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We can solve this question using matching operations (see the video lesson below )
Given: Quantity A: \(\frac{x+5}{x-2}\)
Quantity B: \(\frac{ x+9}{x-1}\)
Since x > 3, we can be certain that (x-2) is positive, which means we can safely multiply both quantities by (x-2) to get: Quantity A: \(x+5\)
Quantity B: \(\frac{ (x-2)(x+9)}{x-1}\)
Similarly, since x > 3, we can be certain that (x-1) is positive, which means we can safely multiply both quantities by (x-1) to get: Quantity A: \((x-1)(x+5)\) Quantity B: \((x-2)(x+9)\)
Expand and simplify both quantities: Quantity A: \(x^2 + 4x - 5\) Quantity B: \(x^2 + 7x - 18\)
Subtract \(x^2\) from both quantities: Quantity A: \(4x - 5\) Quantity B: \(7x - 18\)
Subtract \(4x\) from both quantities: Quantity A: \(-5\) Quantity B: \(3x - 18\)
Add \(18\) to both quantities: Quantity A: \(13\) Quantity B: \(3x\)
Divide both quantities by \(3\): Quantity A: \(\frac{13}{3}≈4.33\) Quantity B: \(x\)
Since \(3<x<6\), it could be the case that x = 4, in which case Quantity A is greater. It could also be the case that x = 5, in which case Quantity B is greater.