Question:

Let $x$ be a real number such that $\frac{3(x+3)}{7} \le \frac{6(x-1)}{5}$. Then the solution set of the inequality is:

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Always simplify the fraction $\frac{87}{27}$ by dividing both numerator and denominator by 3.
Updated On: Apr 28, 2026
  • $(-\infty, \frac{29}{9})$
  • $(\frac{29}{9}, \infty)$
  • $[\frac{29}{9}, \infty)$
  • $(-\infty, \infty)$
  • $(\frac{17}{9}, \infty)$
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The Correct Option is C

Solution and Explanation

Step 1: Analysis
Cross-multiply to remove fractions: $5 \cdot 3(x+3) \le 7 \cdot 6(x-1)$. $15(x+3) \le 42(x-1)$.

Step 2: Calculation

$15x + 45 \le 42x - 42$. $45 + 42 \le 42x - 15x$. $87 \le 27x$.

Step 3: Calculation

$x \ge \frac{87}{27} = \frac{29}{9}$. Final Answer: (C)
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