Find all real number 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