Question

(x-3)/(x-7)≤0

Answer 1

zeros are at 3 and 7
it will be negative between zeros and positive outside
your answer is
\[[3,7)\]

Answer 2

\[\frac{x-3}{x-7} \le 0\] Multiply both sides by the square of the denominator (to make sure it's not negative):
\[\frac{(x-3)(x-7)^{2}}{x-7} \le 0\]\[(x-3)(x-7) \le 0\]
The turning points therefore are \(x=3,7\). Below \(x=3\), \((x-3)(x-7)\) is positive, so x is not less than 3. At \(x=3\), \((x-3)(x-7)\) is negative, so x can be 3. Between 3, and 7, \((x-3)(x-7)\) is negative, so x can be between those. At x=7, \((x-3)(x-7)\) is positive, so x cannot be 3. Above 7, the product is positive, so x cannot be more than 7.
\([3,7)\), or \(3 \le x < 7\)