Find all real numbers x which satisfy the inequality \(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)
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Correct Answer: Option D
Explanation:
\(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)
= \(\frac{x + 1}{3}\) - 1 > \(\frac{x + 4}{5}\)
\(\frac{x + 1}{3}\) - \(\frac{x + 4}{5}\) - 1 > 0
= \(\frac{5x + 5 - 3x - 12}{15}\)
= 2x - 7 > 15
= 2x > 22
= x > 11
\(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)
= \(\frac{x + 1}{3}\) - 1 > \(\frac{x + 4}{5}\)
\(\frac{x + 1}{3}\) - \(\frac{x + 4}{5}\) - 1 > 0
= \(\frac{5x + 5 - 3x - 12}{15}\)
= 2x - 7 > 15
= 2x > 22
= x > 11