we need to solve both linear inequalities independently. The final solution will be the intersection (overlap) of the solution sets from both inequalities, as they are joined by the word "and".
Step 1: Solve the first inequality.
$$\frac{21x-6}{4} - 9 \le 0$$ Add 9 to both sides: $$\frac{21x-6}{4} \le 9$$ Multiply by 4: $$21x - 6 \le 36$$ Add 6 to both sides: $$21x \le 42$$ Divide by 21: $$x \le 2$$ This gives the interval $(-\infty, 2]$.
Step 2: Solve the second inequality.
$$\frac{x-1}{3} + 1 \ge 0$$ Subtract 1 from both sides: $$\frac{x-1}{3} \ge -1$$ Multiply by 3: $$x - 1 \ge -3$$ Add 1 to both sides: $$x \ge -2$$ This gives the interval $[-2, \infty)$.
Step 3: Find the intersection of the two intervals.
We need values of $x$ that satisfy both $x \le 2$ AND $x \ge -2$. Combining these conditions yields: $$-2 \le x \le 2$$ In interval notation, this is represented as $[-2, 2]$.