JAMB - Mathematics (1986 - No. 19)
Find all real numbers x which satisfy the inequality \(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)
x < 11
x < -1
x > 6
x > 11
Giải thích
\(\frac{1}{3}\)(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
= \(\frac{x + 1}{3}\) - 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