The original expression is (x^2 - 9)/(x^2 - 2x - 3)
This can be simplified by factoring to [(x + 3)(x - 3)]/[(x + 1)(x - 3)]
That can further be simplified by cancelling out (x - 3) on top and bottom: (x + 3) / (x + 1)
So, (x^2 - 9)/(x^2 - 2x - 3) = (x + 3) / (x + 1)
But when I plug in numbers (like x = 3), that isn't true.
What am I missing?
PS - this problem was actually a limit problem, but I don't think that should matter? It said "with limits we can cancel the x-3 because x approaches 3, hence x-3 gets close to zero, but not equal to zero". Shouldn't you always be able to cancel an (x-3) that is on the top and bottom of a fraction like that?
This can be simplified by factoring to [(x + 3)(x - 3)]/[(x + 1)(x - 3)]
That can further be simplified by cancelling out (x - 3) on top and bottom: (x + 3) / (x + 1)
So, (x^2 - 9)/(x^2 - 2x - 3) = (x + 3) / (x + 1)
But when I plug in numbers (like x = 3), that isn't true.
What am I missing?
PS - this problem was actually a limit problem, but I don't think that should matter? It said "with limits we can cancel the x-3 because x approaches 3, hence x-3 gets close to zero, but not equal to zero". Shouldn't you always be able to cancel an (x-3) that is on the top and bottom of a fraction like that?